Physics Chapt 19

Chapter 19 The Second Law of ThermodynamicsConceptual Problems 1 • Modern automobile gasoline engines have efficiencies of about 25%. About what percentage of the heat of combustion is not used for work but released as heat? (a) 25%, (b) 50%, (c) 75%, (d) 100%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. The percentage of the heat of combustion (heat absorbed from the hightemperature reservoir) is the ratio of Qc to Qh. We can use the relationship between W, Qh, and Qc ( W " Qh ! Qc ) to find Qc/ Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: Solving for Qc/ Qh yields: #" Q W Qh ! Qc " " 1! c Qh Qh Qh Qc " 1! # Qh Substitute for # to obtain: Qc " 1 ! 0.25 " 0.75 Qh and $c % is correct. 2 • If a heat engine does 100 kJ of work per cycle while releasing 400 kJ of heat, what is its efficiency? (a) 20%, (b) 25%, (c) 80%, (d) 400%, (e) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( W " Qh ! Qc ) to express the efficiency of the heat engine in terms of Qc and W. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: #" W W " " Qh W & Qc 1 Q 1& c W 1867 1868 Chapter 19 Substitute for Qc and W to obtain: 1 " 0.2 400 kJ 1& 100 kJ and $a % is correct. #" 3 • If the heat absorbed by a heat engine is 600 kJ per cycle, and it releases 480 kJ of heat each cycle, what is its efficiency? (a) 20%, (b) 80%, (c) 100%, (d) You cannot tell from the data given. Determine the Concept The efficiency of a heat engine is the ratio of the work done per cycle W to the heat absorbed from the high-temperature reservoir Qh. We can use the relationship between W, Qh, and Qc ( W " Qh ! Qc ) to express the efficiency of the heat engine in terms of Qc and Qh. Use the definition of efficiency and the relationship between W, Qh, and Qc to obtain: Substitute for Qc and Qh to obtain: #" Q W Qh ! Qc " " 1! c Qh Qh Qh 480 kJ " 0.2 600 kJ and $a % is correct. # " 1! 4 • Explain what distinguishes a refrigerator from a !heat pump.! Determine the Concept The job of a refrigerator is to move heat from its cold interior to the warmer kitchen environment. This process moves heat in a direction that is opposite its 'natural' direction of flow, analogous to the use of a water pump to pump water out of a boat. The term heat pump is used to describe devices, such as air conditioners, that are used to cool living and working spaces in the summer and warm them in the winter. 5 An air conditioner’s COP is mathematically identical to that Q of a refrigerator, that is, COPAC " COPref " c . However a heat pump’s COP is W Q defined differently, as COPhp " h . Explain clearly why the two COPs are W defined differently. Hint: Think of the end use of the three different devices. • [SSM] Determine the Concept The COP is defined so as to be a measure of the effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior, Qc. For a heat pump, the ideas is to focus on the heat drawn into the warm interior of the house, Qh. The Second Law of Thermodynamics 1869 6 • Explain why you cannot cool your kitchen by leaving your refrigerator door open on a hot day. (Why does turning on a room air conditioner cool down the room, but opening a refrigerator door does not?) Determine the Concept As described by the second law of thermodynamics, more heat must be transmitted to the outside world than is removed by a refrigerator or air conditioner. The heating coils on a refrigerator are inside the room and so the refrigerator actually heats the room in which it is located. The heating coils on an air conditioner are outside one’s living space, so the waste heat is vented to the outside. 7 • Why do steam-power-plant designers try to increase the temperature of the steam as much as possible? Determine the Concept Increasing the temperature of the steam increases the Carnot efficiency, and generally increases the efficiency of any heat engine. 8 • To increase the efficiency of a Carnot engine, you should (a) decrease the temperature of the hot reservoir, (b) increase the temperature of the cold reservoir, (c) increase the temperature of the hot reservoir, (d) change the ratio of maximum volume to minimum volume. Determine the Concept Because the efficiency of a Carnot cycle engine is given T by # C " 1 ! c , you should increase the temperature of the hot reservoir. $c % is Th correct. 9 •• [SSM] Explain why the following statement is true: To increase the efficiency of a Carnot engine, you should make the difference between the two operating temperatures as large as possible; but to increase the efficiency of a Carnot cycle refrigerator, you should make the difference between the two operating temperatures as small as possible. Determine the Concept A Carnot-cycle refrigerator is more efficient when the temperatures are close together because it is easier to extract heat from an already cold interior if the temperature of the exterior is close to the temperature of the interior of the refrigerator. A Carnot-cycle heat engine is more efficient when the temperature difference is large because then more work is done by the engine for each unit of heat absorbed from the hot reservoir. 10 •• A Carnot engine operates between a cold temperature reservoir of o 27 C and a high temperature reservoir of 127(C. Its efficiency is (a) 21%, (b) 25%, (c) 75%, (d) 79%. 1870 Chapter 19 Determine the Concept The efficiency of a Carnot cycle engine is given by T # C " 1 ! c where Tc and Th (in kelvins) are the temperatures of the cold and hot Th reservoirs, respectively. Substituting numerical values for Tc and Th yields: #C " 1! $b % 300 K " 0.25 400 K is correct. 11 •• The Carnot engine in Problem 10 is run in reverse as a refrigerator. Its COP is (a) 0.33, (b) 1.3, (c) 3.0 (d) 4.7. Determine the Concept The coefficient of performance of a Carnot cycle engine Q run in reverse as refrigerator is given by COPref " c . We can use the relationship W between W, Qc, and Qh to eliminate W from this expression and then use the Q T relationship, applicable only to a device operating in a Carnot cycle, c " c to Qh Th express the refrigerator’s COP in terms of Tc and Th. The coefficient of performance of a refrigerator is given by: Qc W or, because W " Qh ! Qc , COPref " COPref " Qc Qh ! Qc Dividing the numerator and denominator of this fraction by Qc yields: For a device operating in a Carnot cycle: Substitute in the expression for COPref to obtain: COPref " 1 Qh !1 Qc Qc Tc " Qh Th COPref, C " 1 Th !1 Tc Determine the Concept When water vapor condenses. For these two paths. We can use the concept of a state function to choose from among the alternatives given as possible answers to the problem. Two possible paths are (A) an isothermal expansion followed by an adiabatic compression and (B) an adiabatic compression followed by an isothermal expansion. 13 •• An ideal gas is taken reversibly from an initial state Pi. is a state function and its change when the system moves from one state to another depends only on the system’s initial and final states. Thus )S A " )S B . A " )Eint. (b) remains constant. Determine the Concept The two paths are shown on the PV diagram to the right. (c) decreases. )Eint. C " 1 " 3. (d) (d ) is correct. C: COPref. B . (a ) is correct.The Second Law of Thermodynamics 1871 Substitute numerical values and evaluate COPref. (d) may decrease or remain unchanged. Tf. Explain your answer. 12 •• On a humid day. Figure 19-12 shows a thermodynamic cycle for an ideal gas on an ST 14 •• diagram. It is not dependent on the process by which the change occurs. its entropy decreases (the liquid state is a more ordered state than is the vapor state) and the entropy of the universe increases. (b) and (c) S. (c) )SA < )SB. P B Ti Pf Pi B f i A Vi Vf A Tf V (a) Because Eint is a state function and the initial and final states are the same for the two paths. the entropy of the water (a) increases. . Ti to the final state Pf. (a) )Eint A > )Eint B. water vapor condenses on a cold surface.0 400 K !1 300 K $c % is correct. (d) None of the above. During condensation. Vf. Vi. (b) )SA > )SB. like Eint. Identify this cycle and sketch it on a PV diagram. from b to c. Refer to Figure 19-3. respectively: Because Won = 0 for this constantvolume process: Substituting for Q yields: Q T where. The points A. the cycle is the Carnot cycle shown in the adjacent PV diagram. B. and D in Figure 19-13 correspond to points c. consider the entropy change of the gas from point b to an arbitrary point on the path b*c where the entropy and temperature of the gas are S and T. Identify the type of engine represented by this diagram. So. from c*d S is constant while T decreases. in Figure 19-3. )S " . from a to b. and b. To determine how S depends on T along b*c and d*a. The cycle is that of the Otto engine (see Figure 19-3). B*C is a constant-volume process in which the entropy decreases. P B C A D V 15 •• Figure 19-13 shows a thermodynamic cycle for an ideal gas on an SV diagram. S is constant and T increases. There a*b is an adiabatic compression. and from d to a both S and T decrease.1 ! b + T / T . Q is positive. C. d. Determine the Concept Note that A*B is an adiabatic expansion. a.1872 Chapter 19 Determine the Concept The processes A*B and C*D are adiabatic and the processes B*C and D*A are isothermal. respectively. Therefore. )S " )Eint " Qin " Q " C V )T " C V $T ! Tb % C V $T ! Tb % 0 T " C V . c*d is an adiabatic expansion and d*a is a constant-volume cooling. (The Otto cycle is discussed in Section 19-1. because heat is entering the system. b*c is a constant volume heating. C*D is an adiabatic compression and D*A is a constant-volume process that returns the gas to its original state. 16 •• Sketch an ST diagram of the Otto cycle. heat is added to the system and both S and T increase.) Determine the Concept The Otto cycle consists of four quasi-static steps. In this process heat is added to the system and the entropy and volume increase. process 4*1 is adiabatic. .): These results tell us that. dS dT and d 2 S dT 2 . give the slope and concavity of the path. T dS " C V b2 dT T 2 T d S " !2C V b3 2 dT T The first and second derivatives. Determine the Concept Referring to Figure 19-8.The Second Law of Thermodynamics 1873 On path b*c the entropy is given by: 0 T S " S b & )S " S b & C V . Process 2*3 is adiabatic. The concavity of the path is negative for all T. isentropic. and S is constant while V decreases.1 ! d + / T . along path d*a. process 1*2 is an isothermal expansion. so S is constant as V increases. (For an ideal gas CV is a positive constant. Calculate these derivatives assuming CV is constant.1 ! b + / T . An ST diagram for the Otto cycle is shown to the right. the slope of the path is positive and the slope decreases as T increases. along path b*c. Following the same procedure on path d*a gives: 0 T S " S d & C V . that is. Finally. the slope of the path is positive and the slope decreases as T increases. Process 3*4 is an isothermal compression in which S decreases and V also decreases. S d c a b T 17 •• [SSM] Sketch an SV diagram of the Carnot cycle for an ideal gas. T dS " C V d2 dT T 2 T d S " !2C V d3 2 dT T These results tell us that. The concavity of the path is negative for all T. + . The change in entropy of the gas from point 1 (where the temperature is T1) to an arbitrary point on the curve is given by: For an isothermal expansion. we have: The graph of S as a function of V for an isothermal expansion shown to the right was plotted using a spreadsheet program. S V An SV graph for the Carnot cycle (see Figure 19-8) is shown to the right. + . and thus the heat added to the gas.V + / 1. S 2 3 1 4 V 18 •• Sketch an SV diagram of the Otto cycle. + . 0V S " S1 & nR ln. are given by: Substituting for Q yields: )S " Q T1 0V Q " W " nRT1 ln.1874 Chapter 19 During the isothermal expansion (from point 1 to point 2) the work done by the gas equals the heat added to the gas. (The Otto cycle is discussed in Section 19-1. the work done by the gas.V + / 1. 0V )S " nR ln.V + / 1.) . This graph establishes the curvature of the 1*2 and 3*4 paths for the SV graph. Since S " S1 & )S . therefore )S > 0. Make a sketch of this cycle on a PV diagram. Q > 0 and. Finally. Determine the Concept Process A*B is at constant entropy. P C B D A V 20 •• One afternoon. is positive and so. Process B*C is one in which P is constant and S decreases. here. this time with heat leaving the system and )S < 0. however. both Q = 0 and )S = 0 along this path. Process a*b takes place adiabatically and so both Q = 0 and )S = 0 along this path. and your friend replies. Process c*d also takes place adiabatically and so. the mother of one of your friends walks into his room and finds a mess. that is. A sketch of the SV diagram for the Otto cycle follows: S c d b a V 19 •• Figure 19-14 shows a thermodynamic cycle for an ideal gas on an SP diagram. That’s all. Process b*c takes place at constant volume. She asks your friend how the room came to be in such a state. again. Process D*A returns the system to its original state at constant pressure. heat is exhausted from the system and the volume decreases. Is his mother correct to ground him for not cleaning his room. The cycle is shown in the adjacent PV diagram. it is the natural destiny of any closed system to degenerate toward greater and greater levels of entropy. Mom. or is cleaning the room really impossible? Determine the Concept The son is out of line. but besides that.The Second Law of Thermodynamics 1875 Determine the Concept The Otto cycle is shown in Figure 19-3.! Her reply is a sharp !Nevertheless. it is an adiabatic process in which the pressure increases. he’s also wrong. process d*a is a constant-volume process. !But that can’t happen. while )V = 0 along this path. Qin.! Critique your friend’s response. While it is true that systems tend to degenerate to greater levels of . It would violate the second law of thermodynamics. you’d better clean your room!! Your friend retorts. !Well. Process C*D is an adiabatic compression. 1876 Chapter 19 disorder.kit and simplify to obtain: COPbasement COPkit Qh. it is not true that order cannot be brought forth from disorder. It is true that order will not come about from the disordered chaos of his room – unless he applies some elbow grease. and freezer. your friend – on the system in order to reduce the level of chaos and bring about order. Picture the Problem We can use the definition of the coefficient of performance to express the ratio of the coefficient of performance in your basement to the coefficient of performance in the kitchen.kit ! Qc.basement . then we can use the proportion Qh Qc " Th Tc to express the ratio of the coefficients of performance in terms of the temperatures in the kitchen.kit !1 Qc.kit Qc.basement !1 Qc.basement and Qc. If we further assume that the freezer operates in a Carnot cycle.kit Qh.basement " h. The ratio of the coefficients of performance in the basement and kitchen is given by: COPbasement COPkit Qc. Estimation and Approximation 21 • Estimate the change in COP of your electric food freezer when it is removed from your kitchen to its new location in your basement.basement Q ! Qc.basement !1 Qc.kit Qh. What is required is an agent doing work – for example.basement Qc.kit !1 Qc.basement " 1 Qh.basement Qc.kit 1 Because W " Qh ! Qc for a heat engine or refrigerator: COPbasement COPkit Divide the numerators and denominators by Qc. which is 8°C cooler than your kitchen. basement.kit " Qh. His cleanup efforts will be rewarded with an orderly system after a sufficient time for him to complete the task.kit Wc.basement W " c. 325 kPa %$50 m3 % N" $1. 2 + . Picture the Problem The probability that all the molecules in your bedroom are 0V located in the (open) closet is given by p " .kit !1 Tc. If the original volume of the air in your bedroom is V1. the probability p of finding the N molecules. confined to your closet whose volume is V2 is given by: N 0V p". We’ll assume that the volume of your room is about 50 m3 and that the temperature of the air is 20(C. molecules in your bedroom and V1 and V2 are the volumes of your bedroom and closet. because V2 " 10 V1 .252 1 1027 molecules . then Qh Th " and our expression for the Qc Tc ratio of the COPs becomes: Assuming that the temperature in your kitchen is 20(C and that the temperature of the interior of your freezer is !5(C. We can use the ideal-gas law to find the number of molecules N. N N (1) Use the ideal-gas law to express N: N" PV kT Substitute numerical values and evaluate N: $101.The Second Law of Thermodynamics 1877 If we assume that the freezer unit operates in a Carnot cycle.V + / 1.basement COPkit !1 Tc.kit COPbasement " Th.basement 293 K !1 COPbasement 268 K " 1. 2 + where N is the number of air . substitute numerical values and evaluate the ratio of the coefficients of performance: Th. + / 10 . 1 or.47 " 285 K COPkit !1 268 K or an increase of 47% in the performance of the freezer! 22 •• Estimate the probability that all the molecules in your bedroom are located in the (open) closet which accounts for about 10% of the total volume of the room.381110!23 J/K %$293 K % " 1.V + / 1. respectively. normally in your bedroom. 01p". Assume the engine operates according to the Otto cycle and assume 3 = 1. To make the sale.0%1. ) Picture the Problem The maximum efficiency of an automobile engine is given by the efficiency of a Carnot engine operating between the same two temperatures. Wanting to buy the most efficient refrigerator possible. We can use the expression for the Carnot efficiency and the equation relating V and T for a quasi-static adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio. h + . . She decides to return the next day to buy the most efficient refrigerator.252110 27 " 10 !1.0.252110 27 2 10!10 23 •• [SSM] Estimate the maximum efficiency of an automobile engine that has a compression ratio of 8.4 for diatomic gases) and evaluate #C: 1 2 56% $8.0:1. 1. 3 !1 Express the compression ratio r: r" Vc Vh 1 Substituting for r yields: #C " 1 ! # C " 1! r 3 !1 Substitute numerical values for r and 3 (1.V + / c.V + Th Vc / c.1878 Chapter 19 Substitute for N in equation (1) and evaluate p: 01p". and (b) and the highest rate possible for the heat to be released by the refrigerator if the refrigerator uses 600 W of electrical power. 3 !1 TcVc 3 !1 " ThVh 3 !1 Tc to obtain: Th 0V #C " 1 ! . your physics professor comes into your store to buy a new refrigerator. + / 10 .4!1 24 •• You are working as an appliance salesperson during the summer. (The Otto cycle is discussed in Section 19-1. Express the Carnot efficiency of an engine operating between the temperatures Tc and Th: Relate the temperatures Tc and Th to the volumes Vc and Vh for a quasistatic adiabatic compression from Vc to Vh: Substitute for #C " 1 ! Tc Th T V 3 !1 0 V 4 c " h3 !1 " .4. h + . she asks you about the efficiencies of the available models.25211027 " 27 1 101. One day. you need to provide her with the following estimates: (a) the highest COP possible for a household refrigerator. for the sake of finding the upper limit on the COP. then the refrigerator must be able to maintain a temperature difference of about 30 K.1%$600 J/s % " 5. (a) Using its definition.The Second Law of Thermodynamics 1879 Picture the Problem If we assume that the temperature on the inside of the refrigerator is 0(C (273 K) and the room temperature to be about 30(C (303 K). express the COP of a household refrigerator: Apply conservation of energy to the refrigerator to obtain: Substitute for W and simplify to obtain: COP " Qc W (1) W & Qc " Qh 4 W " Qh ! Qc Qc 1 " Qh ! Qc Qh !1 Qc COP " Assume. In (b) we can solve the definition of COP for Qc and differentiate the resulting equation with respect to time to estimate the rate at which heat is being drawn from the refrigerator compartment. that the refrigerator is a Carnot refrigerator and relate the temperatures of the hot and cold reservoirs to Qh and Qc: Substitute for Qh Th " Qc Tc Qh to obtain: Qc COPmax " 1 Th !1 Tc 1 " 9.5 kW dt . We can use the definition of the COP of a refrigerator and the relationship between the temperatures of the hot and cold reservoir and Qh and Qc to find an upper limit on the COP of a household refrigerator.1 303 K !1 273 K (2) Substitute numerical values and evaluate COPmax: COPmax " (b) Solve equation (1) for Qc: Differentiate equation (2) with respect to time to obtain: Substitute numerical values and dQc evaluate : dt Qc " W $COP % dQc dW " $COP % dt dt dQc " $9. the average temperature of the surface of Earth is about 290 K. and (d) 1. (a) Using its definition.02 11014 J/K 6 s 26 •• A 1. (b) Estimate the net rate at which Earth’s entropy is increasing due to this solar radiation. (c) 1000. (a) Estimate the total power of the sunlight hitting Earth. The solar constant (the intensity of sunlight reaching Earth’s atmosphere) is about 1. P " 5 1. Picture the Problem We can use the definition of intensity to find the total power of sunlight hitting the earth and the definition of the change in entropy to find the changes in the entropy of Earth and the Sun resulting from the radiation from the Sun. If a vacuum chamber has the same volume as the box. express the intensity of the Sun’s radiation on Earth in terms of the power P delivered to Earth and Earth’s cross sectional area A: Solve for P and substitute for A to obtain: Substitute numerical values and evaluate P: I" P A P " IA " I5R 2 where R is the radius of Earth. (e) The best vacuums that have been created to date have pressures of about 10–12 torr. (b) 100. Calculate the average time it should take before we observe all N molecules in the left half of the box if N is equal to (a) 10. it would have a 50% chance of being on one side or the other.1880 Chapter 19 25 •• [SSM] The average temperature of the surface of the Sun is about 5400 K.746 11017 W " dt 290 K " 6.37 kW/m 2 6. Picture the Problem If you had one molecule in a box. With two molecules. how long will a physicist have to wait before all of the gas molecules in the vacuum chamber occupy only the left half of it? Compare that to the expected lifetime of the universe. there are four .0-L box contains N molecules of an ideal gas.0 mole. and the positions of the molecules are observed 100 times per second. We don’t care which side the molecules are on as long as they all are on one side.746 1 1017 W " 1.37 kW/m2. which is about 1010 years. so with one molecule you have a 100% chance of it being on one side or the other.75 1 1017 W dS Earth P " dt TEarth (b) Express the rate at which Earth’s entropy SEarth changes due to the flow of solar radiation: Substitute numerical values and dS Earth evaluate : dt dS Earth 1.37 1 10 6 m $ %$ % 2 " 1. 022110 let 23 10 x " 26. if the molecules shuffle 100 times a second. and the reverse). the time it would take them to cover all the combinations and all get on one side 2N or the other is t " .156 1 107 s " 0. so there is a 25% (1 in 4) chance of them both being on a particular side.022110 and take the logarithm of both sides of the equation to obtain: 23 26.34 1 1027 s 1 2 2 1 1020 y (c) Evaluate t for N = 1000 molecules: t" 21000 2 100 s !1 $ % To evaluate 21000 let 10 x " 21000 and take the logarithm of both sides of the equation to obtain: Substitute to obtain: $1000%ln 2 " x ln10 4 x " 301 10301 t" 2 100 s !1 $ % 1y 3.022 110 %ln 2 " x ln10 4 x 2 10 23 .5 1 10 299 s 1 2 2 1 10 291 y (d) Evaluate t for N = 1.022110 t" 2 100 s !1 23 $ % 23 $6.0 L of air at a pressure of 10!12 torr and an assumed temperature of 300 K. Extending this logic.12 s 2 5 s 2$100 s !1 % (b) Evaluate t for N = 100 molecules: t" 2100 2 100 s !1 $ % 1y 3. one on one side and one on the other. which means that.The Second Law of Thermodynamics 1881 possible combinations (both on one side. both on the other. (a) Evaluate t for N = 10 molecules: 210 t" " 5. the probability of N molecules all being on one side of the box is P = 2/2N.156 1 107 s " 6.0 mol =6.022 11023 molecules: To evaluate 26. In (e) we can apply the ideal gas law to find the number 2$100 % of molecules in 1. or a 50% chance of them both being on either side. we can express the efficiency of the engine in terms of the heat Qc . 7 + 2 100 s / 3.0% efficiency does 0.156 110 s .22110 and take the logarithm of both sides of the equation to obtain: Substituting for x yields: 7 N" PV kT !12 $10 N" torr %$133. substitute numerical values and evaluate N: Evaluate t for N = 3. (a) How much heat is absorbed from the hot reservoir during each cycle? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be # " W Qh where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle.22110 t" 2 100 s !1 7 $ % 7 $3.22 1 10 %ln 2 " x ln10 4 x 2 10 7 1010 1y 1 t" !1 2 100 s 3. 23 $ % 2 1010 y 23 (e) Solve the ideal gas law for the number of molecules N in the gas: Assuming the gas to be at room temperature (300 K).22 1 107 molecules 23.156 1 107 s 7 $ % 2 1010 y 7 Express the ratio of this waiting time to the lifetime of the universe tuniverse: t tuniverse or 7 1010 y " 10 2 1010 10 y 7 t 2 1010 tuniverse 7 Heat Engines and Refrigerators [SSM] A heat engine with 20. from conservation of energy. Qh " W & Qc .32 Pa/torr %$1.100 kJ of work 27 • during each cycle.0 L % $1. (b) Because.22110 let 7 10 x " 23.221107 molecules: To evaluate 23.381110!23 J/K %$300 K % " 3.1882 Chapter 19 Substituting for x yields: 1010 1y 0 t2 !1 . 200 Qc " Qh ! W " 500 J ! 100 J " 400 J 28 • A heat engine absorbs 0. (a) Qh absorbed from the hot reservoir during each cycle is given by: (b) Use Qh " W & Qc to obtain: Qh " W " # 100 J " 500 J 0. (b) We can apply conservation of energy to the engine to obtain Qh " W & Qc and solve this equation for the heat Qc released to the cold reservoir during each cycle.50 s.400 kJ of heat from the hot reservoir and does 0. find the power output of this engine.The Second Law of Thermodynamics 1883 released to the cold reservoir during each cycle. Picture the Problem We can use its definition to find the efficiency of the engine and the definition of power to find its power output. (a) What is its efficiency? (b) If each cycle takes 0.120 kJ of work during each cycle. (a) What is its efficiency? (b) How much heat is released to the cold reservoir during each cycle? Picture the Problem (a) The efficiency of the engine is defined to be # " W Qh where W is the work done per cycle and Qh is the heat absorbed from the hot reservoir during each cycle. (a) The efficiency of the heat engine is given by: Substitute numerical values and evaluate #: #" Q W Qh ! Qc " " 1! c Qh Qh Qh # " 1! 60 J " 40% 100 J . (a) The efficiency of the heat engine is given by: (b) Apply conservation of energy to the engine to obtain: Substitute numerical values and evaluate Qc: #" W 120 J " " 30% Qh 400 J Qh " W & Qc 4 Qc " Qh ! W Qc " 400 J ! 120 J " 280 J 29 • A heat engine absorbs 100 J of heat from the hot reservoir and releases 60 J of heat to the cold reservoir during each cycle. The gas is heated at constant volume to P2 = 2.0 kJ #" #" Qh ! Qc Q " 1! c Qh Qh # " 1! 5.00 atm and V1 = 24. The gas is then cooled at constant volume until its .6 L.0 kJ " 38% 8.500 s + " 80 W + / . The cycle begins at P1 = 1. what is its efficiency? Picture the Problem We can apply their definitions to find the COP of the refrigerator and the efficiency of the heat engine. If it is run backward as a heat engine between the same two reservoirs. .1884 Chapter 19 (b) The power output P of this engine is the rate at which it does work: Substitute numerical values and evaluate P: dQh dW d " # Qh " # dt dt dt P" 0 100 J P " $0. 0. (b) The refrigerator is reversible.7 8. It then expands at constant pressure until its volume is 49.0 kJ to a hot reservoir.0 kJ 31 •• [SSM] The working substance of an engine is 1.0 kJ " 1.0 kJ ! 5.40 %.00 mol of a monatomic ideal gas.0 kJ of heat from a cold reservoir and releases 8.00 atm. 30 • A refrigerator absorbs 5. (a) Find the coefficient of performance of the refrigerator.2 L. (a) The COP of a refrigerator is defined to be: Apply conservation of energy to relate the work done per cycle to Qh and Qc: Substitute for W to obtain: COP " Qc W W " Qh ! Qc COP " Qc Qh ! Qc Substitute numerical values and evaluate COP: (b) The efficiency of a heat pump is defined to be: Apply conservation of energy to the heat pump to obtain: Substitute numerical values and evaluate # : COP " W Qh 5. The Second Law of Thermodynamics 1885 pressure is again 1.00 atm. (a) Show this cycle on a PV diagram. All the steps are quasi-static and reversible. + mol 6 K .00 mol%0 8. Hence: The volume doubles while the pressure remains constant between states 2 and 3. 3. find the work done by the gas. the heat absorbed by the gas. and the change in the internal energy of the gas. It is then compressed at constant pressure to its original state.00 atm %$24. (b) Find the efficiency of the cycle.6 L % $1. For each step of the cycle. We can use the 1st law of thermodynamics to find the change in internal energy for each step of the cycle. we can find the efficiency of the cycle from the work done each cycle and the heat that enters the system each cycle. Finally. Hence: The pressure is halved while the volume remains constant between states 3 and 4. / T2 " 2T1 " 600 K " 300 K The pressure doubles while the volume remains constant between states 1 and 2.206 110 !2 L 6 atm . Hence: T3 " 2T2 " 1200 K T4 " 1 T3 " 600 K 2 . We can then find the work done on the gas during each process from the area under each straight-line segment and the heat that enters the system from Q " CV )T and Q " CP )T . (a) The cycle is shown to the right: Apply the ideal-gas law to state 1 to find T1: T1 " P1V1 " nR $1. 2. and 4. Picture the Problem To find the heat added during each step we need to find the temperatures in states 1. J 0 Q23 " C P "T23 " 5 R"T23 " 5 . and J 0 Q41 " C P "T41 " 5 R"T41 " 5 .325 J Won " !W41 " ! P"V41 " !$1.24 kJ 2 2 mol 6 K .99 kJ / L 6 atm .6 L %. / The change in the internal energy of the system as it goes from state 1 to state 2 is given by the 1st law of thermodynamics: Because W12 " 0 : For path 2*3: "Eint " Qin & Won "Eint. 34 " !7. 8.314 + $300 K ! 600 K % " ! 6.1886 Chapter 19 For path 1*2: W12 " P"V12 " 0 and J 0 Q12 " C V "T12 " 3 R"T12 " 3 .5 kJ and J 0 Q34 " "Eint. 8.325 J Won " !W23 " ! P"V23 " !$2.5 kJ ! 4. 34 " C V "T34 " 3 R"T34 " 3 .49 kJ / L 6 atm . / Apply "Eint " Qin & Won to obtain: For path 3*4: W34 " P)V34 " 0 "Eint. 23 " 12.2 L %.12 " Q12 " 3. 8.00 atm %$49.314 + $600 K ! 1200 K % " ! 7.99 kJ " 7.314 + $600 K ! 300 K % " 3.74 kJ 2 2 mol 6 K .6 L ! 49.74 kJ 0 101.48 kJ & 0 " ! 7. + " ! 4.5 kJ 2 2 mol 6 K .314 + $1200 K ! 600 K % " 12.48 kJ 0 101.2 L ! 24.48 kJ 2 2 mol 6 K . 8. + " 2.00 atm %$24. / . / Apply "Eint " Qin & Won to obtain: For path 4*1: "Eint. 32 •• The working substance of an engine is 1. the sum of the changes in the internal energy for the cycle is zero. (2) a compression at constant pressure to its original volume of 10. 41 " !6.74 7.00 atm and a volume of 20.0 V(L) The pressures and volumes at the end points of the adiabatic expansion are related according to: 0V P1V1 " P2V2 4 P1 " .0 L to a pressure of 1.V + / 1.639 2 1 1 3 2 0 0 10. Note also that. P (atm) 2.0 L.49 kJ 2 15% 3.5 !7.5 kJ Remarks: Note that the work done per cycle is the area bounded by the rectangular path. kJ 1*2 2*3 3*4 4*1 0 !4.24 3.74 kJ & 12.00 mol of a diatomic ideal gas.74 12.0 20.75 kJ For easy reference.48 !3.75 Wby Qin (b) The efficiency of the cycle is given by: Substitute numerical values and evaluate #: #" " ! W23 & $! W41 % Q12 & Q23 #" 4. Picture the Problem The three steps in the process are shown on the PV diagram. 2 + P2 . Find the efficiency of this cycle.49 kJ " ! 3. kJ "Eint $" Qin & Won % . We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle. as expected because the system returns to its initial state.99 0 2.99 kJ ! 2.48 !6.49 3. 3 3 3 . The engine operates in a cycle consisting of three steps: (1) an adiabatic expansion from an initial volume of 10. the results of the preceding calculations are summarized in the following table: Process Won . and (3) heating at constant volume to its original pressure.The Second Law of Thermodynamics 1887 Apply "Eint " Qin & Won to obtain: "Eint.24 kJ & 2. kJ Qin .0 L.5 !7. 0 L . + / 10.6 L and a temperature of 400 K performs a cycle consisting of four steps: (1) an isothermal expansion at 400 K to twice its initial volume. and (4) heating at constant volume to its original temperature of 400 K. Picture the Problem We can find the efficiency of the cycle by finding the work done by the gas and the heat that enters the system per cycle.0 atm 6 L & 41.0 L % " !35.00 atm %$10.0 atm 6 L Substitute numerical values in equation (1) and evaluate # : #" 6.639 atm ! 1.0 L ! 20.00 atm %$10.0 L P1 " . cycle " 0 ) to obtain: Q12 " 0 Q23 " C V "T23 " 7 R"T23 " 7 P"V23 2 2 " 7 2 $1.0 atm 6 L " 6.0 atm 6 L " 15% 41atm 6 L 33 •• An engine using 1. 1.00 atm% " 2.00 mol of an ideal gas initially at a volume of 24.0 atm 6 L Q31 " C V "T31 " 5 R"T31 " 5 "PV31 2 2 " 5 2 $2.0 L % " 41.4 $1. Sketch the cycle on a PV diagram and find its efficiency. Assume that Cv = 21.0 J/K.0 atm 6 L Won " "Eint ! Qin " !Qin " Q12 & Q23 & Q31 " 0 ! 35. (2) cooling at constant volume to a temperature of 300 K (3) an isothermal compression to its original volume. .639 atm (1) #" W Qh No heat enters or leaves the system during the adiabatic expansion: Find the heat entering or leaving the system during the isobaric compression: Find the heat entering or leaving the system during the constantvolume process: Apply the 1st law of thermodynamics to the cycle ( )Eint.1888 Chapter 19 Substitute numerical values and evaluate P1: Express the efficiency of the cycle: 0 20. and because the internal energy does not change during step 1 while work is done by the gas.V / A + + . A. 0V W3 " nRTc ln. C.V / A + + .3 & Qh. and 4 represent the four steps through which the gas is taken.5 4 A 1 D 3 C B 2 400 K 300 K 1 0. 2 & Qh. B . W2 = W4 = 0: Because the internal energy of the gas increases in step 4 while no work is done. B. heat enters the system only during these processes: The work done during the isothermal expansion (1) is given by: The work done during the isothermal compression (3) is given by: Because there is no change in the internal energy of the system during step 1.5 0 0 10 20 30 40 V (L) 50 60 Express the efficiency of the cycle: #" W1 & W2 & W3 & W4 W " Qh Qh. B . D + . 2. 2 & Qh.1 & Qh.1 & Qh. 4 W1 & 0 & W3 & 0 W " Qh Qh. Q4 " C V "T " C V $Th ! Tc % . 0V Q1 " W1 " nRTh ln. and D identify the four states of the gas and the numerals 1.3 & Qh.1 & Qh.The Second Law of Thermodynamics 1889 The PV diagram of the cycle is shown to the right. 3. 2 P (atm) 1. 4 Because steps 2 and 4 are constantvolume processes. 4 W & W3 W " 1 Qh Qh.V + / C. the heat that enters the system during this isothermal expansion is given by: The heat that enters the system during the constant-volume step 4 is given by: #" #" (1) 0V W1 " nRT ln. / C. (b) the heat transfer for each part of the cycle. Noting the VB V 1 " 2 and D " . Because the change in internal energy is zero for the isothermal process.V + / A. #" 0V nRTh ln.0 L. / J + mol 6 K .1% J 21.0 K $400 K ! 300 K % 400 K & J 0 $1.00 mol%0 8.1890 Chapter 19 Substituting in equation (1) yields: 0V 0V nRTh ln. and (c) the efficiency of the cycle.00 mol of an ideal monatomic gas initially at a volume of 25.0 L % $1. / 34 •• Figure 19-15 shows the cycle followed by 1. we can find the efficiency of the cycle from its definition.314 + ln $2% mol 6 K . Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas and the heat capacities at constant volume and constant pressure to find the heat flow for the constant-volume and isobaric processes. + Th ! Tc / 2 . " Th ln $2% ! Tc ln $2% " #" C C CV $Th ! Tc % Th ln $2% & V $Th ! Tc % Th ln $2% & V $Th ! Tc % Th & nR nR nR ln $2% Substitute numerical values and evaluate #: #" 400 K ! 300 K " 13. 8. Determine (a) the temperature of each numbered state of the cycle.V + .314 . Finally. " 301 K . B + & C V $Th ! Tc % . B + & nRTc ln. All the processes are quasi-static. D + .00 mol%. (a) Use the ideal-gas law to find the temperature at point 1: T1 " P1V1 " nR $100 kPa %$25.V + / A. substitute and simplify to obtain: VA VC 2 01Th ln $2% & Tc ln. we can use the expression for the work done on or by a gas during an isothermal process to find the heat flow during such a process. 74 kJ & 3.314 + $601 K %ln.74 kJ 2 2 mol 6 K . 8. / " 601 K (b) Find the heat entering the system for the constant-volume process from 1 * 2: J 0 Q12 " C V "T12 " 3 R"T12 " 3 . Determine (a) the temperatures of the other three numbered states of the cycle and (b) the efficiency of the cycle.314 + $301 K ! 601 K % " ! 6. Picture the Problem We can use the ideal-gas law to find the temperatures of each state of the gas. We can find the efficiency of the cycle from its definition. Find the heat leaving the system during the isobaric compression from 3 * 1: J 0 Q31 " C P "T31 " 5 R"T31 " 5 . . 8.V + mol 6 K . / / .24 kJ 2 2 mol 6 K .24 kJ " 0.96 kJ because.96 kJ " 13% 3. / (c) Express the efficiency of the cycle: Apply the 1st law of thermodynamics to the cycle: #" W W " Qin Q12 & Q23 (1) W " 7 Q " Q12 & Q23 & Q31 " 3.0 L J 0 Q23 " nRT2 ln.314 + $601 K ! 301 K % " 3. The temperature of state 1 is 200 K.0 L + " 3. 8. / 2. + mol 6 K .74 kJ & 3. Substitute numerical values in equation (1) and evaluate # : #" 0.314 J . for the cycle. 3 + " $1.0 L % " $1.46 kJ 35 •• An ideal diatomic gas follows the cycle shown in Figure 19-16. .00 mol%.46 kJ + .46 kJ ! 6.00 mol%0 8.The Second Law of Thermodynamics 1891 Use the ideal-gas law to find the temperatures at points 2 and 3: T2 " T3 " P2V2 nR $200 kPa %$25. 25. / Find the heat entering or leaving the system for the isothermal process from 2 * 3: 0V 0 5 0 . "Eint " 0 . 0 atm % 600 K T3 " T2 P3V3 V " T2 3 P2V2 V2 T3 " $600 K % $300 L % " $100 L % 1800 K T4 " T3 P4V4 P " T3 4 P3V3 P3 T4 " $1800 K % $1.0 atm % " $1.0 atm ! 1. express Qin: Qin " Q12 & Q23 " C V )T12 & C P )T23 " 5 nR)T12 & 7 nR)T23 2 2 " $ 5 )T12 & 7 )T23 %nR 2 2 .0 atm % " 400 atm 6 L Apply the ideal-gas law to state 1 to find the product of n and R: nR " P1V1 $1.0 atm % 600 K #" W Qin (1) Use the area of the rectangle to find the work done each cycle: W " "P"V " $300 L ! 100 L %$3.50 L 6 atm/K Noting that heat enters the system between states 1 and 2 and states 2 and 3. (a) Use the ideal-gas law for a fixed amount of gas to find the temperature in state 2 to the temperature in state 1: Substitute numerical values and evaluate T2: Apply the ideal-gas law for a fixed amount of gas to states 2 and 3 to obtain: Substitute numerical values and evaluate T3: Apply the ideal-gas law for a fixed amount of gas to states 3 and 4 to obtain: Substitute numerical values and evaluate T4: (b) The efficiency of the cycle is: PV P PV1 P2V2 1 " 4 T2 " T1 2 2 " T1 2 T1 T2 PV1 P 1 1 T2 " $200 K % $3.0 atm % " $3.1892 Chapter 19 using the area enclosed by the cycle to find the work done per cycle and the heat entering the system between states 1 and 2 and 2 and 3 to determine Qin.0 atm %$100 L % " T1 200 K " 0. 0. P S 2 (2) (3) Th (1) Tc (4) )V = 0 3 Tc 4 V )V = 0 Th 1 T (b) The change in entropy for one Stirling cycle is the sum of the entropy changes during the cycle: "S cycle " "S12 & "S 23 & "S 34 & "S 41 (1) Express the entropy change for the isothermal process from state 1 to state 2: 0V "S12 " nR ln.The Second Law of Thermodynamics 1893 Substitute numerical values and evaluate Qin: L 6 atm 0 Qin " 8 5 $600 K ! 200 K % & 7 $1800 K ! 600 K %9. and (4) cooling of the gas at constant volume. / Substitute numerical values in equation (1) and evaluate # : #" 400 atm 6 L " 15% 2600 atm 6 L 36 ••• Recently. .50 + " 2600 atm 6 L 2 2 K .V + / 1. The cycle of a Stirling engine is as follows: (1) isothermal compression of the working gas (2) heating of the gas at constant volume. (3) an isothermal expansion of the gas. (a) Sketch PV and ST diagrams for the Stirling cycle. 2 + . Picture the Problem (a) The PV and ST cycles are shown below. known as the Stirling engine has been promoted as a means of producing power from solar energy. (b) We can show that the entropy change during one Stirling cycle is zero by adding up the entropy changes for the four processes. an old design for a heat engine. (b) Find the entropy change of the gas for each step of the cycle and show that the sum of these entropy changes is equal to zero. T / c + + . Princeton University Press (1988). 2 + . 1 + " ! nR ln. Picture the Problem We can use the efficiency of a Carnot engine operating between reservoirs at body temperature and typical outdoor temperatures to find an upper limit on the efficiency of an engine operating between these temperatures. 2 + ! C V ln. and a general knowledge of the conditions under which most warm-blooded organisms exist.T / i + + . (a) Calculate the efficiency of a heat engine operating between body temperature (98.6ºF) and a typical outdoor temperature (70ºF). c . c .V + / 2. 0V 0V "S 34 " nR ln. c . Life’s Devices. because V2 = V3 and V1 = V4. h .1894 Chapter 19 Similarly. or. + + .V + . the entropy change for the isothermal process from state 3 to state 4 is: 0V "S 34 " nR ln. / 1. 0T + " !C V ln. 4 + . 0V 0T + ! nR ln.V + / h / 1. give a reason why no warm-blooded organisms have evolved heat engines to increase their internal energies.T + . and compare this to the human body’s efficiency for converting chemical energy into work (approximately 20%).V + / h . Does this efficiency comparison contradict the second law of thermodynamics? (b) From the result of Part (a). f . Nature has never evolved a heat engine!—Steven Vogel. / 1. 2 + & C V ln.T + / h . 37 •• !As far as we know. The change in entropy for a constantvolume process is given by: "S isochoric dQ f nC V dT ": ": T T Ti 0T " nC V ln. c . T For the constant-volume process from state 2 to state 3: For the constant-volume process from state 4 to state 1: Substituting in equation (1) yields: 0T 0V "S cycle " !nR ln.T / h 0T "S 41 " C V ln. 0T "S 23 " C V ln. (a) Express the maximum efficiency of an engine operating between body temperature and 70°F: #C " 1 ! Tc Th . +" 0 + .T .V + / 3. 16% 310 K The fact that this efficiency is considerably less than the actual efficiency of a human body does not contradict the second law of thermodynamics. (b) Most warm-blooded animals survive under roughly the same conditions as humans. Note that no heat enters or leaves the system during the adiabatic processes ab and cd. To find the efficiency of the diesel cycle we can find the heat that enters the system and the heat that leaves the system and use the expression that gives the efficiency in terms of these quantities. Heat enters the system during the isobaric process bc and leaves the system during the isovolumetric process da.The Second Law of Thermodynamics 1895 Use T " 5 9 $tF ! 32% & 273 to obtain: Tbody " 310 K and Troom " 294 K Substitute numerical values and evaluate # C : # C " 1! 294 K " 5. 38 ••• The diesel cycle shown in Figure 19-17 approximates the behavior of a diesel engine. Qda (the constant-volume cooling process) is given by: Substitute for Qh and Qc and simplify using 3 " C P C V to obtain: #" Q W Qh ! Qc " " 1! c Qh Qh Qh Qbc " Qh " C P $Tc ! Tb % Qda " Qc " C V $Td ! Ta % # " 1! $T ! Ta % C V $Td ! Ta % " 1! d C P $Tc ! Tb % 3 $Tc ! Tb % . Express the efficiency of the cycle in terms of Qc and Qh: Express Q for the isobaric warming process bc: Because CV is independent of T. Picture the Problem The working fluid will be modeled as an ideal gas and the process will be modeled as quasistatic. The application of the second law to chemical reactions such as the ones that supply the body with energy have not been discussed in the text but we can note that we don’t get our energy from heat swapping between our body and the environment. internal body temperatures would have to be maintained at an unreasonably high level. we eat food to get the energy that we need. Find the efficiency of this cycle in terms of the volumes Va. process cd is an adiabatic expansion. Rather. process bc is an expansion at constant pressure. Process ab is an adiabatic compression. Vb and Vc. and process da is cooling at constant volume. To make a heat engine work with appreciable efficiency. / a 3 Second Law of Thermodynamics 39 •• [SSM] A refrigerator absorbs 500 J of heat from a cold reservoir and releases 800 J to a hot reservoir.V V + a . # " 1! / 3 $Tc ! Tb % 00 V . V Va + d . / a . / a V ! b Vc 3 !1 + + . Assume that the heat-engine statement of the second law of thermodynamics is false. 3 !1 Because Va = Vd: T 0V ! b.V + . / Noting that Pb = Pc. apply the idealgas law to relate Tb and Tc: Tb Vb " Tc Vc 0 Substitute for the ratio of Tb to Tc and simplify to obtain: 0 Vc . Tc 3 !1 ! Tb 3 !1 + . Va / # " 1! / + + . c ! b + .V + V 3 ! Vb3 " 1 ! / a . 3 . " 1 ! 3 !c1 3Va $Vc ! Vb % 0V V 3. and show how a perfect engine working .1896 Chapter 19 Using an equation for a quasistatic adiabatic process..V # " 1! / a + + . + + . 3 . relate the temperatures Tc and Td to the volumes Vc and Vd: Use equations (1) and (2) to eliminate Ta and Td: Vb3 !1 4 Ta " Tb 3 !1 (1) Va TaVa 3 !1 " TbVb 3 !1 TcVc3 !1 " TdVd3 !1 4 Td " Tc Vc3 !1 (2) Vd3 !1 0 Vc3 !1 Vb3 !1 . . V ! b Vc 3 + + . 3 !1 0 Vc .1 ! . 0 Vb . b Tc . . .1 ! + + / Vc . + !.V Va 6 " 1! / a Vc Va + + . . Va / Tb + Tc + . relate the temperatures Ta and Tb to the volumes Va and Vb: Proceeding similarly.0 Vb . + . .V / a 0 Vc Vb 3.V / a 0 Vb . ! + . c . 3 !1 0 Vc Vc .V V + a . 3 !1 0 Vb . 3 !1 + + . Determine the Concept The work done by the system is the area enclosed by the cycle. Then. Ordinary refrigerator 300 J 300 J 500 J (a) Cold reservoir at temperature Tc (b) (c ) If two curves that represent quasi-static adiabatic processes could 40 •• intersect on a PV diagram. . We could use this perfect heat engine to remove 300 J of energy from the hot reservoir and do 300 J of work on the ordinary refrigerator (see (b) in the diagram). Show that such a cycle violates the second law of thermodynamics.This violates the refrigerator statement of the second law. Thus. (a) What is its efficiency? (b) If it . Suppose the heat-engine statement of the second law is false. in violation of the second law of thermodynamics. a cycle could be completed by an isothermal path between the two adiabatic curves shown in Figure 19-18. . There is no heat transfer in the adiabatic expansion or compression. without exhausting any heat to a cold reservoir. 800 J 300 J 500 J . Determine the Concept The following diagram shows an ordinary refrigerator that uses 300 J of work to remove 500 J of heat from a cold reservoir and releases 800 J of heat to a hot reservoir (see (a) in the diagram). where we assume that we start with the isothermal expansion. It is only in this expansion that heat is extracted from a reservoir. the combination of the perfect heat engine and the ordinary refrigerator would be a perfect refrigerator. 500 J Perfect heat engine Perfect refrigerator . we would completely convert heat to mechanical energy. Hot reservoir at temperature Th .The Second Law of Thermodynamics 1897 with this refrigerator can violate the refrigerator statement of the second law of thermodynamics. transferring 500 J of heat from the cold reservoir to the hot reservoir without requiring any work (see (c) in the diagram). Then a 'perfect' heat engine could remove energy from the hot reservoir and convert it completely into work with 100 percent efficiency. . Carnot Cycles 41 • [SSM] A Carnot engine works between two heat reservoirs at temperatures Th = 300 K and Tc = 200 K. . 3 J " 66. We can find the COP of the engine working as a refrigerator from its definition.3% Th 300 K W " # C Qh " $0.333%$100 J % " 33.3 J Qc " Qh ! W " 100 J ! 33. relate the work done each cycle to the heat absorbed from the hot reservoir: (c) Apply conservation of energy to relate the heat given off each cycle to the heat absorbed and the work done: (d) Using its definition. (a) Express the efficiency of the engine in terms of the heat absorbed from the high-temperature reservoir and the heat exhausted to the lowtemperature reservoir: #" Q W Qh ! Qc " 1! c " Qh Qh Qh 200 J " 20.7 J " " 2 . where #C is the Carnot efficiency. (a) The efficiency of the Carnot engine depends on the temperatures of the hot and cold reservoirs: (b) Using the definition of efficiency.1898 Chapter 19 absorbs 100 J of heat from the hot reservoir during each cycle. how much work does it do each cycle? (c) How much heat does it release during each cycle? (d) What is the COP of this engine when it works as a refrigerator between the same two reservoirs? Picture the Problem We can find the efficiency of the Carnot engine using # " 1 ! Tc / Th and the work done per cycle from # " W / Qh .0 W 33.3 J 42 • An engine absorbs 250 J of heat per cycle from a reservoir at 300 K and releases 200 J of heat per cycle to a reservoir at 200 K. express and evaluate the refrigerator’s coefficient of performance: # C " 1! Tc 200 K " 1! " 33. (a) What is its efficiency? (b) How much additional work per cycle could be done if the engine were reversible? Picture the Problem We can find the efficiency of the engine from its definition and the additional work done if the engine were reversible from W " # CQh . We can apply conservation of energy to find the heat rejected each cycle from the heat absorbed and the work done each cycle.7 J " 67 J COP " Qc 66.0% 250 J " 1! . in violation of the second law. it releases 140 J per cycle of heat to the cold reservoir. thus replacing the heat removed by the refrigerator. / "W " 83.2.3 J .1 ! c . Working as a heat engine. deliver 200 J to the hot reservoir. the heat removed from the hot reservoir by this engine.3 J ! 50 J " 33 J 43 •• A reversible engine working between two reservoirs at temperatures Th and Tc has an efficiency of 30%. with #2 > 0. A second engine working between the same two reservoirs also does 100 J of work per cycle. then Qh2. Now take the second engine and run it between the same reservoirs. The end result of all this is that the second engine can run the refrigerator. The two systems working together then convert heat into mechanical energy without rejecting any heat to a cold reservoir. operate between the same two heat reservoirs and use it to drive the refrigerator. the two engines working together would violate the refrigerator statement of the second law. is greater than 30%.The Second Law of Thermodynamics 1899 (b) Express the additional work done if the engine is reversible: Relate the work done by a reversible engine to its Carnot efficiency: Substitute numerical values and evaluate W: Substitute numerical values in equation (1) and evaluate )W: )W " WCarnot ! WPart $a % (1) 0 T W " # C Qh " . T h / +Qh + . Determine the Concept Let the first engine be run as a refrigerator. Then it will remove 140 J from the cold reservoir. A second engine working between the same two reservoirs also releases 140 J per cycle to the cold reservoir. 44 •• A reversible engine working between two reservoirs at temperatures Th and Tc has an efficiency of 20%. is 140 J/(1 ! #2) > 200 J. replacing the heat taken from the cold reservoir. and the work done by this engine is W = #2Qh2 > 60 J. it does 100 J of work per cycle. the two engines working together would violate the heat-engine statement of the second law. and let it eject 140 J into the cold reservoir. and require 60 J of energy to operate. Determine the Concept If the reversible engine is run as a refrigerator. and do additional mechanical work. Now let the second engine. Working as a heat engine. If #2. 300 K + $250 J % " 83. the efficiency of this engine. Show that if the second engine has an efficiency greater than 30%.1 ! + . 0 200 K W " . . Show that if the efficiency of the second engine is greater than 20%. it will require 100 J of mechanical energy to take 400 J of heat from the cold reservoir and deliver 500 J to the hot reservoir. this engine will remove less than 500 J from the hot reservoir in the process of doing 100 J of work.0 K Tc " 1! " 74.1900 Chapter 19 Because #2 > 0. how much work does it do? (c) How much heat does it release to the low-temperature reservoir during each cycle? (d) What is the coefficient of performance of this engine when it works as a refrigerator between these two reservoirs? Picture the Problem We can use the definitions of the efficiency of a Carnot engine and the coefficient of performance of a refrigerator to find these quantities. So running engine described in Part (a) to operate the refrigerator with a COP > 2 will result in the transfer of heat from the cold to the hot reservoir without doing any net mechanical work in violation of the second law. The work done each cycle by the Carnot engine is given by W " # CQh and we can use the conservation of energy to find the heat rejected to the low-temperature reservoir.3% 300 K Th . 100 J of heat are absorbed and 150 J are released to the hot reservoir. 45 •• A Carnot engine works between two heat reservoirs as a refrigerator. then 50 J of work will remove more than 100 J of heat from the cold reservoir and put more than 150 J of heat into the hot reservoir. (a) What is the efficiency of the Carnot engine when it works as a heat engine between the same two reservoirs? (b) Show that no other engine working as a refrigerator between the same two reservoirs can have a COP greater than 2. (a) The efficiency of the Carnot engine is given by: #C " W 50 J " " 33% Qh 150 J (b) If the COP > 2. (a) The efficiency of a Carnot engine depends on the temperatures of the hot and cold reservoirs: #C " 1! 77. Picture the Problem We can use the definition of efficiency to find the efficiency of the Carnot engine operating between the two reservoirs.2. The net result is then that no net work is done by the two systems working together.0 K. 46 •• A Carnot engine works between two heat reservoirs at temperatures Th = 300 K and Tc = 77.00. During each cycle. in violation of the refrigerator statement of the second law. but a finite amount of heat is transferred from the cold reservoir to the hot reservoir. (a) What is its efficiency? (b) If it absorbs 100 J of heat from the hot reservoir during each cycle. It is then compressed at constant pressure back to its original state. We can use Q " C V )T and Q " C P )T to find the heat entering and leaving during the constant-volume and isobaric processes and the first law of thermodynamics to find the work done each cycle.35 W 74. express and evaluate the refrigerator’s coefficient of performance: W " # C Qh " $0. we can use its definition to find the efficiency of the cycle and the definition of the Carnot efficiency to find the efficiency of a Carnot engine operating between the extreme temperatures. (c) the efficiency of this cycle. (a) Apply the ideal-gas law for a fixed amount of gas to relate the temperature at point 3 to the temperature at point 1: PV1 P3V3 1 " T1 T3 or. V T3 " T1 3 V1 PV T PV1 P2V2 1 " 4 P2 " 1 1 2 T1 T2 V2T1 (1) Apply the ideal-gas law for a fixed amount of gas to relate the pressure at point 2 to the temperatures at points 1 and 2 and the pressure at 1: .3 J Qc " Qh ! W " 100 J ! 74. The gas is heated at constant volume to T2 = 150ºC and is then expanded adiabatically until its pressure is again 1. and pressures at the end points of each process in the given cycle.3 J COP " 47 •• [SSM] In the cycle shown 1 in Figure 19-19. 1.00 atm and a temperature of 0. volumes. because P1 = P3. Once we’ve calculated these quantities. (b) the heat absorbed or released by the system during each step.743%$100 J % " 74.0ºC.00 atm. Find (a) the temperature after the adiabatic expansion. Picture the Problem We can use the ideal-gas law for a fixed amount of gas and the equations of state for an adiabatic process to find the temperatures.The Second Law of Thermodynamics 1901 (b) Express the work done each cycle in terms of the efficiency of the engine and the heat absorbed from the high-temperature reservoir: (c) Apply conservation of energy to obtain: (d) Using its definition.3 J " 26 J Qc 26 J " " 0. and (d) the efficiency of a Carnot cycle operating between the temperature extremes of this cycle.00 mol of an ideal diatomic gas is initially at a pressure of 1. 00 atm % " 1. 8.1902 Chapter 19 Because V1 = V2: P2 " P1 T2 423 K " $1. / " ! 2.1. 1 + 1 . Evaluating W yields: W " 7 Q " Q12 & Q23 & Q31 " 3. 1 0 1.314 + $273 K ! 373 K % 2 mol 6 K .4 L %. Substitute numerical values in equation (1) and evaluate T3 and t3: T3 " $273 K % 30. 1atm + " 30.4 L and t 3 " T3 ! 273 " 100(C (b) Process 1*2 takes place at constant volume (note that 3 = 1. because )Eint.12 kJ & 0 ! 2. cycle " 0 (the system begins and ends in the same state) and Won " !Wby the gas " Qin .4 L.6 L / . evaluate V3: 0 P -3 PV13 " P3V33 4 V3 " V1 .12 kJ Q23 " 0 Process 2*3 takes place adiabatically: Process 3*1 is isobaric (note that CP = CV + R): Q31 " C P "T31 " 7 R"T12 2 J 0 " 7 .314 + $423 K ! 273 K % 2 mol 6 K .55 atm T1 273 K 1 Apply an equation for an adiabatic process to relate the pressures and volumes at points 2 and 3: Noting that V1 = 22. 8.6 L " 373 K 22.4 corresponds to a diatomic gas and that CP – CV = R): Q12 " CV "T12 " 5 R"T12 2 J 0 " 5 .P + / 3.4 V3 " $22.91 kJ " 0. + .55 atm . / " 3.91 kJ (c) The efficiency of the cycle is given by: Apply the first law of thermodynamics to the cycle: #" W Qin (2) "Eint " Qin & Won or.21 kJ . (a) The efficiency of the steam engine as a percentage of the maximum possible efficiency is given by: The efficiency of a Carnot engine operating between temperatures Fc and Th is: Substituting for #max yields: # steam engine 0.00 h.52% Th 543 K # steam engine 0.300 " " 74.5% 423 K Th # C " 1! 48 •• You are part of a team that is completing a mechanical-engineering project. how much heat does the engine release to its surroundings in 1. Your team has measured its efficiency to be 30.21 kJ " 6.0%.00 h? Picture the Problem We can find the maximum efficiency of the steam engine by calculating the Carnot efficiency of an engine operating between the given temperatures. Your team built a steam engine that takes in superheated steam at 270ºC and discharges condensed steam from its cylinder at 50.300 " # max # max # max " 1 ! Tc 323 K " 1! " 40.7% 3.740# max (b) Relate the heat Qc discharged to the engine’s surroundings to Qh and the efficiency of the engine: #" W Qh ! Qc " 4 Qc " $1 ! # %Qh Qh Qh .The Second Law of Thermodynamics 1903 Substitute numerical values in equation (2) and evaluate # : (d) Express and evaluate the efficiency of a Carnot cycle operating between 423 K and 273 K: #" 0.0ºC. (a) How does this efficiency compare with the maximum possible efficiency for your engine? (b) If the useful power output of the engine is known to be 200 kW.4052 or # steam engine " 0. We can apply the definition of efficiency to find the heat discharged to the engine’s surroundings in 1.12 kJ 273 K Tc " 1! " 35.05% # max 0. / 0. in January. The temperature of the air in the air handler inside the house is to be 40ºC. ! 1+ . relate the efficiency of the engine to the heat intake of the engine and the work it does each cycle: Substitute for Qh in the expression for Qc and simplify to obtain: W P)t Qh " # " # Qc " $1 ! # % P"t # 01 " . the COP of the heat pump will be only 60 percent of the ideal value. you are designing a heat pump that is capable of delivering heat at the rate of 20 kW to a house. 200 + $3600 s % " 1. What is the minimum power of the engine when the COP is 60 percent of the ideal value? Picture the Problem We can use the definition of the COPHP and the Carnot efficiency of an engine to express the maximum efficiency of the refrigerator in terms of the reservoir temperatures.00 h % " .1904 Chapter 19 Using its definition. ! 1+ P"t /# . The house is located where.26 313 K ! 263 K " 6.00 h % : kJ 0 1 -0 Qc $1. the average outside temperature is –10ºC. (a) What is maximum possible COP for a heat pump operating between these temperatures? (b) What must be the minimum power of the electric motor driving the heat pump? (c) In reality.3 .300 . / *Heat Pumps 49 • [SSM] As an engineer. We can apply the definition of power to find the minimum power needed to run the heat pump. (a) Express the COPHP in terms of Th and Tc: COPHP " " Qh Qh " W Qh ! Qc Th 1 1 " " Q T Th ! Tc 1! c 1! c Qh Th Substitute numerical values and evaluate COPHP: COPHP " 313 K " 6.68 GJ s . Substitute numerical values and evaluate Qc $1. Substitute numerical values and evaluate Pmin: Pmin " 20 kW " 5. max % (c) The minimum power of the engine is given by: Pmin dQc " dt " # HP where #HP is the efficiency of the heat pump. how much heat can it absorb from the food compartment in 1.00 min under these conditions? Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. Th.0ºC and it releases heat into a room at 20. P.00 min if the foodcompartment temperature of the refrigerator is 0.3 kW $0.26 dQc dt # $COPHP.26% 50 • A refrigerator is rated at 370 W. and )t. (a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP: Express the COP in terms of Th and Tc and simplify to obtain: Qc " $COP %W " $COP %P)t COP " " Qc Q Q !W " c " h # Qh W # Qh 1! # # " 1 # !1 " 1 !1 Tc 1! Th " Tc Th ! Tc .60%$6.2 kW 6.0ºC? (b) If the COP of the refrigerator is 70% of that of a reversible refrigerator.The Second Law of Thermodynamics 1905 (b) The COPHP is also given by: COPHP " Pout Pout 4 Pmotor " Pmotor COPHP Substitute numerical values and evaluate Pmotor: Pmotor " 20 kW " 3. (a) What is the maximum amount of heat it can absorb from the food compartment in 1. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc. Th. (a) What is the maximum amount of heat it can absorb for the food compartment in 1.1.0ºC and it releases heat into a room at 35ºC? (b) If the COP of the refrigerator is 70% of that of a reversible pump. (a) Express the amount of heat the refrigerator can remove in a given period of time as a function of its COP: Express the COP in terms of Th and Tc and simplify to obtain: Qc " $COP %W " $COP %P)t COP " " " Qc Q Q !W " c " h #Qh W #Qh 1! # # " 1 # !1 Tc 1 !1 " T Th ! Tc 1! c Th Substituting for COP yields: 0 Tc Qc " . Substitute numerical values and evaluate Qc: 273 K 60 s 0 0 Qc " . and )t. P.70 %$303 kJ % " 0.00 min if the temperature in the compartment is 0.30 MJ min . T ! T + P"t + / h c.00 min 1 + " 303 kJ " 0. .1906 Chapter 19 Substituting for COP yields: 0 Tc Qc " . Picture the Problem We can use the definition of the COP to relate the heat removed from the refrigerator to its power rating and operating time. how much heat can it absorb from the food compartment in 1. . + $370 W %. + .00 min? Is the COP for the refrigerator greater when the temperature of the room is 35ºC or 20ºC? Explain. By expressing the COP in terms of Tc and Th we can write the amount of heat removed from the refrigerator as a function of Tc. / 293 K ! 273 K .21 MJ 51 • A refrigerator is rated at 370 W. / (b) If the COP is 70% of the efficiency of an ideal pump: ' Qc " $0. T ! T + P"t / h c. how long will you have to wait in order for the room’s air to warm (take the specific heat of air to be 1. You will use the pump on chilly winter nights to increase the air temperature in your bedroom. and the temperature at the air handler in the room is 112°F. We’re given that the coefficient of performance of the heat pump is half the coefficient of performance of an ideal heat pump: Substituting for COPHP yields: "t " . and )t is the time required to warm the bedroom. whose COP is half the COP of a reversible heat pump. 2. Picture the Problem We can use the definition of the coefficient of performance of a heat pump and the relationship between the work done per cycle and the pump’s power consumption to find your waiting time. If the pump’s electric power consumption is 750 W.50 m. walls. Also assume that the heat capacity of the floor. Th ! Tc + / . 52 ••• You are installing a heat pump. Your bedroom’s dimensions are 5.00 min 1 min . / / 308 K ! 273 K .12 MJ Because the temperature difference increases when the room is warmer.70 %$173 kJ % " 0.00 m × 3.005 kJ/(kg·°C)? Assume you have good window draperies and good wall insulation so that you can neglect the release of heat through windows. The outside temperature is 35°F . The coefficient of performance of the heat pump is defined as: Qh Q Qh " h 4 "t " $COPHP %P W P"t where Qh is the heat required to raise the temperature of your bedroom. ceiling. ceilings and floors. 2Qh 0 Th . + " 173 kJ " 0. The air temperature should increase from 63°F to 68°F.17 MJ + $370 W %. walls and furniture are negligible.1. P is the power consumption of the heat pump. the COP decreases. COPHP " COPHP " Qh 1 " COPmax W 2 0 Th + " 1.50 m × 2. T ! T +P + / h c. (b) If the COP is 70% of the efficiency of an ideal pump: ' Qc " $0. .The Second Law of Thermodynamics 1907 Substitute numerical values and evaluate Qc: 273 K 60 s 0 0 Qc " . The change in entropy of the water is given by: "S H 2 O " Qabsorbed by H 2 O T The heat absorbed by the water as it vaporizes is the product of its mass and latent heat of vaporization: Substituting for Qabsorbed yields: by H 2 O Qabsorbed " mLv " <VLv by H 2 O "S H 2 O " <VLv T . / / "t " " 56 s 317 K 0 . 2 <Vc"T 0 Th .1. + $750 W % / 317 K ! 275 K .1908 Chapter 19 The heat required to warm the room is related to the volume of the room. and the desired increase in temperature: Substitute for Qh to obtain: Qh " mc"T " <Vc"T where < is the density of air and c is its specific heat capacity. See Table 18-2 its change in entropy is positive and given by "S H 2 O " T for the latent heat of vaporization of water. What is the change in entropy of the water associated with its change of state from liquid to gas? Picture the Problem Because the water absorbed heat in the vaporization process Qabsorbed by H 2 O .50 m 1 2. the density of air. You return just in time to see the last drop converted into steam.00 L of boiling water. The pan originally held 1. 5 F( 1 2. + m . "t " Substitute numerical values and evaluate )t: 0 kg J -0 5 C( 0 + .50 m %.00 m 1 3. T ! T +P + / h c. .293 3 + $5.1005 + . kg 6 C( . / 9 F( . Entropy Changes 53 • [SSM] You inadvertently leave a pan of water boiling away on the hot stove. " ! 22. the change in entropy of the liquid water is negative.0ºC. Note that.00 + $1.18. Because heat is removed from liquid water when it freezes. while the entropy of the water decreases.0°C? Picture the Problem We can use the definition of entropy change to find the change in entropy of the liquid water as it freezes.5 + .00 mol of liquid water at 0. The water.1. Show that even though the entropy of the water decreases.00 L %. / / . initially liquid at 0. " 373 K kJ " 6. The change in entropy of the water is given by:: The heat removed from the water as it freezes is the product of its mass and latent heat of fusion: "S H 2 O " Qremoved T from H 2 O Qremoved " !mLf from H 2 O or. because m " nM H 2 O . . Assume the walls of the freezer are maintained at –10ºC. 2257 + . kg + / / .05 K "S H 2O 54 • What is the change in entropy of 1. g+ mol . the net entropy of the universe increases.0ºC that freezes to ice at 0. 333.The Second Law of Thermodynamics 1909 Substitute numerical values and evaluate "S H 2O : 0 kg kJ 0 .015 + .00 mol%. L. See Appendix C for the molar mass of water and Table 18-2 for the latent heat of fusion of water.0 J " K 273 K "S H2O 55 • Consider the freezing of 50. Qremoved " ! nM H 2 O Lf from H 2 O Substitute numerical values and evaluate "S H 2O : Jg -0 0 ! $1.0 g of water once it is placed in the freezer compartment of a refrigerator. Picture the Problem The change in the entropy of the universe resulting from the freezing of this water and the cooling of the ice formed is the sum of the entropy changes of the water-ice and the freezer. is frozen into ice and cooled to –10ºC. the entropy of the freezer increases. 1910 Chapter 19 The change in entropy of the universe resulting from this freezing and cooling process is given by: Express )S water : Express )S freezing : )S u " )S water & )S freezer (1) )S water " )S freezing & )S cooling )S freezing " ! Qfreezing Tfreezing (2) (3) where the minus sign is a consequence of the fact that heat is leaving the water as it freezes. freezer . f . > B ! Lf 0T " m@ & C p ln. express )S freezer : )S freezer " " )Qice )Qcooling ice & Tfreezer Tfreezer mC p )T mLf & Tfreezer Tfreezer Substitute for )S water and )S freezer in equation (1): )S u " 0T ! mLf & mCp ln.T @ Tfreezing / i A . + + . f .T Tfreezing / i Noting that the freezer gains heat (at 263 K) from the freezing water and cooling ice.mLf +& & + T Tfreezer . Relate Qfreezing to the latent heat of fusion and the mass of the water: Substitute in equation (3) to obtain: )S freezing " ! mLf Tfreezing + + . f .T Tfreezing / i mCp )T . f . Qfreezing " mLf Express )S cooling : 0T )S cooling " mC p ln.T / i )S water " Substitute in equation (2) to obtain: 0T ! mLf & mCp ln.Lf & C p )T ? +& = + Tfreezer = . 5 110 3 + . kg 6 K + .V / f 0V + 4 Q " !nRT ln.0 L J 0 "S gas " !$2. 56 • In this problem. i .0 40. The work done on the gas is given by: Substitute for Q in equation (1) to obtain: 0V "S gas " !nR ln. + + .The Second Law of Thermodynamics 1911 Substitute numerical values and evaluate )Su: B 3 J @ 333. i + .00 mol of an ideal gas at 400 K expand quasistatically and isothermally from an initial volume of 40.0500 kg % @! .0 263 K + + ln. / . @ A ? J J 0 + $273 K ! 263 K % = & . .314 + ln.5 K mol 6 K .0 L + " 11. kg / = " 2. 2100 333.5 110 kg 0 J . + . because )Su > 0. kg 6 K .00 mol%.V . because )Eint = 0 for an isothermal expansion of a gas.40 J/K & = 263 K = > and. 8. the entropy of the universe increases.V / f Substitute numerical values and evaluate )S: J . & . / / .0 L to a final volume of 80. 80. (a) The entropy change of the gas is given by: Apply the first law of thermodynamics to the isothermal process to express Q in terms of Won: "S gas " Q T (1) Q " "Eint ! Won or. 2100 "S u " $0. (a) What is the entropy change of the gas? (b) What is the entropy change of the universe for this process? Picture the Problem We can use the definition of entropy change and the 1st law of thermodynamics to express )S for the ideal gas as a function of its initial and final volumes. 273 K + 273 K @ / . Q " !Won 0V Won " nRT ln.0 L. / f + + . i . 2. During step 1 the system absorbs 300 J of heat from a reservoir at 300 K. (During steps 2. during which the total work done by the system is 100 J. 4 and 6 the system undergoes adiabatic processes in which the temperature of the system changes from one reservoir’s temperature to that of the next. Heat is rejected by the two high-temperature reservoirs and absorbed by the cold reservoir: Solving for T3 yields: "Ssystem " 0 1 complete cycle "S1 & "S 2 & "S 3 & "S system " 0 or Q1 Q2 Q3 & & &0 " 0 T1 T2 T3 ! 300 J ! 200 J 400 J & & "0 300 K 400 K T3 T3 " 267 K 58 •• In this problem. (a) Because S is a state function of the system. during step 3 the system absorbs 200 J of heat from a reservoir at 400 K. what is the temperature T3? Picture the Problem We can use the fact that the system returns to its original state to find the entropy change for the complete cycle. 2. What is (a) the entropy change of the gas and (b) the entropy change of the universe? .1912 Chapter 19 (b) Because the process is reversible: )S u " 0 Remarks: The entropy change of the environment of the gas is !11.) (a) What is the entropy change of the system for the complete cycle? (b) If the cycle is reversible. Because the entropy change for the complete cycle is the sum of the entropy changes for each process. and because the system’s final state is identical to its initial state: (b) Relate the entropy changes for each of the three heat reservoirs and the system for one complete cycle of the system: Substitute numerical values.0 L. The gas undergoes a free adiabatic expansion to twice its initial volume. 57 •• [SSM] A system completes a cycle consisting of six quasi-static steps.00 mol of an ideal gas initially has a temperature of 400 K and a volume of 40. we can find the temperature T3 from the entropy changes during the 1st two processes and the heat released during the third. and during step 5 it absorbs heat from a reservoir at temperature T3.5 J/K. The Second Law of Thermodynamics 1913 Picture the Problem The initial and final temperatures are the same for a free expansion of an ideal gas. Thus, the entropy change )S for a free expansion from Vi to Vf is the same as )S for an isothermal process from Vi to Vf. We can use the definition of entropy change and the 1st law of thermodynamics to express )S for the ideal gas as a function of its initial and final volumes. (a) The entropy change of the gas is given by: Apply the first law of thermodynamics to the isothermal process to express Q: "S gas " Q T (1) Q " "Eint ! Won or, because )Eint = 0 for a free expansion of a gas, Q " !Won 0V Won " nRT ln. i .V / f 0V + 4 Q " !nRT ln. i + .V , / f + + , + + , The work done on the gas is given by: Substitute for Q in equation (1) to obtain: 0V "S gas " !nR ln. i .V / f Substitute numerical values and evaluate )S: J - 0 40.0 L J 0 "S gas " !$2.00 mol%. 8.314 + ln. . 80.0 L + " 11.5 K + mol 6 K , / / , (b) The change in entropy of the universe is the sum of the entropy changes of the gas and the surroundings: For the change in entropy of the surroundings we use the fact that, during the free expansion, the surroundings are unaffected: The change in entropy of the universe is the change in entropy of the gas: "S u " "S gas & "S surroundings "S surroundings " Qrev 0 " "0 T T "S u " 11.5 J K 59 •• A 200-kg block of ice at 0.0ºC is placed in a large lake. The temperature of the lake is just slightly higher than 0.0ºC, and the ice melts very 1914 Chapter 19 slowly. (a) What is the entropy change of the ice? (b) What is the entropy change of the lake? (c) What is the entropy change of the universe (the ice plus the lake)? Picture the Problem Because the ice gains heat as it melts, its entropy change is positive and can be calculated from its definition. Because the temperature of the lake is just slightly greater than 0(C and the mass of water is so much greater than that of the block of ice, the absolute value of the entropy change of the lake will be approximately equal to the entropy change of the ice as it melts. (a) The entropy change of the ice is given by: Substitute numerical values and evaluate "Sice : "S ice " mLf T . + $200 kg %0 333.5 kJ . + / 273 K kg , "S ice " " 244 kJ K (b) Relate the entropy change of the lake to the entropy change of the ice: (c) The entropy change of the universe due to this melting process is the sum of the entropy changes of the ice and the lake: "Slake 2 !"Sice " ! 244 kJ K "S u " "Sice & "Slake Because the temperature of the lake is slightly greater than that of the ice, the magnitude of the entropy change of the lake is less than 244 kJ/K and the entropy change of the universe is greater than zero. The melting of the ice is an irreversible process and )S u C 0 . 60 •• A 100-g piece of ice at 0.0ºC is placed in an insulated calorimeter with negligible heat capacity containing 100 g of water at 100ºC. (a) What is the final temperature of the water once thermal equilibrium is established? (b) Find the entropy change of the universe for this process. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a melting process and for constant-pressure processes to find the entropy change of the universe (the entropy change of the piece of ice plus the entropy change of the water in the insulated container). The Second Law of Thermodynamics 1915 (a) Apply conservation of energy to obtain: 7Q i ice i "0 or Qmelting & Qwarming ! Qcooling " 0 water water Substitute to relate the masses of the ice and water to their temperatures, specific heats, and the final temperature of the water: . + . $100 g %0 333.5 kJ - & $100 g %0 4.18 . + . / kg , kJ +t kg 6 C( + , / 0 kJ + $100(C ! t % " 0 ! $100 g %. 4.18 . kg 6 C( + / , Solving for t yields: (b) The entropy change of the universe is the sum of the entropy changes of the ice and the water: Using the expression for the entropy change for a constant-pressure process, express the entropy change of the melting ice and warming icewater: Substitute numerical values to obtain: . + $0.100 kg %0 333.5 kJ . + / 273 K kg , t " 10.1(C )S u " )Sice & )S water )Sice " )S melting ice & )S warming water " 0T mLf & mcP ln. f .T Tf / i + + , "Sice " 0 J kJ - 0 283 K + ln. & $0.100 kg %. 4.18 + . 273 K + " 137 K . + kg 6 K , / / , Find the entropy change of the cooling water: 0 kJ - 0 283 K J + ln. "S water " $0.100 kg %. 4.18 . + . 373 K + " !115 K + kg 6 K , / / , Substitute for )Sice and )Swater and evaluate the entropy change of the universe: "S u " 137 J J J ! 115 " 22 K K K 1916 Chapter 19 Remarks: The result that )Su > 0 tells us that this process is irreversible. 61 •• [SSM] A 1.00-kg block of copper at 100ºC is placed in an insulated calorimeter of negligible heat capacity containing 4.00 L of liquid water at 0.0ºC. Find the entropy change of (a) the copper block, (b) the water, and (c) the universe. Picture the Problem We can use conservation of energy to find the equilibrium temperature of the water and apply the equations for the entropy change during a constant pressure process to find the entropy changes of the copper block, the water, and the universe. (a) Use the equation for the entropy change during a constant-pressure process to express the entropy change of the copper block: Apply conservation of energy to obtain: 0T )S Cu " mCu cCu ln. f .T / i + + , (1) 7Q i i "0 or Qcopper block & Qwarming water " 0 Substitute to relate the masses of the block and water to their temperatures, specific heats, and the final temperature Tf of the water: . $1.00 kg %0 0.386 . / kJ + $Tf ! 373 K % kg 6 K + , kg - 0 kJ 0 + $Tf ! 273 K % " 0 & $4.00 L %.1.00 + . 4.18 . L ,/ kg 6 K + / , Solve for Tf to obtain: Tf " 275.26 K Substitute numerical values in equation (1) and evaluate "S Cu : 0 kJ - 0 275.26 K J +ln. "S Cu " $1.00 kg %. 0.386 . + . 373 K + " ! 117 K + kg 6 K , / / , (b) The entropy change of the water is given by: 0T )S water " mwater c water ln. f .T / i + + , 00 kg %. express and evaluate the entropy change of the lead: 0T "S Pb " mPb cPb ln. We can apply the equation for the entropy change during a constant pressure process to find the entropy changes of the piece of lead. 273 K + " 138 K + kg 6 K .00-kg piece of lead at 100ºC is dropped into a lake at 10ºC.26 K + ln. f .69 K + + kg 6 K . Express the entropy change of the universe in terms of the entropy changes of the lead and the water in the lake: )S u " )S Pb & )S w (1) Using the equation for the entropy change during a constant-pressure process. (c) Substitute for "S Cu and "S water and evaluate the entropy change of the universe: "S u " "SCu & "S water " !117 " 20 J K J J & 138 K K Remarks: The result that )Su > 0 tells us that this process is irreversible. + .0 275. the temperature of the lake will increase only slightly and we can reasonably assume that its final temperature is 10(C. the water in the lake.128 + ln. 0. 4. 373 K + " !70.The Second Law of Thermodynamics 1917 Substitute numerical values and evaluate "S water : 0 J kJ .18 . / / .00 kg %. "S w " The entropy change of the water in the lake is given by: Qw QPb mPb cPb "TPb " " Tw Tw Tw . Picture the Problem Because the mass of the water in the lake is so much greater than the mass of the piece of lead. and the universe. . .T / i 0 kJ .0 283 K J + " $2. find the entropy change of the universe. / / . 62 •• If a 2. + . "S water " $4. 00 kg %0 0. (a) The entropy change of the universe is the sum of the entropy changes of the two reservoirs: Substitute numerical values and evaluate )Su: )S u " )S h & )Sc " ! 01 1" !Q. / 283 K " 81. kJ + $90 K % kg 6 K + .41 " 11 K K K Entropy and 'Lost' Work 63 •• [SSM] A a reservoir at 300 K absorbs 500 J of heat from a second reservoir at 400 K. (a) What is the change in entropy of the universe. .41 J/K Substitute numerical values in equation (1) and evaluate )Su: "S u " !70.T T + c . " 0.1918 Chapter 19 Substitute numerical values and evaluate )Sw: "S w " $2. / h Q Q & Th Tc 0 1 1 "S u " $! 500 J % .and low-temperature reservoirs. ! + .128 .42 J/K (b) Relate the heat that could have been converted into work to the maximum efficiency of an engine operating between the two reservoirs: The maximum efficiency of an engine operating between the two reservoir temperatures is the efficiency of a Carnot device operating between the reservoir temperatures: W " # max Qh # max " # C " 1 ! Tc Th . The maximum amount of the 500 J of heat that could be converted into work can be found from the maximum efficiency of an engine operating between the two reservoirs.69 J J J & 81. . and (b) how much work is lost during the process? Picture the Problem We can find the entropy change of the universe from the entropy changes of the high. 400 K ! 300 K + + / . 3 L J J 0 "S u " !$1. because "Sgas during "S u " "S gas during free expansion " isothermal compresion !Q T + + . Picture the Problem Although no heat is lost by the gas in the adiabatic free expansion. In the isothermal reversible process that returns the gas to its original state.1 ! .1 ! c +Qh . the gas releases heat to the surroundings. / " 1. 24.00 mol%. the entropy change of the universe is zero.763 + K. .V / i + + .V / i 0V "S u " ! nR ln.763 K " 5.73 kJ . / / .00 mol of an ideal gas at 300 K undergoes a free adiabatic expansion from V1 = 12. It is then compressed isothermally and reversibly back to its original state. (a) Relate the entropy change of the universe to the entropy changes of the gas during 1 complete cycle: "S u " "S gas during free expansion & "S gas during "0. Consequently.6 L. 8. / 0 300 K W " . f . (a) What is the entropy change of the universe for the complete cycle? (b) How much work is lost in this cycle? (c) Show that the work lost is T)Su. T + h . However. f . isothermal compresion or. The work done by the gas during its isothermal compression is given by: Substituting for "S gas in the expression for "S u yields: 0V Wby " !Won " !Q " !nRT ln. 1. 400 K + $500 J % " 125 J + / .314 +ln. (1) Substitute numerical values and evaluate )Su: J . the process is irreversible and the entropy of the gas increases.6 L + " 5. Substitute numerical values and evaluate W: 64 •• In this problem. because the process is reversible. the net entropy change is the negative of that of the gas in the isothermal compression.The Second Law of Thermodynamics 1919 Substitute for #max to obtain: 0 T W " .76 K + mol 6 K .3 L to V2 = 24. 5. (b) Use Equation 19-22 to find the amount of energy that becomes unavailable for doing work during this process: J0 Wlost " T"S u " $300 K %.0 12. f .1920 Chapter 19 (c) Use equation (1) to express the product of T and )Su: 0 0V T"S u " T . ..0 s !1 Wcycle Qh. ! nR ln. Wcycle " 200 W " 20.cycle " Qh. It operates at 10. / i + + . / i / " Wby gas during its -0V + + " ! nRT ln.V .cycle " # " 20.cycle ! W " 67 J ! 20 J " 47 J During each cycle.30 Qc. Application of the first law of thermodynamics will yield the heat given off each cycle. isothermal compression General Problems 65 • A heat engine with an output of 200 W has an efficiency of 30%.0 J 10. (a) How much work is done by the engine during each cycle? (b) How much heat is absorbed from the hot reservoir and how much is released to the cold reservoir during each cycle? Picture the Problem We can use the definition of power to find the work done each cycle and the definition of efficiency to find the heat that is absorbed each cycle. f ++ . (a) What is the efficiency of this engine? (b) What is the ratio of its efficiency to that of a Carnot engine working between the same reservoirs? (This ratio is called the second law efficiency. a heat engine operating between two heat 66 • reservoirs absorbs 150 J from the reservoir at 100ºC and releases 125 J to the reservoir at 20ºC.) Picture the Problem We can use their definitions to find the efficiency of the engine and that of a Carnot engine operating between the same reservoirs. (a) Use the definition of power to relate the work done in each cycle to the frequency of each cycle: Substitute numerical values and evaluate Wcycle: (b) Express the heat absorbed in each cycle in terms of the work done and the efficiency of the engine: Apply the 1st law of thermodynamics to find the heat given off in each cycle: Wcycle " P"t " P f where f is the frequency of the engine.0 cycles/s.V .0 J " 67 J 0. W " # Qh " $0. (a) Express the efficiency of the engine in terms of the efficiency of a Carnot engine working between the same reservoirs: Substitute numerical values and evaluate # : (b) Use the definition of efficiency to find the work done in each cycle: (c) Apply the first law of thermodynamics to the cycle to obtain: # " 0. # " 0.1 ! .777 # C 21. .85.7% 150 J #C " 1! Tc 293 K " 1! " 21.cycle " Qh.85# C " 0.1 ! .510 " 51% 500 K + .45% 67 • [SSM] An engine absorbs 200 kJ of heat per cycle from a reservoir at 500 K and releases heat to a reservoir at 200 K. Its efficiency is 85 percent of that of a Carnot engine working between the same reservoirs.10 MJ Qc.510%$200 kJ % " 102 kJ " 0.85.The Second Law of Thermodynamics 1921 (a) The efficiency of the engine is given by: Substitute numerical values and evaluate #: (b) Find the efficiency of a Carnot engine operating between the same reservoirs: Express the ratio of the two efficiencies: #" Q W Qh ! Qc " " 1! c Qh Qh Qh # " 1! 125 J " 16. / 0 Tc + Th + .67% " " 0.67% " 16. / 0 200 K + " 0.cycle ! W " 200 kJ ! 102 kJ " 98 kJ Estimate the change in entropy of the universe associated with an 68 • Olympic diver diving into the water from the 10-m platform. (a) What is the efficiency of this engine? (b) How much work is done in each cycle? (c) How much heat is released to the low-temperature reservoir during each cycle? Picture the Problem We can use the definition of efficiency to find the work done by the engine during each cycle and the first law of thermodynamics to find the heat released to the low-temperature reservoir during each cycle.45% Th 373 K # 16. The change in entropy of the universe associated with a dive is given by: Qadded to water Twater where Qadded to water is the energy entering "S u " "S water " the water as a result of the kinetic energy of the diver as he enters the water.0 kW on a day when the outside temperature is –7ºC. At what rate does this house contribute to the increase in the entropy of the universe? Picture the Problem The change in entropy of the universe is the change in entropy of the house plus the change in entropy of the environment. because "S house " 0 . The energy added to the water is the change in the gravitational potential energy of the diver: Substitute numerical values and evaluate "S u : "S u " mgh Twater $75 kg %$9.81 m/s 2 %$10 m % "S u " $25 & 273%K 2 25 J K 69 • To maintain the temperature inside a house at 20ºC. and the state of the house does not change.1922 Chapter 19 Picture the Problem Assume that the mass of the diver is 75 kg and that the temperature of the water in the pool is 25(C. We can find the change in entropy of the house by exploiting the given information that the temperature inside the house is maintained at a constant temperature. "S u " "Ssurroundings Q Tsurroundings R"t Tsurroundings "S surroundings " " . the electric power consumption of the electric baseboard heaters is 30. Entropy is a state function. The energy added to the water in the pool is the change in the gravitational potential energy of the diver during the dive. Therefore the entropy of the house does not change: Heat is absorbed by the surroundings at the same rate R that energy is delivered to the house: "S u " "S house & "Ssurroundings or. We can find the change in entropy of the surrounding by dividing the heat added by the temperature. from this value. Heat is absorbed by the liquid sodium in the core.The Second Law of Thermodynamics 1923 Substitute for )Ssurroundings yields: "S u " R"t Tsurroundings 4 "S u R " "t Tsurroundings Substitute numerical values and evaluate )Su/)t: "S u 30. and the water in the river flows by at a temperature of 25ºC. Heat is released into the river. In this plant. (a) The Carnot efficiency of a plant operating between temperatures Tc and Th is given by: Substitute numerical values and evaluate #C: (c) The power that must be supplied.00 GW of power. liquid sodium circulates between the reactor core and a heat exchanger located in the superheated steam that drives the turbine.00 GW 0. at 40. The rate at which heat is being released to the river is related to the requisite flow rate of the river by dQ dt " c)T< dV dt . Because of these laws.4% efficiency. the power that is wasted. (a) What is the highest efficiency that this plant can have? (b) How much heat is released into the river every second? (c) How much heat must be released by the core to supply 1.404 500 K Poutput Psupplied " # max " 1. What is the minimum flow rate that the water in the Hobbes River must have? Picture the Problem We can use the expression for the Carnot efficiency of the plant to find the highest efficiency this plant can have.00 GW of power and. We can then use this efficiency to find the power that must be supplied to the plant to generate 1. to produce an output of 1. generates 1. the plant is not allowed to heat the river by more than 0.00 GW of electrical power? (d) Assume that new environmental laws have been passed to preserve the unique wildlife of the river.50ºC.48 GW Pwasted " Psupplied ! Pgenerated (b) Relate the wasted power to the power generated and the power supplied: .0 kW W " " 113 K "t 266 K 70 •• Calvin Cliffs Nuclear Power Plant.404 " 2. and released by the liquid sodium (and into the superheated steam) in the heat exchanger. located on the Hobbes River.00 GW is given by: # max " # C " 1 ! Tc Th # max " 1 ! 298 K " 0. The temperature of the superheated steam is 500 K. you decide he has made an error in the measurement of his exhausted-heat value.48 1 10 9 " 7.50 K %.1924 Chapter 19 Substitute numerical values and evaluate Pwasted : (d) Express the rate at which heat is being dumped into the river: Pwasted " 2.11 10 5 L/s 71 •• An inventor comes to you to explain his new invention. (b) After careful analysis of the data in his prospectus folder.0 W.3% 125 W & 25. and releases heat to the air at the rate of only 25. does work at the rate of 125 W.48 GW dQ dm d " c)T " c)T $<V % dt dt dt dV " c)T< dt dV dQ dt " dt c)T< Solve for the flow rate dV/dt of the river: Substitute numerical values (see Table 19-1 for the specific heat of water) and evaluate dV/dt: J dV s " dt 0 kg J 0 . It is a novel heat engine using water vapor as the working substance. + kg . He claims that the water vapor absorbs heat at 100°C.00 GW " 1. m .10 3 3 + . What is the minimum rate of exhausting heat that would make you consider believing him? Picture the Problem We can use the inventor’s data to calculate the thermal efficiency of his steam engine and then compare this value to the efficiency of a Carnot engine operating between the same temperatures. / / 1. when the air temperature is 25°C.1% 373 K Th dW dW dt # " dt " dQh dW dQc & dt dt dt 125 W " " 83. (a) The Carnot efficiency of an engine operating between these temperatures is: The thermal efficiency of the inventor’s device. (a) Explain to him why he cannot be correct. 4180 + $0.0 W . in terms of the rate at which it expels heat to the air and does work is: # C " 1! Tc 298 K " 1! " 20.48 GW ! 1. his data is not consistent with what is known about the thermodynamics of engines.# dt / C .201 . He must have made a mistake in his analysis of his data!or he is a con man looking for people to swindle. What is the efficiency of the cyclic process ABCDA? .201%$150 W % 2 15 W dQc dQh dW " ! .0 W of work are done per cycle. ! 1+ + .dW dQc 0 2 " . Setting the efficiency of his steam engine equal to half the Carnot efficiency of the engine yields: dW dW dt 1 # " dt " 2 C dQh dW dQc & dt dt dt .a dt dt dt reasonable value for dQc/dt is: Because dQc " 150 W ! 15 W " 135 W dt The cycle represented in Figure 19-12 (next to Problem 19-14) is for 72 •• 1. (b) The maximum efficiency of a steam engine that has ever been achieved is about 50% of the Carnot efficiency of an engine operating between the same temperatures. respectively. ! 1+ $125 W % 2 1100 W dt / 0.The Second Law of Thermodynamics 1925 You should explain to him that.00 mol of an ideal monatomic gas. a value totally inconsistent with the inventor’s claims for his engine. dW dW 1 dQh 1 # " dt 4 " 2 #C 2 C dQh dt dt dt Ignoring his claim that 25. The temperatures at points A and B are 300 and 750 K. let’s assume that his device does take in 150 W of energy each cycle and find how much work would do with an efficiency half that of a Carnot engine: Substituting numerical values yields: dW " dt 1 2 $0. dt Solve for dQc/dt to obtain: Assuming that the inventor has measured the work done per cycle by his invention correctly: dQc 0 2 ". because the efficiency he claims for his invention is greater than the efficiency of a Carnot engine operating between the same two temperatures. we can find the heat that would be converted to work by a Carnot engine operating between the given temperatures and subtract that amount of work from 1.00 kJ of heat to a reservoir at 300 K? Explain your choice. (b) Find the change in entropy of the universe for process (1): "S1 " "Q 500 J " " 1. transformed into heat in process (1). The Carnot efficiency of the cycle is given by: Substitute numerical values and evaluate #C: #C " 1! # C " 1! Tc Th 300 K " 60. (a) For process (2): The efficiency of a Carnot engine operating between temperatures Th and Tc is given by: W2.00 kJ to find the energy that is lost. In Part (b) we can use its definition to find the change in entropy for each process.max " Wrecovered " # CQin # C " 1! Tc Th +Qin + . or 750 J are lost.00 kJ of heat could be converted into work by an ideal cyclic process? (b) What is the change in entropy of the universe for each process? Picture the Problem All 500 J of mechanical energy are lost. i.67 J/K T 300 K . For process (2).1 ! +$1..1926 Chapter 19 Picture the Problem Because the cycle represented in Figure 19-12 is a Carnot cycle. Hint: How much of the 1. T h / Substitute for # C to obtain: 0 300 K Wrecovered " .00 kJ % " 250 J / 400 K . or (2) A reservoir at 400 K releasing 1.0% 750 K 73 •• [SSM] (a) Which of these two processes is more wasteful? (1) A block moving with 500 J of kinetic energy being slowed to rest by sliding (kinetic) friction when the temperature of the environment is 300 K.1 ! c . and hence 0 T Wrecovered " . Process (2) is more wasteful of available work. its efficiency is that of a Carnot engine operating between the temperatures of its isotherms.e. 00 kJ %. 300 K ! 400 K + . a monatomic gas. Picture the Problem Denote the three states of the gas as 1. and 3 with 1 being the initial state. is initially at a pressure of 16 atm. / " 0. . 2. ! + .833 J/K 74 •• Helium. we find the efficiency of the cycle from the work done each cycle and the heat that enters the system during the isothermal expansion. and 3. isobaric. It is quasi-statically expanded at constant temperature until its volume is 4. and a temperature of 600 K. We can use the ideal-gas law and the equation of state for an adiabatic process to find the temperatures.0 L and is then quasi-statically compressed at constant pressure until its volume and temperature are such that a quasi-static adiabatic compression will return the gas to its original state. (a) Sketch this cycle on a PV diagram. (a) The PV diagram of the cycle is shown to the right.0 L. 4. (d) Find the efficiency of the cycle.0 L V1 " $16 atm %.0 L + " 4.T T + h . . 2. (b) Apply the ideal-gas law to the isothermal expansion 1*2 to find P2: P2 " P1 0 1. we can use the equations for the work done during isothermal. Finally.The Second Law of Thermodynamics 1927 )Q )Q & Th Tc Express the change in entropy of the universe for process (2): )S 2 " )S h & )Sc " ! 01 1" )Q . and pressures at points 1.0 atm + V2 / . a volume of 1. / c Substitute numerical values and evaluate )S2: 0 1 1 "S 2 " $1. To find the work done during each cycle. (b) Find the volume and temperature after the compression at constant pressure. and adiabatic processes. + . volumes. (c) Find the work done during each step of the cycle. 2.1928 Chapter 19 Apply an equation for an adiabatic process to relate the pressures and volumes at 1 and 3: Substitute numerical values and evaluate V3: 0PPV1 " P3V3 4 V3 " V1 .67 !1 3 !1 0 1.294 L %9 2 " !10. 1 + .824 atm 6 L ! 10.4 110 2 K " 344 K (c) Express the work done each cycle: For the process 1*2: W " W12 & W23 & W31 (1) 0V 0V W12 " nRT1 ln.00 L % " !6. / 1.3 L 1 1.0 L " $16 atm %$1. " 2.67 " 2.V + / 3. .V + / 1.294 L + + / . .0 L + + / .P + / 3. 2 + " P1V1 ln.18 atm 6 L For the process 2*3: W23 " P2 "V23 " $4.67) to relate the temperatures and volumes at 1 and 3: Substitute numerical values and evaluate T3: T3V3 3 !1 " T1V1 3 !1 0V 4 T3 " T1 . 1. 1.18 atm 6 L ! 6.0 atm %$2. 3 3 13 0 16 atm V3 " $1. 4. " 22.V + .116 atm 6 L " 5 atm 6 L . " 3.0 atm + + / .294 L Apply an equation for an adiabatic process (3 =1. .0 L % ! $4. 1 + 1 .0 L %ln.0 L %.0 atm %$2. 0 4.24 atm 6 L " 5.24 atm 6 L Substitute numerical values in equation (1) and evaluate W: W " 22. 2 + .824 atm 6 L For the process 3*1: W31 " !C V "T31 " ! 3 nR$T1 ! T3 % " ! 3 $P1V1 ! P3V3 % 2 2 " ! 3 8$16 atm %$1.294 L ! 4.0 L T3 " $600 K %. relate Tc to the work done by and the heat input to the real heat engine. Using its definition. Picture the Problem We can express the temperature of the cold reservoir as a function of the Carnot efficiency of an ideal engine and. and heat is released to a reservoir at a temperature Tc.00 atm %$4. Q in / + + . where Tc < 120ºC.00 L % " 4.1 ! . relate the efficiency of a Carnot engine working between the same reservoirs to the temperature of the cold reservoir: Relate the efficiency of the heat engine to that of a Carnot engine working between the same temperatures: Substitute for # C to obtain: #C " 1! Tc 4 Tc " Th $1 ! # C % Th #" 2W W " 1 #C 4 #C " 2 Qin Qin 0 2W Tc " Th .The Second Law of Thermodynamics 1929 (d) Use its definition to express the efficiency of the cycle: Substitute numerical values and evaluate #: #" W W W " " Qin Q12 W12 5.116 atm 6 L 2 20% 22. The work done by the gas in expanding the balloon is: W " P"V " $1.18 atm 6 L #" 75 •• [SSM] A heat engine that does the work of blowing up a balloon at a pressure of 1.00 kJ from a reservoir at 120ºC.00 L.00 atm 6 L . find the temperature Tc. given that the efficiency of the heat engine is half that of a Carnot engine. If the efficiency of the heat engine is 50% of the efficiency of a Carnot engine working between the same two reservoirs. The volume of the balloon increases by 4.00 atm absorbs 4. 00 atm 6 L 1 ++ atm 6 L . 78 •• Show that the coefficient of performance of a Carnot engine run as a refrigerator is related to the efficiency of a Carnot engine operating between the same two temperatures by # C 1 COPC " Tc Th . T 1! # C h COPc " " / ." h Because # C " T W " 1! c : Qh Th #C #C #C and # C 1 COPc " Tc Th 1 77 •• A freezer has a temperature Tc = –23ºC.00 kJ . 4.1930 Chapter 19 Substitute numerical values and evaluate Tc: 0 101.325 J . 2 . + 4. Using the definition of the COP. relate the heat removed from the cold reservoir to the work done each cycle: Apply energy conservation to relate Qc. + . and W: Substitute for Qc to obtain: COP " Qc W Qc " Qh ! W Qh ! W W 1! COP " Divide the numerator and denominator by Qh and simplify to obtain: W Q !W Qh " COP " h W W Qh 0 T 1 ! . + / .0 . The air in the kitchen has a temperature Th = 27ºC. Qh. The freezer is not perfectly insulated and some heat .1 ! / Tc " $393 K % " 313 K .1 ! c . Picture the Problem We can use the definitions of the COP and #C to show that their relationship is # C 1 COPC " TC Th .Tc + + T . Differentiation of this expression with respect to time will yield an expression for the power of the motor that is needed to maintain the temperature in the freezer. relate the heat that must be removed from the freezer to the work done by the motor: Differentiate this expression with respect to time to express the power of the motor: Express the maximum COP of the motor: Substitute for COPmax to obtain: COP " Qc Q 4W " c W COP P" dW dQc dt " dt COP COPmax " P" Tc )T dQc )T dt Tc Substitute numerical values and evaluate P: 0 50 K P " $50 W %. 2. and isothermal processes of the cycle. adiabatic. . The segment BC is an adiabatic expansion and the segment CA is an isothermal compression. (The PV diagram is not drawn to scale. Using the definition of the COP. Find the power of the motor that is needed to maintain the temperature in the freezer. and volumes at points A.The Second Law of Thermodynamics 1931 leaks through the walls of the freezer at a rate of 50 W. B.) At A the pressure and temperature are 5.00 atm and 600 K. Picture the Problem We can use the definition of the COP to express the work the motor must do to maintain the temperature of the freezer in terms of the rate at which heat flows into the freezer. pressures. The volume at B is twice the volume at A. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle? (f) How much heat is absorbed or released by the gas in each segment of this cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures. 78 •• In a heat engine.00 mol of a diatomic gas are taken through the cycle ABCA as shown in Figure 19-20. . and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the constantpressure. 250 K + " 10 W + / . 4 L $2.00 atm 1 atm 1L " 1.45 atm 6 L 1 101.38 L " 39.98 kJ .325 kPa 5.T + / C.325 J atm 6 L " 9. 600 K + " 222.1. Hence: Apply the ideal-gas law to this constant-pressure process to obtain: (c) Because the process C*A is isothermal: (d) Apply an equation for an adiabatic process (3 = 1.4!1 VC " $39. / " 223 L 1 (e) The work done by the gas during the constant-pressure process AB is given by: Substitute numerical values and evaluate WAB: WAB " PA $VB ! VA % " PA $2VA ! VA % " PAVA WAB " $5.969 110 ! 2 m 3 1 !3 3 " 19.3 !1 4 VC " VB .69 L % " 39. + . + .9754 110 3 J " 9.4) to find the volume of the gas at C: Substitute numerical values and evaluate VC: VB " 2$19. TB " TA VB 2V " $600 K % A VA VA " 1200 K TC " TA " 600 K 1 TBVB 3 !1 " TCVC 3 !1 0 TB .7 L (b) We’re given that VB " 2VA .38 L %. 0 1200 K .314 . / " 101.1932 Chapter 19 (a) Apply the ideal-gas law to find the volume of the gas at A: VA " nRTA PA J + $600 K % mol 6 K .77 L .69 L % " 98.00 mol%0 8.00 atm %$19.69 L 10 m " 19. 00 mol of a diatomic gas are carried through the cycle ABCDA shown in Figure 19-21. . 8.2 kJ because )Eint. / " 24.00 mol%0 8.00 mol%. (The PV diagram is not drawn to scale. (a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle. . 8.00 mol%.314 + $600 K % ln. A . 222. BC ! Qin. " !24. The pressure at D is 1. The volume at B is twice the volume at A.494 110 J 2 mol 6 K .V / C 0 19. BC " !ncV )TBC " ! 5 nR)TBC 2 Substitute numerical values and evaluate WBC: J 0 4 WBC " ! 5 $2.2 kJ (f) The heat absorbed during the constant-pressure expansion AB is: QAB " ncP "TA ! B " 7 nR"TA ! B " 2 " 34.00 atm and 600 K.) The segment AB represents an isothermal expansion. QCA " WCA & )Eint. BC " )Eint. / J + $1200 K ! 600 K % mol 6 K .77 L + + + mol 6 K .The Second Law of Thermodynamics 1933 Apply the first law of thermodynamics to express the work done on the gas during the adiabatic expansion BC: WBC " )Eint. .20 kJ " ! 24. CA " WCA " ! 24.00 atm. The pressure and temperature at A are 5.69 L J 0 + " $2.314 .9 kJ The work done by the gas during the isothermal compression CA is: 0V WCA " nRTC ln. 2.92 kJ " 34.9 kJ The heat absorbed during the adiabatic expansion BC is: Use the first law of thermodynamics to find the heat absorbed during the isothermal compression CA: QBC " 0 7 2 $2. the segment BC an adiabatic expansion. 79 •• [SSM] In a heat engine. / / .314 + $600 K ! 1200 K % " 2. BC ! 0 " )Eint. CA " 0 for an isothermal process. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB to find the pressure at B: PB " PA VA V " $5.00 atm % A 2VA VB 101.50 atm 1 " 253 kPa (b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC to express the temperature at C: Use the ideal-gas law to find the volume of the gas at B: TC " TB PCVC PBVB (1) VB " nRTB PB $2.39 L% W " WAB & WBC & WCD & WDA . B + . pressures. We can then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle.4 to find the volume occupied by the gas at C: Substitute numerical values in equation (1) and evaluate TC: 0P VC " VB .00 atm%$75. 1 1.50 atm " $39. 1.78 L % $2.00 atm + + / . C. " 75.78 L TC " $600 K % " 462 K (c) The work done by the gas in one cycle is given by: 13 J + $600 K % mol 6 K .00 mol%0 8.39 L %. and D.325 kPa " 253.P + / C. / 253.1934 Chapter 19 Picture the Problem We can use the ideal-gas law to find the unknown temperatures. .50 atm%$39.3 kPa 1atm " 2.4 $1.314 . " " 39.3 kPa 0 2.39 L Use the equation of state for an adiabatic process and 3 = 1. and volumes at points B. 737 kJ The work done during the isobaric compression CD is: WCD " PC $VD ! VC % " $1. / A + " 6.314 + $462 K ! 600 K % 2 2 mol 6 K . A + .915 kJ + .00 mol%. B. and volumes at points A. 8. / " 5. 2. and C and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the isobaric.00 atm %$19.325 J atm 6 L Express and evaluate the work done during the constant-volume process DA: Substitute numerical values and evaluate W: WDA " 0 W " 6. and isothermal processes of the cycle. / .00 mol%.The Second Law of Thermodynamics 1935 The work done during the isothermal expansion AB is: 0V WAB " nRTA ln. B .915 kJ & 5.972 kJ " 6. (The PV diagram is not drawn to scale.680 kJ & 0 " 6.78 L % " !56. (a) What is the volume of the gas at A? (b) What are the volume and temperature of the gas at B? (c) What is the temperature of the gas at C? (d) What is the volume of the gas at C? (e) How much work is done by the gas in each of the three segments of the cycle? (f) How much heat is absorbed by the gas in each segment of the cycle? Picture the Problem We can use the ideal-gas law to find the unknown temperatures.) At A the pressure and temperature are 5.09 atm 6 L 1 " !5. The segment BC is an adiabatic expansion and the segment CA is an isothermal compression. V mol 6 K .97 kJ 80 •• In a heat engine. .680 kJ 101.V / A 0 2V J 0 + " $2.314 + $600 K % ln.00 atm and 600 K.7 L ! 75. pressures.00 mol of a monatomic gas are taken through the cycle ABCA as shown in Figure 19-20. The work done during the adiabatic expansion BC is: J 0 WBC " !C V "TBC " ! 5 nR"TBC " ! 5 $2. The volume at B is twice the volume at A. adiabatic. 8.737 kJ ! 5. 00 mol%0 8.4 L " 111 L (e) The work done by the gas during the isobaric process AB is given by: Substitute numerical values and evaluate WAB: WAB " PA $VB ! VA % " PA $2VA ! VA % " PAVA WAB " $5. + . / 101. " J + $600 K % mol 6 K .314 . 3 1 TBVB 3 !1 " TCVC 3 !1 0 1200 K . 600 K + + / .3 !1 4 VC " VB . .39 L " 39.39 L %.2 VC " $39.4 L Apply the ideal-gas law to this isobaric process to find the temperature at B: (c) Because the process CA is isothermal: (d) Apply an equation for an adiabatic process (3 = 5/3) to express the volume of the gas at C: Substitute numerical values and evaluate VC: TB " TA 2V VB " $600 K % A VA VA " 1200 K TC " TA " 600 K 0 TB .325 J atm 6 L " 9.325 kPa 5.45 atm 6 L 1 101.69 L % " 98.00 atm 1 atm " 19.00 atm %$19.T + / C.975 kJ " 9.69 L % " 39.7 L (b) We’re given that: VB " 2VA " 2$19.98 kJ . " 111.69 L " 19.1936 Chapter 19 (a) Apply the ideal-gas law to find the volume of the gas at A: VA " nRTA PA $2. 3 kJ because )Eint = 0 for an isothermal process.314 . CA " WCA " ! 17. BC " )Eint.9 kJ Express and evaluate the heat absorbed during the adiabatic expansion BC: Use the first law of thermodynamics to express and evaluate the heat absorbed during the isothermal compression CA: QBC " 0 5 2 $2. BC ! 0 " !17. the segment BC an adiabatic expansion.00 atm. The volume at B is twice the volume at A. . BC " !$ncV )TBC % " ! 3 nR)TBC 2 Substitute numerical values and evaluate WBC: J 0 WBC " ! 3 $2. 8.00 mol%0 8. 111.0 kJ The work done by the gas during the isothermal compression CA is: 0V WCA " nRTC ln. + $600 K % ln.3 kJ (f) The heat absorbed during the isobaric expansion AB is: Qin.00 mol%. BC ! Qin.314 .00 atm and 600 K. / " 15. A .29 kJ ! 17. / J + $1200 K ! 600 K % mol 6 K .The Second Law of Thermodynamics 1937 Apply the first law of thermodynamics to express the work done by the gas during the adiabatic expansion BC: WBC " )Eint.) The segment AB represents an isothermal expansion.00 mol%0 8. 81 •• In a heat engine.4 L + + mol 6 K .V / C 0 19. " )Eint. AB " ncP "TAB " 5 nR"TAB " 2 " 24. (The PV diagram is not drawn to scale. / .314 + $600 K ! 1200 K % " 14. + . The pressure and temperature at A are 5.69 L J + " $2. / . QCA " WCA & "Eint.00 mol of a monatomic gas are carried through the cycle ABCDA shown in Figure 19-21. 2.97 kJ 2 mol 6 K . The pressure at D is 1. 50 atm " $39.9 K " 416 K (c) The work done by the gas in one cycle is given by: W " WAB & WBC & WCD & WDA (2) .1938 Chapter 19 (a) What is the pressure at B? (b) What is the temperature at C? (c) Find the total work done by the gas in one cycle.50 atm %$39. " 68. (a) Apply the ideal-gas law for a fixed amount of gas to the isothermal process AB: PB " PA VA V " $5.325 kPa 1atm " 2. B + .00 atm % A 2VA VB 101. C.50 atm 1 " 253. 1. Picture the Problem We can use the ideal-gas law to find the unknown temperatures.3 kPa " 253 kPa (b) Apply the ideal-gas law for a fixed amount of gas to the adiabatic process BC: Use the ideal-gas law to find the volume at B: TC " TB PCVC PBVB (1) VB " nRTB PB $2.39 L %.3 kPa 0 2. and volumes at points B.314 .26 L 13 J + $600 K % mol 6 K . " " 39.26 L % $2.P + / C.39 L Use the equation of state for an adiabatic process and 3 = 5/3 to find the volume occupied by the gas at C: Substitute numerical values in equation (1) and evaluate TC: 0P VC " VB .00 atm + + / . pressures. 35 TC " $600 K % $1. and D and then find the work done by the gas and the efficiency of the cycle by using the expressions for the work done on or by the gas and the heat that enters the system for the various thermodynamic processes of the cycle. . / 253.00 atm %$68.39 L % " 415.00 mol%0 8. 7 L ! 68. We can use the ideal-gas law to find the highest temperature of the gas during its cycle and use this temperature to express the efficiency of a Carnot engine.) Picture the Problem We can express the efficiency of the Otto cycle using the result from Example 19-2.9 K ! 600 K % 2 mol 6 K . / " 4. / A + " 6.920 kJ 101.V mol 6 K . 8.592 kJ ! 4.56 atm 6 L 1 " !4.59 kJ 82 •• Compare the efficiency of the Otto cycle to the efficiency of the Carnot cycle operating between the same maximum and minimum temperatures. (The Otto cycle is discussed in Section 19-1. The work done during the adiabatic expansion BC is: WBC " !C V "TBC " ! 3 nR"TBC 2 J 0 " ! 3 $2.00 mol%.915 kJ + . 8. A + .920 kJ & 0 " 6. / .314 + $600 K % ln.314 + $415. .The Second Law of Thermodynamics 1939 The work done during the isothermal expansion AB is: 0V WAB " nRTA ln. Finally.592 kJ The work done during the isobaric compression CD is: WCD " PC $VD ! VC % " $1. B .00 atm %$19.V / A 0 2V J 0 + " $2. We can apply the relation TV 3 !1 " constant to the adiabatic processes of the Otto cycle to relate the end-point temperatures to the volumes occupied by the gas at these points and eliminate the temperatures at c and d.26 L % " !48.00 mol%. we can compare the efficiencies by examining their ratio.325 J atm 6 L The work done during the constantvolume process DA is: Substitute numerical values in equation (2) to obtain: WDA " 0 W " 6.915 kJ & 4. V + / b. 0V ! Ta . Va = Vd and Vc = Vb. Note that. 3 !1 0V ! Ta . a + . d .V + Tb / a.V + / b.V + / b. Tb is not the highest temperature. 0V Tc " Td . d . a + . 3 !1 3 !1 0V Tc ! Tb " Td . Apply the ideal-gas law to c and b to obtain an expression for the cycle’s highest temperature Tc: P Pc Pb 4 Tc " Tb c C Tb " Pb Tc Tb . a + . 3 !1 3 !1 0V " $Td ! Ta %. Va .V / c 3 !1 + + . Substitute in equation (1) and simplify to obtain: # Otto " 1 ! Td ! Ta 0 $Td ! Ta %. 3 !1 0V Tc ! Tb " Td . a + . Substitute to obtain: 0V Tb " Ta .1940 Chapter 19 The efficiency of the Otto engine is given in Example 19-2: Td ! Ta Tc ! Tb # Otto " 1 ! (1) where the subscripts refer to the various points of the cycle as shown in Figure 19-3. a + .V + / b. 3 !1 0V T " 1! .V + / b. Apply the relation TV 3 !1 " constant to the adiabatic process a*b to obtain: Apply the relation TV 3 !1 " constant to the adiabatic process c*d to obtain: Subtract the first of these equations from the second to obtain: In the Otto cycle.V / b 3 !1 + + .V / c + + . while Ta is the lowest temperature of the cycle. b + " 1! a . The Second Law of Thermodynamics 1941 The efficiency of a Carnot engine operating between the maximum and minimum temperatures of the Otto cycle is given by: Express the ratio of the efficiency of a Carnot engine to the efficiency of an Otto engine operating between the same temperatures: Ta Tc # Carnot " 1 ! # Carnot # Otto Ta Tc " C 1 because Tc > Tb. is the Brayton cycle. often used in refrigeration. can be written as # " 1 ! r $1! 3 % 3 . and transitions to temperatures T2. P 2 Qh . Ta 1! Tb 1! Hence. (a) The Brayton heat engine cycle is shown to the right. (a) Sketch this cycle on a PV diagram.(3) an adiabatic expansion. all we must do is consider the heat flow in and out of the engine during the isobaric transitions. Because the adiabatic transitions in the cycle do not have heat flow associated with them. which involves (1) an adiabatic compression. where r is the pressure ratio Phigh/Plow of the maximum and minimum pressures in the cycle. and (4) an isobaric compression back to the original state. (c) Show that this $T3 ! T2 % efficiency. (2) an isobaric (constant pressure) expansion. # Carnot C # Otto 83 ••• [SSM] A common practical cycle. 4 V (1) Qc (b) The efficiency of a heat engine is given by: #" Q ! Qc W " h Qin Qin . Assume the system begins the adiabatic compression at temperature T1. 3 1 . Heat Qh enters the gas during the isobaric transition from state 2 to state 3 and heat Qc leaves the gas during the isobaric transition from state 4 to state 1. (b) Show that the $T ! T % efficiency of the overall cycle is given by # " 1 ! 4 1 . Picture the Problem The efficiency of the cycle is the ratio of the work done to the heat that flows into the engine. The paths 1*2 and 3*4 are adiabatic. T3 and T4 after each leg of the cycle. for an adiabatic transition.P / high + + . low . 3 !1 3 T3 3 !1 3 3 !1 3 Subtract T1 from T4 and simplify to obtain: 0P T4 ! T1 " .1942 Chapter 19 During the constant-pressure expansion from state 1 to state 2 heat enters the system: During the constant-pressure compression from state 3 to state 4 heat enters the system: Substituting in equation (1) and simplifying yields: Q23 " Qh " nC P "T " nC P $T3 ! T2 % Q41 " !Qc " ! nC P "T " ! nC P $T1 ! T4 % #" " $T3 ! T2 % & $T1 ! T4 % $T3 ! T2 % $T ! T % " 1! 4 1 $T3 ! T2 % (c) Given that.P / high + + . 0P T3 ! . low . low . + + .P / high 0P " .P / high + + .P / high + + . T2 3 !1 3 $T3 ! T2 % . low . for the adiabatic transition from state 3 to state 4: T3 " constant P 3 !1 nC P $T3 ! T2 % ! $! nC P $T1 ! T4 %% nC P $T3 ! T2 % T1 T " 32!1 3 !1 Plow Phigh 3 3 0P 4 T1 " . use the ideal-gas law to eliminate V and obtain: Let the pressure for the transition from state 1 to state 2 be Plow and the pressure for the transition from state 3 to state 4 be Phigh. low . 3 !1 3 T2 0P T4 " . Then for the adiabatic transition from state 1 to state 2: Similarly. TV 3 !1 " constant . 0 atm. it adiabatically expands until it returns to its initial state at temperature T1. 3 !1 3 Substitute in the result of Part (b) and simplify to obtain: 0P # " 1 ! . Picture the Problem The efficiency of the Brayton refrigerator cycle is the ratio of the heat that enters the system to the work done to operate the refrigerator. . The cylinder containing the refrigerant (a monatomic gas) has an initial volume and pressure of 60 mL and 1.The Second Law of Thermodynamics 1943 Dividing both sides of the equation by T3 ! T2 yields: T4 ! T1 0 Plow ". (b) $T4 ! T1 % . Then the gas is adiabatically compressed.0. T3 ! T2 . until its temperature is T3. After the expansion at constant pressure. And then it is compressed at constant pressure until its temperature T2. 3 !1 3 0 Phigh " 1! .P / low + + .P / high " 1 ! $r % 3 where r " + + . the cycle begins at temperature T1 and expands at constant pressure until its temperature T4. In this case. Show that the coefficient of performance is COPB = $T3 ! T2 ! T4 & T1 % (c) Suppose your 'Brayton cycle refrigerator' is run as follows. What is the coefficient of performance for your refrigerator? (d) To absorb heat from the food compartment at the rate of 120 W. Finally. Phigh / + + . what is the rate at which electrical energy must be supplied to the motor of this refrigerator? (e) Assuming the refrigerator motor is actually running for only 4. The pressure ratio r = Phigh/Plow for the cycle is 5. low . Assume 15 cents per kWh of electric energy and thirty days in a month. the volume and temperature are 75 mL and –25°C.0 h each day. . 1!3 3 1!3 Phigh Plow 84 ••• Suppose the Brayton cycle engine (see Problem 83) is run in reverse as a refrigerator in your kitchen. all we must do is consider the heat flow in and out of the refrigerator during the isobaric transitions. (a) Sketch this cycle on a PV diagram. Because the adiabatic transitions in the cycle do not have heat flow associated with them. how much does it add to your monthly electric bill. and 4. We’re given that the temperature in state 4 is: For the constant-pressure transition from state 1 to state 4. 3 1 Qh Qh " W Qh ! Qc .1944 Chapter 19 (a) The Brayton refrigerator cycle is shown to the right.V + V1 V4 / 4. + $248 K % " 198 K / 75 mL . The paths 1*2 and 3*4 are adiabatic. the quotient T/V is constant: Substitute numerical values and evaluate T1: T4 " !25(C & 273 K " 248 K 0V T1 T4 " 4 T1 " . 1 +T4 . 0 60 mL T1 " . Heat Qc enters the gas during the constantpressure transition from state 1 to state 4 and heat Qh leaves the gas during the constant-pressure transition from state 3 to state 2. 4 V (1) Qc (b) The coefficient of performance of the Brayton cycle refrigerator is given by: During the constant-pressure compression from state 3 to state 2 heat leaves the system: During the constant-pressure expansion from state 1 to state 4 heat enters the system: Substituting in equation (1) and simplifying yields: COPB " Q32 " !Qh " !nC P "T " !nC P $T2 ! T3 % Q14 " Qc " nC P "T " nC P $T4 ! T1 % COPB " nC P $T4 ! T1 % ! nC P $T4 ! T1 % ! nC P $T2 ! T3 % " " $T4 ! T1 % ! $T4 ! T1 % ! $T2 ! T3 % T4 ! T1 T3 ! T2 ! T4 & T1 (c) The COPB requires the temperatures corresponding to states 1. . 2. 3. P 2 Qh . TV 3 !1 " constant . 3 + 3 !1 . if the frequency of the AC power input is f. 1.67 !1 0P . fQc dW " dt COPB Express the heat Qc that is drawn from the cold reservoir: Substituting for Qc yields: Qc " nC P "T " nC P $T4 ! T1 % fnC P $T4 ! T1 % dW " dt COPB n" P4V4 RT4 Use the ideal-gas law to express the number of moles of the gas: .1 W" Qc COPB The rate at which energy must be supplied to this refrigerator is given by: 1 dQc dW " dt COPB dt or. for an adiabatic transition.67 $198 K % .P + / 1.3 T2 " .The Second Law of Thermodynamics 1945 Given that.P + P3 P4 / 4.67 $248 K % " 473 K 3 !1 1. 2 + T1 " $5% 1. use the ideal-gas law to eliminate V and obtain: For the adiabatic transition from state 4 to state 3: T3 " constant P 3 !1 0P T33 T3 " 34!1 4 T3 " .67 !1 3 !1 3 T4 Substitute numerical values and evaluate T3: Similarly. for the adiabatic transition from state 2 to state 1: T3 " $5% 1. " 378 K Substitute numerical values in the expression derived in Part (a) and evaluate COPB: (d) From the definition of COPB: COPB " 248 K ! 198 K 473 K ! 378 K ! 248 K & 198 K " 1. and TV 3 !1 = a constant to show that the entropy change for a quasi-static adiabatic expansion that proceeds from state (V1. C P " 5 R .325 kPa %.11%$248 K % dt 5 2 !1 " 0. show explicitly that the entropy change is zero for a quasi-static adiabatic expansion from state (V1. V2 + / . T2). / 1. 3 " CP CV .0 h 1 207 W 1 1 30 d 2 $4 kWh d Using )S " Cv ln $T2 T1 % ! nR ln $V2 V1 % (Equation 19-16) for the entropy change of an ideal gas.T + . / " 207 W " $1.21 kW (e) The monthly cost of operation is given by Monthly Cost " Cost Per Unit of Power 1 Power Consumption " rate 1 daily consumption 1 number of days per month Substitute numerical values and evaluate the monthly cost of operation: Monthly Cost " 85 ••• $0. Picture the Problem We can use nR " CP ! CV . 75 mL 1 $60 s %$101. Express the entropy change for a general process that proceeds from state 1 to state 2: For an adiabatic process: 0T 0V )S " CV ln. 2 + .1946 Chapter 19 Because the gas is monatomic.V + / 1. T2 0 V1 ". + T1 . T1) to state (V2.T2) is zero. 3 !1 . L + dW . 2 + & nR ln.T1) to state (V2. Substitute for n and CP to 2 obtain: 5 2 dW " dt f " Substitute numerical values and evaluate dW/dt: 5 2 fP4V4 $T4 ! T1 % $COPB %T4 P4V4 R$T4 ! T1 % RT4 COPB 0 10 !3 m 3 +$248 K ! 198 K % .15 4. / 1. + & nR ln. then the entropy of the universe could decrease. + @nR & . + = 3 !1 . ! ln = @ V2 = @ > A Use the relationship between CP and CV to obtain: Substituting for nR and 3 and simplifying yields: nR " CP ! CV . + " ln.V + . 2 + @CP ! CV ! . Because Th > Tc.? 0C 0 V -B )S " ln. p ! 1+CV = . is heat Qh is taken from the hot reservoir and no heat is rejected to the cold reservoir. which is negative. that is. Su < 0. Have you just proved that this statement is equivalent to the refrigerator and heat-engine statements? Picture the Problem (a) Suppose the refrigerator statement of the second law is violated in the sense that heat Qc is taken from the cold reservoir and an equal mount of heat is transferred to the hot reservoir and W = 0.V + . = )S " CV ln.. = " ln0 V2 . i. .? @ CV ln. (b) Show that if the heat-engine statement of the second law were not true. then the entropy change of the universe is )Su = !Qh/Th + 0.V + .e.@ 0 V1 / 2.The Second Law of Thermodynamics 1947 T2 and simplify to obtain: T1 Substitute for 3 !1 B 0 V1 .V + 0V / 2 . + " ln. Qc = 0. the entropy of the universe would decrease.C / 1 . The entropy change of the universe is then )Su = Qc/Th ! Qc/Tc. / 1 .V + + . then the entropy of the universe could decrease.A . the entropy of the universe would decrease. 2 + @nR & V . (c) A third statement of the second law is that the entropy of the universe cannot decrease.8nR ! $3 ! 1%C 9 . Again. ln 2 = V1 @ = A > B 0 -? $3 ! 1%CV ln. (b) In this case.V +@ V1 = / 1.V + 0 V2 0 V2 .@ / 2.V + = V / 1. V1 + = @ . > / V " 0 86 ••• (a) Show that if the refrigerator statement of the second law of thermodynamics were not true. Show that the net efficiency of the combination is given by # net " # 1 & # 2 ! # 1# 2 . such that the heat released by the first engine is used as the heat absorbed by the second engine. Picture the Problem We can express the net efficiency of the two engines in terms of W1. as shown in Figure 19-22. respectively. Show that the net efficiency of T the combination is given by # net " 1 ! c . The efficiencies of the engines are #1 and #2. such that the heat released by the first engine is used as the heat absorbed by the second engine as shown in Figure 19-22. and Qh and then use #1 = W1/Qh and #2 = W2/Qm to eliminate W1. (Note that this result means that two Th . W2. Express the net efficiency of the two heat engines connected in series: Express the efficiencies of engines 1 and 2: Solve for W1 and W2 and substitute to obtain: Express the efficiency of engine 1 in terms of Qm and Qh: Substitute for Qm/Qh and simplify to obtain: # net " W1 & W2 Qh #1 " W1 W and # 2 " 2 Qh Qm # net " #1Qh & # 2Qm Qh " #1 & Qm #2 Qh #1 " 1 ! Qm Q 4 m " 1 ! #1 Qh Qh # net " # 1 & $1 ! # 1 %# 2 " # 1 & # 2 ! # 1# 2 88 ••• Suppose that two heat engines are connected in series. W2. The statement )Su D 0 is more restrictive. and Qm. 87 ••• Suppose that two heat engines are connected in series. Suppose that each engine is an ideal reversible heat engine. Qh. The heat-engine and refrigerator statements in conjunction with the Carnot efficiency are equivalent to )Su D 0. Engine 1 operates between temperatures Th and Tm and Engine 2 operates between Tm and Tc. but these statements do not specify the minimum amount of heat rejected or work that must be done. where Th > Tm > Tc.1948 Chapter 19 (c) The heat-engine and refrigerator statements of the second law only state that some heat must be rejected to a cold reservoir and some work must be done to transfer heat from the cold to the hot reservoir. and Qm. W2. W2. Qh./ . T T T T " 1! m & m ! c " 1! c Th Th Th Th 89 ••• [SSM] The English mathematician and philosopher Bertrand Russell (1872-1970) once said that if a million monkeys were given a million typewriters and typed away at random for a million years.1 ! + Th . Finally. they would produce all of Shakespeare’s works.0 Tc & . Let us limit ourselves to the following fragment of Shakespeare (Julius Caesar III:ii): . we can substitute the expressions for the efficiencies of the ideal reversible engines to obtain # net " 1 ! Tc Th .) Picture the Problem We can express the net efficiency of the two engines in terms of W1. Express the efficiencies of ideal reversible engines 1 and 2: #1 " 1 ! and Tm Th Tc Tm (1) #2 " 1! The net efficiency of the two engines connected in series is given by: Express the efficiencies of engines 1 and 2: Solve for W1 and W2 and substitute in equation (3) to obtain: Express the efficiency of engine 1 in terms of Qm and Qh: Substitute for Qm to obtain: Qh (2) # net " #1 " W1 & W2 Qh (3) W1 W and # 2 " 2 Qh Qm # net " #1Qh & # 2Qm Qh " #1 & Qm #2 Qh #1 " 1 ! Q Qm 4 m " 1 ! #1 Qh Qh # net " #1 & $1 ! #1 %# 2 Substitute for #1 and #2 and simplify to obtain: # net " 1 ! Tm 0 Tm .The Second Law of Thermodynamics 1949 reversible heat engines operating !in series! are equivalent to one reversible heat engine operating between the hottest and coldest reservoirs. Tm + / . and Qh and then use #1 = W1/Qh and #2 = W2/Qm to eliminate W1. + . Th + . 5 %$330 % 1 " 1 " 10 !495 495 10 T" $330 s %$10495 % 106 0 1y " 3. Assuming the monkeys type at random. Romans. $ % 2 10484 y T TRussell or T 2 10 478 TRussell " 10484 y " 10478 6 10 y . not to praise him. And. (You can even assume that the monkeys are immortal. disregarding capitalization.5 to obtain: Assuming the monkeys can type at a rate of 1 character per second. The good is oft interred with the bones. Even with this small fragment. . 3. it will take a lot longer than a million years! By what factor (roughly speaking) was Russell in error? Make any reasonable assumptions you want. comma. This fragment is 330 characters (including spaces) long. So let it be with Caesar. And grievously hath Caesar answered it . express the probability P that one monkey will write out this passage: Use the approximation 30 2 1000 " 101. and exclamation point) used in the English language. it were a grievous fault. there are then 30330 different possible arrangements of the character set to form a fragment this long.16 1 107 s + + / .1950 Chapter 19 Friends. if so. it would take about 330 s to write a passage of length equal to the quotation from Shakespeare.) Picture the Problem There are 26 letters and four punctuation marks (space. countrymen! Lend me your ears. We can use this number of possible arrangements to express the probability that one monkey will write out this passage and then an estimate of a monkey’s typing speed to approximate the time required for one million monkeys to type the passage from Shakespeare. . The evil that men do lives on after them. so we have a grand total of 30 characters to choose from. . The noble Brutus hath told you that Caesar was ambitious. period. I come to bury Caesar. Find the time T required for a million monkeys to type this particular passage by accident: Express the ratio of T to Russell’s estimate: P" 1 30330 P" 10 $1.30 1 10 491 s .