Chapter 20Thermal Properties and Processes Conceptual Problems 1 • Why does the mercury level of a thermometer first decrease slightly when the thermometer is first placed in warm water? Determine the Concept The glass bulb warms and expands first, before the mercury warms and expands. 2 • A large sheet of metal has a hole cut in the middle of it. When the sheet is heated, the area of the hole will (a) not change, (b) always increase, (c) always decrease, (d) increase if the hole is not in the exact center of the sheet, (e) decrease only if the hole is in the exact center of the sheet. Determine the Concept The heating of the sheet causes the average separation of its molecules to increase. The consequence of this increased separation is that the area of the hole always increases. ) (b is correct. 3 • [SSM] Why is it a bad idea to place a tightly sealed glass bottle that is completely full of water, into your kitchen freezer in order to make ice? Determine the Concept Water expands greatly as it freezes. If a sealed glass bottle full of water is placed in a freezer, as the water freezes there will be no room for the expansion to take place. The bottle will be broken. 4 • The windows of your physics laboratory are left open on a night when the temperature of the outside dropped well below freezing. A steel ruler and a wooden ruler were left on the window sill, and when you arrive in the morning they are both very cold. The coefficient of linear expansion of wood is about 5 × 10 −6 K −1 . Which ruler should you use to make the most accurate length measurements? Explain your answer. Determine the Concept You should use the wooden ruler. Because the coefficient of expansion for wood is about half that for metal, the metal ruler will have shrunk considerably more than will have the wooden ruler. 5 • Bimetallic strips are used both for thermostats and for electrical circuit breakers. A bimetallic strip consists of a pair of thin strips of metal that have different coefficients of linear expansion and are bonded together to form one doubly thick strip. Suppose a bimetallic strip is constructed out of one steel strip and one copper strip, and suppose the bimetallic strip is curled in the shape of a circular arc with the steel strip on the outside. If the temperature of the strip is decreased, will the strip straighten out or curl more tightly? 1863 Chapter 20 1864 Determine the Concept The strip will curl more tightly. Because the coefficient of linear expansion for copper (17 × 10 −6 K −1 ) is greater than the coefficient of linear expansion for steel (11 × 10 −6 K −1 ), the length of the copper strip will decrease more than the length of the steel strip−resulting in a tighter curl. 6 • Metal A has a coefficient of linear expansion that is three times the coefficient of linear expansion of metal B. How do their coefficients of volume expansion β compare? (a) β A = β B , (b) β A = 3β B , (c) β A = 6β B , (d) β A = 9β B , (e) You cannot tell from the data given. Determine the Concept We know that the coefficient of volume expansion is three times the coefficient of linear expansion and so can use this fact to express the ratio of A β to B β . Express the coefficient of volume expansion of metal A in terms of its coefficient of linear expansion: A A 3α β = Express the coefficient of volume expansion of metal B in terms of its coefficient of linear expansion: B B 3α β = Dividing the first of these equations by the second yields: B A B A B A 3 3 α α α α β β = = Because B A 3α α = : 3 3 B B B A = = α α β β ⇒ ( ) b is correct. 7 • The summit of Mount Rainier is 14 410 ft above sea level. Mountaineers say that you cannot hard boil an egg at the summit. This statement is true because at the summit of Mount Rainier (a) the air temperature is too low to boil water, (b) the air pressure is too low for alcohol fuel to burn, (c) the temperature of boiling water is not hot enough to hard boil the egg, (d) the oxygen content of the air is too low to support combustion, (e) eggs always break in climbers′ backpacks. Determine the Concept Actually, an egg can be hard boiled, but it takes quite a bit longer than at sea level. ) (c is the best response. 8 • Which gases in Table 20-3 cannot be condensed by applying pressure at 20ºC? Explain your answer. Determine the Concept Gases that cannot be liquefied by applying pressure at 20°C are those for which T c < 293 K. These are He, Ar, Ne, H 2 , O 2 , NO. Thermal Properties and Processes 1865 9 •• [SSM] The phase diagram in Figure 20-15 can be interpreted to yield information on how the boiling and melting points of water change with altitude. (a) Explain how this information can be obtained. (b) How might this information affect cooking procedures in the mountains? Determine the Concept (a) With increasing altitude, decreases; from curve OC, the temperature of the liquid-gas interface decreases as the pressure decreases, so the boiling temperature decreases. Likewise, from curve OB, the melting temperature increases with increasing altitude. (b) Boiling at a lower temperature means that the cooking time will have to be increased. 10 •• Sketch a phase diagram for carbon dioxide using information from Section 20-3. Determine the Concept The following phase diagram for carbon dioxide was constructed using information in Section 20-3. atm , P K , T 5.1 216.6 304.2 Gas Gas Solid Vapor Liquid point Critical 11 •• Explain why the carbon dioxide on Mars is found in the solid state in the polar regions even though the atmospheric pressure at the surface of Mars is only about 1 percent of the atmospheric pressure at the surface of Earth. Determine the Concept At very low pressures and temperatures, carbon dioxide can exist only as a solid or gas (or vapor above the gas). The atmosphere of Mars is 95 percent carbon dioxide. Mars, on average, is warm enough so that the atmosphere is mostly gaseous carbon dioxide. The polar regions are cold enough to enable solid carbon dioxide (dry ice) to exist, even at the low pressure. 12 •• Explain why decreasing the temperature of your house at night in winter can save money on heating costs. Why doesn’t the cost of the fuel consumed to heat the house back to the daytime temperature in the morning equal Chapter 20 1866 the savings realized by cooling it down in the evening and keeping it cool throughout the night? Determine the Concept The amount of heat lost by the house is proportional to the difference between the temperature inside the house and that of the outside air. Hence, the rate at which the house loses heat (that must be replaced by the furnace) is greater at night when the temperature of the house is kept high than when it is allowed to cool down. 13 •• [SSM] Two solid cylinders made of materials A and B have the same lengths; their diameters are related by d A = 2d B . When the same temperature difference is maintained between the ends of the cylinders, they conduct heat at the same rate. Their thermal conductivities are therefore related by which of the following equations? (a) k A = k B /4, (b) k A = k B /2, (c) k A = k B , (d) k A = 2k B , (e) k A = 4k B B Picture the Problem The rate at which heat is conducted through a cylinder is given by x T kA dt dQ I Δ Δ = = / / (see Equation 20-7) where A is the cross- sectional area of the cylinder. The heat current in cylinder A is the same as the heat current in cylinder B: B A I I = Substituting for the heat currents yields: L T A k L T A k Δ = Δ B B A A ⇒ A B B A A A k k = Because d A = 2d B : ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = B B B A 4A A k k ⇒ B 4 1 A k k = ) (a is correct. 14 •• Two solid cylinders made of materials A and B have the same diameter; their lengths are related by L A = 2L B . When the same temperature difference is maintained between the ends of the cylinders, they conduct heat at the same rate. Their thermal conductivities are therefore related by which of the following equations? (a) k A = k B /4, (b) k A = k B /2, (c) k A = k B , (d) k A = 2k B , (e) k A = 4k B . B Determine the Concept We can use the expression for the heat current in a conductor, Equation 20-7, to relate the heat current in each cylinder to its thermal conductivity, cross-sectional area, temperature difference, and length. Thermal Properties and Processes 1867 The heat current in cylinder A is the same as the heat current in cylinder B: B A I I = Substituting for the heat currents yields: B B A A L T A k L T A k Δ = Δ ⇒ B A B A L L k k = Because L A = 2L B : B B B B A 2 2 k L L k k = = ⇒ ( ) d is correct. 15 •• If you feel the inside of a single pane window during a very cold day, it is cold, even though the room temperature can be quite comfortable. Assuming the room temperature is 20.0°C and the outside temperature is 5.0°C, Construct a plot of temperature versus position starting from a point 5.0 m in behind the window (inside the room) and ending at a point 5.0 m in front of the window. Explain the heat transfer mechanisms that occur along this path. Determine the Concept The temperatures on both sides of the glass are almost the same. Because glass is an excellent conductor of heat, there need not be a huge temperature difference. Thus the temperature must drop quickly as you near the pane on the warm side and the same on the outside. This is sketched qualitatively in the following diagram. Convection and radiation are primarily responsible for heat transfer on the inside and outside, and it is mainly conduction through the glass. Conduction through the interior and exterior air is minimal. C , ° T 20 5 Inside Outside m , x 5 0 16 •• During the thermal retrofitting of many older homes in California, it was found that the 3.5-in-deep spaces between the wallboards and the outer sheathing were filled with just air (no insulation). Filling the space with insulating material certainly reduces heating and cooling costs; although, the insulating material is a better conductor of heat than air is. Explain why adding the insulation is a good idea. Determine the Concept The tradeoff is the reduction of convection cells between the walls by putting in the insulating material, versus a slight increase in conductivity. The net reduction in convection results in a higher R value. Chapter 20 1868 Estimation and Approximation 17 •• [SSM] You are using a cooking pot to boil water for a pasta dish. The recipe calls for at least 4.0 L of water to be used. You fill the pot with 4.0 L of room temperature water and note that this amount of water filled the pot to the brim. Knowing some physics, you are counting on the volume expansion of the steel pot to keep all of the water in the pot while the water is heated to a boil. Is your assumption correct? Explain. If your assumption is not correct, how much water runs over the sides of the pot due to the thermal expansion of the water? Determine the Concept The volume of water overflowing is the difference between the change in volume of the water and the change in volume of the pot. See Table 20-1 for the coefficient of volume expansion of water and the coefficient of linear expansion of steel. Express the volume of water that overflows when the pot and the water are heated: ( ) T V T V T V V V V Δ Δ Δ Δ Δ 0 steel O H 0 steel 0 O H pot O H ovefrlow 2 2 2 β β β β − = − = − = Because the coefficient of volume expansion of steel is three times its coefficient of linear expansion: steel steel 3α β = Substituting for and O H 2 β steel β yields: ( ) T V V Δ 3 0 steel O H overflow 2 α α − = Substitute numerical values and evaluate : overflow V ( ) ( )( )( ) mL 56 C 20 C 100 L 4.0 K 10 11 3 K 10 207 . 0 1 6 1 3 overflow = ° − ° × − × = − − − − V Your assumption was not correct and 56 mL of water overflowed. 18 •• Liquid helium is stored in containers fitted with 7.00-cm-thick ″superinsulation″ consisting of numerous layers of very thin aluminized Mylar sheets. The rate of evaporation of liquid helium in a 200-L container is about 0.700 L per day when the container is stored at room temperature (20ºC). The density of liquid helium is 0.125 kg/L and the latent heat of vaporization is 21.0 kJ/kg. Estimate the thermal conductivity of the superinsulation. Picture the Problem We can express the heat current through the insulation in terms of the rate of evaporation of the liquid helium and in terms of the temperature gradient across the superinsulation. Equating these equations will lead to an expression for the thermal conductivity k of the superinsulation. Note that the boiling temperature of liquid helium is 4.2 K. Thermal Properties and Processes 1869 Express the heat current in terms of the rate of evaporation of the liquid helium: dt dm L I v = Express the heat current in terms of the temperature gradient across the superinsulation and the conductivity of the superinsulation: x T kA I Δ Δ = Equate these expressions and solve for k to obtain: T A dt dm x L k Δ Δ = v Using the definition of density, express the rate of loss of liquid helium: dt dV dt dm ρ = Substitute for dt dm to obtain: T A dt dV x L k Δ Δ = ρ v Express the ratio of the area of the spherical container to its volume: 3 3 4 2 4 r r V A π π = ⇒ 3 2 36 V A π = Substituting for A yields: T V dt dV x L k Δ Δ = 3 2 v 36π ρ Substitute numerical values and evaluate k: ( ) ( ) ( ) K m W 10 12 . 3 K 289 m 10 200 36 s 86400 m 10 0.700 m kg 125 m 10 7.00 kg kJ 21.0 6 3 2 3 3 3 3 3 2 ⋅ × = × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − − − π k 19 •• [SSM] Estimate the thermal conductivity of human skin. Picture the Problem We can use the thermal current equation for the thermal conductivity of the skin. If we model a human body as a rectangular parallelepiped that is 1.5 m high × 7 cm thick × 50 cm wide, then its surface area is about 1.8 m 2 . We’ll also assume that a typical human, while resting, produces energy at the rate of 120 W, that normal internal and external temperatures are 33°C and 37°C, respectively, and that an average skin thickness is 1.0 mm. Chapter 20 1870 Use the thermal current equation to express the rate of conduction of thermal energy: I = kA ΔT Δx ⇒ x T A I k Δ Δ = Substitute numerical values and evaluate k: ( ) K m mW 17 m 10 0 . 1 C 33 C 37 m 8 . 1 W 120 3 2 ⋅ = × ° − ° = − k 20 •• You are visiting Finland with a college friend and have met some Finnish friends. They talk you into taking part in a traditional Finnish after-sauna exercise which consists of leaving the sauna, wearing only a bathing suit, and running out into the mid-winter Finnish air. Estimate the rate at which you initially lose energy to the cold air. Compare this rate of initial energy loss to the resting metabolic rate of a typical human under normal temperature conditions. Explain the difference. Picture the Problem We can use the Stefan-Boltzmann law to estimate the rate at which you lose energy when you first step out of the sauna. If we model a human body as a rectangular parallelepiped that is 1.5 m high × 7 cm thick × 50 cm wide, then its surface area is about 1.8 m 2 . Assume that your skin temperature is initially 37°C (310 K), that the mid-winter outside temperature is −10°C (263 K), and that the emissivity of your skin is 1. Use the Stefan-Boltzmann law to express the net rate at which you radiate energy to the cold air: ( ) 4 air 4 skin net T T A P − = εσ Substitute numerical values and evaluate P net : ( ) ( ) ( ) ( ) ( ) W 450 K 63 2 K 310 m 8 . 1 K m W 10 670 . 5 1 4 4 2 4 2 8 net = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ × = − P This result is almost four times greater than the basal metabolic rate of 120 W. We can understand the difference in terms of the temperature of your skin when you first step out of the sauna and the fact that radiation loses are dependent on the fourth power of the absolute temperature. Remarks: The emissivity of your skin is, in fact, very close to 1. 21 •• Estimate the rate of heat conduction through a 2.0-in-thick wooden door during a cold winter day in Minnesota. Include the brass doorknob. What is the ratio of the heat that escapes through the doorknob to the heat that escapes through the whole door? What is the total overall R-factor for the door, including the knob? The thermal conductivity of brass is ( ) K m W 85 ⋅ . Thermal Properties and Processes 1871 Picture the Problem We can use the thermal current equation (Equation 20-7) to estimate the rate of heat loss through the door and its knob. We’ll assume that the door is made of oak with an area of 2.0 m 2 , that the knob is made of brass, that the conducting path has a diameter of 3.0 cm, and that the inside temperature is 20°C. Take the outside temperature to be −20°C. The reciprocal of the equivalent R- factor is the sum of the reciprocals of the R-values of the door and the knob. The total thermal current through the door is the sum of the thermal currents through the door and the knob: knob knob knob door door door knob door tot Δ Δ Δ Δ x T A k x T A k I I I + = + = Assume that to obtain: x x x Δ Δ Δ knob door = = ( ) x T A k A k I Δ Δ knob knob door door tot + = Substitute numerical values and evaluate I tot : ( ) ( ) ( ) ( ) kW 28 . 0 in 39.37 m 1 in 0 . 2 C 20 C 20 m 10 0 . 3 4 K m W 85 m 0 . 2 K m W 15 . 0 2 2 2 tot = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° − − ° ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = − π I Express the ratio of the thermal currents through the knob and the door: knob knob knob door door door door knob Δ Δ Δ Δ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = x T A k x T A k I I Because the temperature difference across the door and the knob are the same and we’ve assumed that the thickness of the door and length of the knob are the same: knob knob door door door knob A k A k I I = Substitute numerical values to obtain: ( ) ( ) 0 . 5 m 10 0 . 3 4 K m W 85 m 0 . 2 K m W 15 . 0 2 2 2 door knob = × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = − π I I Relate the equivalent R-factor to the R-factors of the door and knob; knob door eq 1 1 1 R R R + = Chapter 20 1872 Substitute for R door and R knob to obtain: x A k A k A k x A k x R Δ Δ 1 Δ 1 1 knob knob door door knob knob door door eq + = + = Solving for R eq yields: knob knob door door eq Δ A k A k x R + = Substitute numerical values and evaluate R eq : ( ) ( ) ( ) W K 14 . 0 m 10 0 . 3 4 K m W 85 m 0 . 2 K m W 15 . 0 in 39.37 m 1 in 0 . 2 2 2 2 eq = × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = − π R 22 •• Estimate the effective emissivity of Earth, given the following information. The solar constant, which is the intensity of radiation incident on Earth from the Sun, is about 1.37 kW/m 2 . Seventy percent of this energy is absorbed by Earth, and Earth’s average surface temperature is 288 K. (Assume that the effective area that is absorbing the light is πR 2 , where R is Earth’s radius, while the blackbody-emission area is 4πR 2 .) Picture the Problem The amount of heat radiated by Earth must equal the solar flux from the Sun, or else the temperature on Earth would continually increase. The emissivity of Earth is related to the rate at which it radiates energy into space by the Stefan-Boltzmann law . 4 r AT e P σ = Using the Stefan-Boltzmann law, express the rate at which Earth radiates energy as a function of its emissivity e and temperature T: 4 r A'T e P σ = ⇒ 4 r A'T P e σ = where A′ is the surface area of Earth. Use its definition to express the intensity of the radiation received by Earth: A P I absorbed = where A is the cross-sectional area of Earth. For 70% absorption of the Sun’s radiation incident on Earth: ( ) A P I r 70 . 0 = Substitute for P r and A in the expression for e and simplify to obtain: ( ) ( ) ( ) 4 4 2 2 4 4 70 . 0 4 70 . 0 70 . 0 T I T R I R A'T AI e σ σ π π σ = = = Thermal Properties and Processes 1873 Substitute numerical values and evaluate e: ( )( ) ( )( ) 61 . 0 K 288 K W/m 10 670 . 5 4 kW/m 37 . 1 70 . 0 4 4 2 8 2 = ⋅ × = − e 23 •• Black holes are highly condensed remnants of stars. Some black holes, together with a normal star, form binary systems. In such systems the black hole and the normal star orbit about the center of mass of the system. One way black holes can be detected from Earth is by observing the frictional heating of the atmospheric gases from the normal star that fall into the black hole. These gases can reach temperatures greater than 1.0 × 10 6 K. Assuming that the falling gas can be modeled as a blackbody radiator, estimate λ max for use in an astronomical detection of a black hole. (Remark: This is in the X-ray region of the electromagnetic spectrum.) Picture the Problem The wavelength at which maximum power is radiated by the gas falling into a black hole is related to its temperature by Wien’s displacement law. Wien’s displacement law relates the wavelength at which maximum power is radiated by the gas to its temperature: T K mm 898 . 2 max ⋅ = λ Substitute for T and evaluate λ max : nm 9 . 2 K 10 0 . 1 K mm 898 . 2 6 max = × ⋅ = λ 24 ••• Your cabin in northern Michigan has walls that consist of pine logs that have average thicknesses of about 20 cm. You decide to finish the interior of the cabin to improve the look and to increase the insulation of the exterior walls. You choose to buy insulation with an R-factor of 31 to cover the walls. In addition, you cover the insulation with 1.0-in-thick gypsum wallboard. Assuming heat transfer is only due to conduction, estimate the ratio of thermal current through the walls during a cold winter night before the renovation to the thermal current through the walls following the renovation. Picture the Problem We can use the thermal current equation to find the thermal current per square meter through the walls of the cabin both before and after the walls have been insulated. See Table 20-5 for the R-factor of gypsum wallboard and Table 20-4 for the thermal conductivity of white pine. The rate at which heat is conducted through the walls is given by the thermal current equation: before before before Δ Δ Δ R T x T A k I = = With the insulation in place: after after after Δ Δ Δ R T x T A k I = = Chapter 20 1874 Divide the second of these equations by the first to obtain: after before before after before after Δ Δ R R R T R T I I = = before R is the R-factor for pine and is the sum of the R-factors for pine, the insulating material, and the gypsum board: after R pine before R R = and gypsum 31 pine eq after R R R R R + + = = Substitute for and to obtain: before R after R gypsum 31 pine pine before after R R R R I I + + = Because pine pine k x R Δ = : gypsum 31 pine pine before after R R k x k x I I + + Δ Δ = Substitute numerical values and evaluate the ratio : before after / I I % 24 Btu F ft h 32 . 0 in 0.375 in 1 Btu F ft h 31 F ft h in Btu 78 . 0 cm 2.54 in 1 cm 20 F ft h in Btu 78 . 0 cm 2.54 in 1 cm 20 2 2 2 2 before after = ° ⋅ ⋅ × + ° ⋅ ⋅ + ° ⋅ ⋅ ⋅ × ° ⋅ ⋅ ⋅ × = I I 25 ••• [SSM] You are in charge of transporting a liver from New York, New York to Los Angeles, California for a transplant surgery. The liver is kept cold in a Styrofoam ice chest initially filled with 1.0 kg of ice. It is crucial that the liver temperature is never warmer than 5.0°C. Assuming the trip from the hospital in New York to the hospital in Los Angeles takes 7.0 h, estimate the R-value the Styrofoam walls of the ice chest must have. Picture the Problem The R factor is the thermal resistance per unit area of a slab of material. We can use the thermal current equation to express the thermal resistance of the styrofoam in terms of the maximum amount of heat that can enter the chest in 7.0 h without raising the temperature above 5.0°C. We’ll assume that the surface area of the ice chest is 1.0 m 2 and that the ambient temperature is 25°C Thermal Properties and Processes 1875 The R-factor needed for the styrofoam walls of the ice chest is the product of their thermal resistance and area: RA R = f (1) Use the thermal current equation to express R: tot tot Δ Δ Δ Δ Δ Q t T t Q T I T R = = = Substitute for R in equation (1) to obtain: tot f Δ Δ Q t T A R = The total heat entering the chest in 7 h is given by: O H O H ice f ice water ice warm ice melt tot 2 2 ΔT c m L m Q Q Q + = + = Substitute for Q tot and simplify to obtain: ( ) O H O H f ice f 2 2 Δ Δ Δ T c L m t T A R + = Substitute numerical values and evaluate R f : ( ) ( ) ( ) Btu ft h F 8 C 5 K kg kJ 18 . 4 kg kJ 5 . 333 kg 1 Btu J 35 . 1054 h 7 C 5 F 9 C 20 m 10 29 . 9 ft 1 m 1.0 2 2 2 2 2 f ⋅ ⋅ ° ≈ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ° ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ° ° × ° ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × × = − R Thermal Expansion 26 •• You have inherited your grandfather’s grandfather clock that was calibrated when the temperature of the room was 20ºC. Assume that the pendulum consists of a thin brass rod of negligible mass with a compact heavy bob at its end. (a) During a hot day, the temperature is 30ºC, does the clock run fast or slow? Explain. (b) How much time does it gain or lose during this day? Picture the Problem We can determine whether the clock runs fast or slow from the expression for the period of a simple pendulum and the dependence of its length on the temperature. We can use the expression for the period of a simple pendulum and the equation describing its length as a function of temperature to find the time gained or lost in a 24-h period. (a) Express the period of the pendulum in terms of its length: g L T π 2 P = Chapter 20 1876 Because L T ∝ P and L is temperature dependent, the clock runs slow. (b) The period of the pendulum when the temperature is 20°C is given by: g L T 20 20 2π = (1) When the temperature increases to 30°C, the period of the pendulum increases due to the increase in its length: ( ) C 20 C 20 1 1 2 2 t T g t L g L T Δ + = Δ + = = α α π π The daily fractional gain or loss is given by : 20 20 20 T T T T T − = Δ = Δ τ τ Substituting for T and simplifying yields: 1 1 1 C 20 20 C 20 − Δ + = − Δ + = Δ t T T t T α α τ τ Solve for Δτ to obtain: ( )τ α τ 1 1 C − Δ + = Δ t Substitute numerical values and Δτ: ( )( ) ( ) s 2 . 8 h s 3600 h 24 1 C 20 C 30 K 10 19 1 1 6 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × − ° − ° × + = Δ − − τ 27 •• [SSM] You need to fit a copper collar tightly around a steel shaft that has a diameter of 6.0000 cm at 20ºC. The inside diameter of the collar at that temperature is 5.9800 cm. What temperature must the copper collar have so that it will just slip on the shaft, assuming the shaft itself remains at 20ºC? Picture the Problem Because the temperature of the steel shaft does not change, we need consider just the expansion of the copper collar. We can express the required temperature in terms of the initial temperature and the change in temperature that will produce the necessary increase in the diameter D of the copper collar. This increase in the diameter is related to the diameter at 20°C and the increase in temperature through the definition of the coefficient of linear expansion. Express the temperature to which the copper collar must be raised in terms of its initial temperature and the increase in its temperature: T T T Δ + = i Thermal Properties and Processes 1877 Apply the definition of the coefficient of linear expansion to express the change in temperature required for the collar to fit on the shaft: D D D D T α α Δ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ Δ = Δ Substitute for ΔT to obtain: D D T T α Δ + = i Substitute numerical values and evaluate T: ( )( ) C 220 cm 5.9800 K 10 17 cm 9800 . 5 cm 0000 . 6 C 20 1 6 ° = × − + ° = − − T 28 •• You have a copper collar and a steel shaft. At 20°C, the collar has an inside diameter of 5.9800 cm and the steel shaft has diameter of 6.0000 cm. The copper collar was heated. When its inside diameter exceeded 6.0000 cm is was slipped on the shaft. The collar fitted tightly on the shaft after they cooled to room temperature. Now, several years later, you need to remove the collar from the shaft. To do this you heat them both until you can just slip the collar off the shaft. What temperature must the collar have so that the collar will just slip off the shaft? Picture the Problem Because the temperatures of both the steel shaft and the copper collar change together, we can find the temperature change required for the collar to fit the shaft by equating their diameters for a temperature increase ΔT. These diameters are related to their diameters at 20°C and the increase in temperature through the definition of the coefficient of linear expansion. Express the temperature to which the collar and the shaft must be raised in terms of their initial temperature and the increase in their temperature: T T T Δ + = i (1) Express the diameter of the steel shaft when its temperature has been increased by ΔT: ( ) T D D Δ + = ° steel C steel,20 steel 1 α Express the diameter of the copper collar when its temperature has been increased by ΔT: ( ) T D D Δ + = ° Cu C Cu,20 Cu 1 α Chapter 20 1878 If the collar is to fit over the shaft when the temperature of both has been increased by ΔT: ( ) ( ) T D T D Δ + = Δ + ° ° steel C steel,20 Cu C Cu,20 1 1 α α Solving for ΔT yields: steel C steel,20 Cu C Cu,20 C Cu,20 C steel,20 α α ° ° ° ° − − = Δ D D D D T Substitute in equation (1) to obtain: steel C steel,20 Cu C Cu,20 C Cu,20 C steel,20 i α α ° ° ° ° − − + = D D D D T T Substitute numerical values and evaluate T: ( )( ) ( )( ) C 580 /K 10 11 cm 6.0000 /K 10 17 cm 5.9800 cm 5.9800 cm 6.0000 C 20 6 6 ° = × − × − + ° = − − T 29 •• A container is filled to the brim with 1.4 L of mercury at 20ºC. As the temperature of container and mercury is increased to 60ºC, a total of 7.5 mL of mercury spill over the brim of the container. Determine the linear expansion coefficient of the material that makes up the container. Picture the Problem The linear expansion coefficient of the container is one- third its coefficient of volume expansion. We can relate the changes in volume of the mercury and the container to their initial volumes, temperature change, and coefficients of volume expansion, and, because we know the amount of spillage, obtain an equation that we can solve for β c . Relate the linear expansion coefficient of the container to its coefficient of volume expansion: c 3 1 c β α = (1) Express the difference in the change in the volume of the mercury and the container in terms of the spillage: mL 5 . 7 c Hg = Δ − Δ V V (2) Express using the definition of the coefficient of volume expansion: Hg V Δ T V V Δ = Δ Hg Hg Hg β Express using the definition of the coefficient of volume expansion: c V Δ T V V Δ = Δ c c c β Thermal Properties and Processes 1879 Substitute for and in equation (2) to obtain: Hg V Δ c V Δ mL 5 . 7 c c Hg Hg = Δ − Δ T V T V β β Solving for β c yields: T V T V Δ mL 5 . 7 Δ c Hg Hg c − = β β or, because V = V Hg = V c , T V T V T V Δ mL 5 . 7 Δ mL 5 . 7 Δ Hg Hg c − = − = β β β Substitute for c β in equation (1) to obtain: T V T V Δ 3 mL 5 . 7 Δ 3 mL 5 . 7 Hg Hg 3 1 c − = − = α β α Substitute numerical values and evaluate α c : ( ) ( )( ) 1 6 1 3 3 1 c K 10 15 K 40 L 4 . 1 3 mL 5 . 7 K 10 18 . 0 − − − − × = − × = α 30 •• A car has a 60.0-L steel gas tank filled to the brim with 60.0-L of gasoline when the temperature of the outside is 10ºC. The coefficient of volume expansion for gasoline at 20°C is 0.950 × 10 −3 K − 1. How much gasoline spills out of the tank when the outside temperature increases to 25ºC? Take the expansion of the steel tank into account. Picture the Problem The amount of gas that spills is the difference between the change in the volume of the gasoline and the change in volume of the tank. We can find this difference by expressing the changes in volume of the gasoline and the tank in terms of their common volume at 10°C, their coefficients of volume expansion, and the change in the temperature. Express the spill in terms of the change in volume of the gasoline and the change in volume of the tank: tank gasoline spill V V V Δ − Δ = Relate to the coefficient of volume expansion for gasoline: gasoline V Δ T V V Δ = Δ gasoline gas β Chapter 20 1880 Relate to the coefficient of linear expansion for steel: tank V Δ T V V Δ = Δ tank tank β or, because β steel = 3α steel , T V V Δ = Δ steel tank 3α Substitute for and and simplify to obtain: gasoline V Δ tank V Δ ( ) steel gasoline steel gas spill 3 3 α β α β − Δ = Δ − Δ = T V T V T V V Substitute numerical values and evaluate : spill V ( )( )[ ( )] L 8 . 0 K 10 11 3 K 10 950 . 0 C 0 1 C 25 L 60.0 1 6 1 3 spill ≈ × − × ° − ° = − − − − V 31 ••• What is the tensile stress in the copper collar of Problem 27 when its temperature returns to 20ºC? Picture the Problem We can use the definition of Young’s modulus to express the tensile stress in the copper in terms of the strain it undergoes as its temperature returns to 20°C. We can show that ΔL/L for the circumference of the collar is the same as Δd/d for its diameter. Using Young’s modulus, relate the stress in the collar to its strain: where L 20°C is the circumference of the collar at 20°C. Express the circumference of the collar at the temperature at which it fits over the shaft: Express the circumference of the collar at 20 °C: Substitute for and and simplify to obtain: T L C 20° L C 20 Strain Stress ° Δ = × = L L Y Y T T d L π = C 20 C 20 ° ° = d L π C 20 C 20 C 20 C 20 Stress ° ° ° ° − = − = d d d Y d d d Y T T π π π Thermal Properties and Processes 1881 Substitute numerical values and evaluate the stress: ( ) 2 12 2 10 N/m 10 7 . 3 cm 5.9800 cm 9800 . 5 cm 0000 . 6 N/m 10 11 Stress − − × = − × = The van der Waals Equation, Liquid-Vapor Isotherms, and Phase Diagrams 32 • (a) Calculate the volume of 1.00 mol of an ideal gas at a temperature of 100ºC and a pressure of 1.00 atm. (b) Calculate the temperature at which 1.00 mol of steam at a pressure of 1.00 atm has the volume calculated in Part (a). Use a = 0.550 Pa⋅m 6 /mol 2 and b = 30.0 cm 3 /mol. Picture the Problem We can apply the ideal-gas law to find the volume of 1.00 mol of steam at 100°C and a pressure of 1.00 atm and then use the van der Waals equation to find the temperature at which the steam will this volume. (a) Solving the ideal-gas law for the volume gives: P nRT V = Substitute numerical values and evaluate V: ( )( )( ) L 6 . 30 m 10 L 1 m 10 06 . 3 atm kPa 101.325 atm 00 . 1 K 373 K J/mol 314 . 8 mol 00 . 1 3 3 3 2 = × × = × ⋅ = − − V (b) Solve van der Waals equation for T to obtain: ( ) nR bn V V an P T − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = 2 2 Substitute numerical values and evaluate T: ( )( ) ( ) ( ) ( ) ( ) ( ) K 375 K J/mol 8.314 mol 1.00 mol 00 . 1 /mol m 10 30.0 m 10 3.06 m 10 3.06 mol 1.00 /mol m Pa 0.550 kPa 325 . 101 3 6 3 2 2 3 2 2 2 6 = ⋅ × − × × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × ⋅ + = − − − T Chapter 20 1882 33 •• [SSM] Using Figure 20-16, find the following quantities. (a) The temperature at which water boils on a mountain where the atmospheric pressure is 70.0 kPa, (b) the temperature at which water boils in a container where the pressure inside the container is 0.500 atm, and (c) the pressure at which water boils at 115ºC. Picture the Problem Consulting Figure 20-16, we see that: (a) At 70.0 kPa, water boils at approximately C 90° . (b) At 0.500 atm (about 51 kPa), water boils at approximately C 78° . (c) The pressure at which water boils at 115°C is approximately kPa 127 . 34 •• The van der Waals constants for helium are a = 0.03412 L 2 ⋅atm/mol 2 and b = 0.0237 L/mol. Use these data to find the volume in cubic centimeters occupied by one helium atom. Then, estimate the radius of the helium atom. Picture the Problem Assume that a helium atom is spherical. Then we can find its volume from the van der Waals equation and its radius from 3 3 4 r V π = . In the van der Waals equation, b is the volume of 1 mol of molecules. For He, 1 molecule = 1 atom. Use Avogadro’s number to express b in cm 3 /atom: atom cm 10 94 . 3 mol atoms 10 6.022 L cm 10 mol L 0237 . 0 3 23 23 3 3 − × = × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = b The volume of a spherical helium atom is given by: 3 3 4 r V π = ⇒ 3 3 4 3 4 3 π π b V r = = Substitute numerical values and evaluate r: ( ) nm 211 . 0 4 cm 10 94 . 3 3 3 3 23 = × = − π r Conduction 35 • [SSM] A 20-ft × 30-ft slab of insulation has an R factor of 11. At what rate is heat conducted through the slab if the temperature on one side is a constant 68ºF and the temperature of the other side is a constant 30ºF? Thermal Properties and Processes 1883 Picture the Problem We can use its definition to express the thermal current in the slab in terms of the temperature differential across it and its thermal resistance and use the definition of the R factor to express I as a function of ΔT, the cross- sectional area of the slab, and R f . Express the thermal current through the slab in terms of the temperature difference across it and its thermal resistance: R T I Δ = Substitute to express R in terms of the insulation’s R factor: f f / R T A A R T I Δ = Δ = Substitute numerical values and evaluate I: ( )( )( ) h kBtu 2.1 Btu F ft h 11 F 30 F 68 ft 30 ft 20 2 = ° ⋅ ⋅ ° − ° = I 36 •• A copper cube and an aluminum, cube each with 3.00-cm-long edges, are arranged as shown in Figure 20-17. Find (a) the thermal resistance of each cube, (b) the thermal resistance of the two-cube combination, (c) the thermal current I, and (d) the temperature at the interface of the two cubes. Picture the Problem We can use kA x R Δ = to find the thermal resistance of each cube and the fact that they are in series to find the thermal resistance of the two-cube system. We can use R T I Δ = to find the thermal current through the cubes and the temperature at their interface. See Table 20-4 for the thermal conductivities of copper and aluminum. (a) Using its definition, express the thermal resistance of each cube: kA x R Δ = Substitute numerical values and evaluate the thermal resistance of the copper cube: ( ) K/W 0831 . 0 K/W 08313 . 0 cm 3.00 K m W 401 cm 3.00 2 Cu = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = R Chapter 20 1884 Substitute numerical values and evaluate the thermal resistance of the aluminum cube: ( ) K/W 141 . 0 K/W 1406 . 0 cm 3.00 K m W 37 2 cm 3.00 2 Al = = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = R (b) Because the cubes are in series, their thermal resistances are additive: K/W 0.224 K/W 0.2237 K/W 0.1406 K/W 0.08313 Al Cu = = + = + = R R R (c) Using its definition, find the thermal current: kW 0.36 W 6 . 357 K/W 0.2237 C 20 C 100 Δ = = ° − ° = = R T I (d) Express the temperature at the interface between the two cubes: Cu interface Δ C 100 T T − ° = Express the temperature differential across the copper cube: Cu Cu Cu Cu IR R I T = = Δ Substitute for ΔT Cu to obtain: Cu interface C 100 IR T − ° = Substitute numerical values and evaluate T interface : ( )( ) C 70 K/W 08313 . 0 W 357.6 C 100 interface ° ≈ − ° = T 37 •• Two metal cubes, one copper and one aluminum, with 3.00-cm-long edges, are arranged in parallel, as shown in Figure 20-18. Find (a) the thermal current in each cube, (b) the total thermal current, and (c) the thermal resistance of the two-cube combination. Picture the Problem We can use Equation 20-9 to find the thermal current in each cube. Because the currents are additive, we can find the equivalent resistance of the two-cube system from total eq I T R Δ = . (a) The thermal current in each cube is given by Equation 20-9: x T kA R T I Δ Δ = Δ = Thermal Properties and Processes 1885 Substitute numerical values and evaluate the thermal current in the copper cube: ( ) kW 96 . 0 W 4 . 962 cm 3.00 C 20 C 100 cm 3.00 K m W 401 2 Cu = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ° − ° ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = I Substitute numerical values and evaluate the thermal current in the aluminum cube: ( ) kW 57 . 0 W 8 . 568 cm 3.00 C 20 C 100 cm 3.00 K m W 37 2 2 Al = = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ° − ° ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = I (b) Because the cubes are in parallel, their total thermal currents are additive: kW 1.5 kW 1.531 W 8 . 68 5 W 4 . 62 9 Al Cu = = + = + = I I I (c) Use the relationship between the total thermal current, temperature differential and thermal resistance to find R eq : K/W 0.052 kW 1.531 C 20 C 100 Δ total eq = ° − ° = = I T R 38 •• The cost of air conditioning a house is approximately proportional to the rate at which heat is absorbed by the house from its surroundings divided by the coefficient of performance (COP) of the air conditioner. Let us denote the temperature difference between the inside temperature and the outside temperature as ΔT. Assuming that the rate at which heat is absorbed by a house is proportional to ΔT and that the air conditioner is operating ideally, show that the cost of air conditioning is proportional to (ΔT) 2 divided by the temperature inside the house. Picture the Problem The cost of operating the air conditioner is proportional to the energy used in its operation. We can use the definition of the COP to relate the rate at which the air conditioner removes heat from the house to rate at which it must do work to maintain a constant temperature differential between the interior and the exterior of the house. To obtain an expression for the minimum rate at which the air conditioner must do work, we’ll assume that it is operating with the maximum efficiency possible. Doing so will allow us to derive an expression for the rate at which energy is used by the air conditioner that we can integrate to obtain the energy (and hence the cost of operation) required. Relate the cost C of air conditioning the energy W required to operate the air conditioner: uW C = (1) where u is the unit cost of the energy. Chapter 20 1886 Express the rate dQ/dt at which heat flows into a house provided the house is maintained at a constant temperature: T k dt dQ P Δ = = where ΔT is the temperature difference between the interior and exterior of the house. Use the definition of the COP to relate the rate at which the air conditioner must remove heat dW/dt to maintain a constant temperature: dt dW dt dQ = COP ⇒ dt dQ dt dW COP 1 = Express the maximum value of the COP: T T Δ = c max COP where house the inside c T T = is the temperature of the cold reservoir. Letting COP = COP max , substitute to obtain an expression for the minimum rate at which the air conditioner must do work in order to maintain a constant temperature: T T dt dQ dt dW Δ c = Substituting for dQ/dt gives: ( ) 2 c c T T k T T T k dt dW Δ = Δ Δ = Separate variables and integrate this equation to obtain: ( ) ( ) t T T k dt' T T k W t Δ Δ = Δ = ∫ Δ 2 c 0 2 c Substitute in equation (1) to obtain: ( ) t T T k u C Δ Δ 2 c = ⇒ ( ) c 2 Δ T T C ∝ 39 •• A spherical shell of thermal conductivity k has inside radius r 1 and outside radius r 2 (Figure 20-19). The inside of the shell is held at a temperature T 1 , and the outside of the shell is held at temperature T 2 , with T 1 < T 2 . In this problem, you are to show that the thermal current through the shell is given by I = − 4πkr 1 r 2 r 2 − r 1 T 2 −T 1 ( ) where I is positive if heat is transferred in the +r direction. Here is a suggested procedure for obtaining this result: (1) obtain an expression for the thermal current I through a thin spherical shell of radius r and thickness dr when there is a temperature difference dT across the thickness of the shell; (2) explain why the thermal current is the same through each such thin shell; (3) express the thermal current I through such a shell element in terms of the area A = 4πr 2 , the thickness Thermal Properties and Processes 1887 dr, and the temperature difference dT across the element; and (4) separate variables (solve for dT in terms of r and dr) and integrate. Picture the Problem We can follow the step-by-step instructions given in the problem statement to obtain the differential equation describing the variation of T with r. Integrating this equation will yield an equation that we can solve for the current I. (1) The thermal current through a thin spherical shell surface of area A and thickness dr due to a temperature gradient dr dT is given by: dr dT kA I = (2) Conservation of energy requires that the thermal current through each shell be the same. (3) For a shell of area A = 4π r 2 : dr dT kr I 2 4π = (4) Separating the variables yields: 2 4 r dr k I dT π = Integrate from r = r 1 to r = r 2 and simplify to obtain: ∫ ∫ = 2 1 2 1 2 4 r r T T r dr k I dT π and ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ − = − 2 1 1 2 1 1 4 1 4 2 1 r r k I r k I T T r r π π Solving for I gives: ( ) 1 2 1 2 2 1 4 T T r r r kr I − − − = π Radiation 40 • Calculate λ max (the wavelength at which the emitted power is maximum) for a human skin. Assume the human skin is a blackbody emitter with a temperature of 33ºC. Picture the Problem We can apply Wein’s displacement law to find the wavelength at which the power is a maximum. Chapter 20 1888 Wein’s law relates the maximum wavelength of the radiation to the temperature of its source: T K mm 898 . 2 max ⋅ = λ Substitute numerical values and evaluate λ max : m 47 . 9 K 33 K 273 K mm 2.898 max μ λ = + ⋅ = 41 • [SSM] The universe is filled with radiation that is believed to be remaining from the Big Bang. If the entire universe is considered to be a blackbody with a temperature equal to 2.3 K, what is the λ max (the wavelength at which the power of the radiation is maximum) of this radiation? Picture the Problem We can use Wein’s law to find the peak wavelength of this radiation. Wein’s law relates the maximum wavelength of the background radiation to the temperature of the universe: T K mm 2.898 max ⋅ = λ Substituting for T gives: mm .3 1 K 3 . 2 K mm 2.898 max = ⋅ = λ 42 • What is the range of temperatures for star surfaces for which λ max (the wavelength at which the power of the emitted radiation is maximum) is in the visible range? Picture the Problem The visible portion of the electromagnetic spectrum extends from approximately 400 nm to 700 nm. We can use Wein’s law to find the range of temperatures corresponding to these wavelengths. Wein’s law relates the maximum wavelength of the radiation to the temperature of its source: T K mm 2.898 max ⋅ = λ Solving for T yields: max K mm 2.898 λ ⋅ = T For λ max = 400 nm: K 7250 nm 400 K mm 2.898 = ⋅ = T For λ max = 700 nm: K 4140 nm 700 K mm 2.898 = ⋅ = T Thermal Properties and Processes 1889 The range of temperatures is K 7250 K 4140 ≤ ≤ T . 43 • The heating wires of a 1.00-kW electric heater are red hot at a temperature of 900ºC. Assuming that 100 percent of the heat released is due to radiation and that the wires act as blackbody emitters, what is the effective area of the radiating surface? (Assume a room temperature of 20ºC.) Picture the Problem We can apply the Stefan-Boltzmann law to find the net power radiated by the wires of its heater to the room. Relate the net power radiated to the surface area of the heating wires, their temperature, and the room temperature: ( ) 4 0 4 net T T A e P − = σ ⇒ ( ) 4 0 4 net T T e P A − = σ Substitute numerical values and evaluate A: ( ) ( ) ( ) [ ] 2 4 4 4 2 8 cm 5 . 93 K 293 K 1173 K m W 10 5.6703 1 kW 1.00 = − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ × = − A 44 •• A blackened, solid copper sphere that has a radius equal to 4.0 cm hangs in an evacuated enclosure whose walls have a temperature of 20ºC. If the sphere is initially at 0ºC, find the initial rate at which its temperature changes, assuming that heat is transferred by radiation only. (Assume the sphere is a blackbody emitter.) Picture the Problem The rate at which the copper sphere absorbs radiant energy is given by and, from the Stephan-Boltzmann law, dt mcdT dt dQ / / = ( ) 4 0 4 net T T A e P − = σ where A is the surface area of the sphere, T 0 is its temperature, and T is the temperature of the walls. We can solve the first equation for dT/dt and substitute P net for dQ/dt in order to find the initial rate at which the temperature of the sphere changes. Relate the rate at which the sphere absorbs radiant energy to the rate at which its temperature changes: dt dT mc dt dQ P = = net Solving for dt dT and substituting for mc gives: c r P Vc P mc P dt dT ρ π ρ 3 3 4 net net net = = = Chapter 20 1890 Apply the Stefan-Boltzmann law to relate the net power radiated to the sphere to the difference in temperature of the walls and the blackened copper sphere: ( ) ( ) 4 0 4 2 4 0 4 net 4 T T e r T T A e P − = − = σ π σ Substitute for to obtain: net P ( ) ( ) c r T T e c r T T e r dt dT ρ σ ρ π σ π 4 0 4 3 3 4 4 0 4 2 3 4 − = − = Substitute numerical values and evaluate dT/dt: ( ) ( ) ( ) [ ] ( ) K/s 10 2 . 2 K kg kJ 0.386 m kg 10 8.93 m 10 4.0 K 273 K 293 K m W 10 5.6703 1 3 3 3 3 2 4 4 4 2 8 − − − × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × × − ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ × − = dt dT 45 •• The surface temperature of the filament of an incandescent lamp is 1300ºC. If the electric power input is doubled, what will the new temperature be? Hint: Show that you can neglect the temperature of the surroundings. Picture the Problem We can apply the Stephan-Boltzmann law to express the net power radiated by the incandescent lamp to its surroundings. Express the rate at which energy is radiated to the surroundings: ( ) ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − = 4 0 4 4 0 4 net 1 T T AT e T T A e P σ σ Evaluate 4 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ T T : 3 4 4 0 10 1 K 1573 K 293 − × ≈ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ T T Because 4 0 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ T T is so small, we can neglect the temperature of the surroundings. Hence: 4 net AT e P σ ≈ ⇒ 4 1 net ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = A e P T σ Express the temperature T ′when the electric power input is doubled: 4 1 net 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = A e P T' σ Thermal Properties and Processes 1891 Dividing the second of these equations by the first and solving for T gives: ( ) 4 1 2 = T T' ⇒ ( ) T T' 4 1 2 = Substitute numerical values and evaluate T ′ ( ) ( ) C 1598 K 1871 K 1573 2 4 1 ° = = = T' 46 •• Liquid helium is stored at its boiling point (4.2 K) in a spherical can that is separated by an evacuated region of space from a surrounding shield that is maintained at the temperature of liquid nitrogen (77 K). If the can is 30 cm in diameter and is blackened on the outside so that it acts as a blackbody emitter, how much helium boils away per hour? Picture the Problem We can differentiate Q = mL, where L is the latent heat of boiling for helium, with respect to time to obtain an expression for the rate at which the helium boils away. Express the rate at which the helium boils away in terms of the rate at which its container absorbs radiant energy: ( ) ( ) ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − = − = − = = 4 0 4 2 4 0 4 2 4 0 4 net 1 T T T L d e L T T d e L T T A e L P dt dm σπ σπ σ If T 0 << T, then: 4 2 T L d e dt dm σπ ≈ Substitute numerical values and evaluate dm/dt: ( ) ( ) ( ) g/h 97 h s 3600 s kg 10 2.68 K 77 kg kJ 21 m 30 . 0 K m W 10 5.6703 1 5 4 2 4 2 8 = × × = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ × ≈ − − π dt dm Chapter 20 1892 General Problems 47 • A steel tape is placed around Earth at the equator when the temperature is 0ºC. What will the clearance between the tape and the ground (assumed to be uniform) be if the temperature of the tape increases to 30ºC? Neglect the expansion of Earth. Picture the Problem The distance by which the tape clears the ground equals the change in the radius of the circle formed by the tape placed around Earth at the equator. Express the change in the radius of the circle defined by the steel tape: T R R Δ = Δ α where R is the radius of Earth, α is the coefficient of linear expansion of steel, and ΔT is the increase in temperature. Substitute numerical values and evaluate ΔR: ( )( )( ) km 1 . 2 m 10 10 . 2 C 0 C 30 K 10 11 m 10 6.37 Δ 3 1 6 6 = × = ° − ° × × = − − R 48 •• Show that change in the density ρ of an isotropic material due to an increase in temperature ΔT is given by Δρ = –βρΔT. Picture the Problem We can differentiate the definition of the density of an isotropic material with respect to T and use the definition of the coefficient of volume expansion to express the rate at which the density of the material changes with respect to temperature. Once we have an expression for dρ in terms of dT, we can apply a differential approximation to obtain Δρ in terms of ΔT. Using its definition, relate the density of the material to its mass and volume: V m = ρ Using its definition, relate the volume of the material to its coefficient of volume expansion: T V V Δ = Δ β Thermal Properties and Processes 1893 Differentiate ρ with respect to T and simplify to obtain: ρβ β ρ β ρ ρ − = − = − = = V V V V V m dT dV dV d dT d 2 2 or dT d ρβ ρ − = Using the differential approximations ρ ρ Δ ≈ d and yields: T dT Δ ≈ T Δ Δ ρβ ρ − = 49 •• [SSM] The solar constant is the power received from the Sun per unit area perpendicular to the Sun’s rays at the mean distance of Earth from the Sun. Its value at the upper atmosphere of Earth is about 1.37 kW/m 2 . Calculate the effective temperature of the Sun if it radiates like a blackbody. (The radius of the Sun is 6.96 × 10 8 m.). Picture the Problem We can apply the Stefan-Boltzmann law to express the effective temperature of the Sun in terms of the total power it radiates. We can, in turn, use the intensity of the Sun’s radiation in the upper atmosphere of Earth to approximate the total power it radiates. Apply the Stefan-Boltzmann law to relate the energy radiated by the Sun to its temperature: 4 r AT e P σ = ⇒ 4 r A e P T σ = Express the surface area of the sun: 2 S 4 R A π = Relate the intensity of the Sun’s radiation in the upper atmosphere to the total power radiated by the sun: 2 4 R P I r π = ⇒ I R P r 2 4π = where R is the Earth-Sun distance. Substitute for P r and A in the expression for T and simplify to obtain: 4 2 S 2 4 2 S 2 4 4 R e I R R e I R T σ π σ π = = Chapter 20 1894 Substitute numerical values and evaluate T: ( ) ( ) ( ) K 5800 m 10 6.96 K m W 10 5.67 1 m kW 1.35 m 10 1.5 4 2 8 2 8 2 2 11 = × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = − T 50 •• As part of your summer job as an engineering intern at an insulation manufacturer, you are asked to determine the R factor of insulating material. This particular material comes in 1 2 -in sheets. Using this material, you construct a hollow cube that has 12-in long edges. You place a thermometer and a 100-W heater inside the box. After thermal equilibrium has been attained, the temperature inside the box is 90ºC when the temperature outside the box is 20ºC. Determine the R factor of the material. Picture the Problem We can solve the thermal-current equation for the R factor of the material. Use the equation for the thermal current to express I in terms of the temperature gradient across the insulation: x T kA I Δ Δ = Rewrite this expression in terms of the R factor of the material: f f R T A A R T kA x T I Δ = Δ = Δ Δ = Solving for the R factor gives: I T A I T A R Δ = Δ = side one f 6 Substitute numerical values and evaluate R: ( ) Btu h ft F 2 . 2 s 3600 h 1 Btu J 1054 m ft 10.76 K 5 F 9 s J m K 0.390 W m K 0.39 C 20 C 90 W 100 in m 10 2.54 in 12 6 2 2 2 2 2 2 2 f ⋅ ⋅ ° = × × × ° × ⋅ = ⋅ = ° − ° ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × × = − R Thermal Properties and Processes 1895 51 • (a) From the definition of β, the coefficient of volume expansion (at constant pressure), show that β = 1/T for an ideal gas. (b) The experimentally determined value of β for N 2 gas at 0ºC is 0.003673 K –1 . By what percent does this measured value of β differ from the value obtained by modeling N 2 as an ideal gas? Picture the Problem We can use the definition of the coefficient of volume expansion with the ideal-gas law to show that β = 1/T. (a) Use the definition of the coefficient of volume expansion to express β in terms of the rate of change of the volume with temperature: dT dV V 1 = β For an ideal gas: P nRT V = and P nR dT dV = Substitute for dT dV to obtain: T P nR V 1 1 = = β (b) Express the percent difference between the experimental value and the theoretical value: % 3 . 0 K 273 1 K 273 1 K 0.003673 1 1 1 th th exp < − = − − − − β β β 52 •• A rod of length L A , made from material A, is placed next to a rod of length L B , made of material B. The rods remain in thermal equilibrium with each other. (a) Show that even though the lengths of each rod will change with changes in the ambient temperature, the difference between the two lengths will remain constant if the lengths L A and L B are chosen such that L A /L B = α B /α A , where α A and α B are the coefficients of linear expansion, respectively. (b) If material B is steel, material A is brass, and L A = 250 cm at 0ºC, what is the value of L B ? Picture the Problem Let ΔL be the difference between L B and L B A and express L B and L A in terms of the coefficients of linear expansion of materials A and B. Requiring that ΔL be constant will lead us to the condition that L A /L B = α B /α A . Chapter 20 1896 (a) Express the condition that ΔL does not change when the temperature of the materials changes: constant A B = − = Δ L L L (1) Using the definition of the coefficient of linear expansion, substitute for L B and L A : ( ) ( ) ( ) ( ) ( ) T L L L T L L L L T L L T L L L Δ − + Δ = Δ − + − = Δ + − Δ + = Δ A A B B A A B B A B A A A B B B α α α α α α or ( ) 0 A A B B = Δ − T L L α α The condition that ΔL remain constant when the temperature changes by ΔT is: 0 A A B B = − L L α α Solving for the ratio of L A to L B yields: A B B A α α = L L (b) From (a) we have: steel brass brass B A A steel B α α α α L L L L = = = Substitute numerical values and evaluate L B : ( ) cm 430 K 10 11 K 10 19 cm 250 1 6 1 6 B = × × = − − − − L 53 •• [SSM] On the average, the temperature of Earth’s crust increases 1.0ºC for every increase in depth of 30 m. The average thermal conductivity of Earth’s crust material is 0.74 J/m⋅s⋅K. What is the heat loss of Earth per second due to conduction from the core? How does this heat loss compare with the average power received from the Sun (which is about 1.37 kW/m 2 )? Picture the Problem We can apply the thermal-current equation to calculate the heat loss of Earth per second due to conduction from its core. We can also use the thermal-current equation to find the power per unit area radiated from Earth and compare this quantity to the solar constant. Express the heat loss of Earth per unit time as a function of the thermal conductivity of Earth and its temperature gradient: x T kA dt dQ I Δ Δ = = (1) or x T k R dt dQ Δ Δ = 2 E 4π Thermal Properties and Processes 1897 Substitute numerical values and evaluate dQ/dt: ( ) kW 10 3 . 1 m 30 C 1.0 K s m J 0.74 m 10 37 . 6 4 10 2 6 × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ° ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⋅ × = π dt dQ Rewrite equation (1) to express the thermal current per unit area: x T k A I Δ Δ = Substitute numerical values and evaluate I/A: ( ) 2 W/m 0.0247 m 30 C 1.0 K s J/m 0.74 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ° ⋅ ⋅ = A I Express the ratio of I/A to the solar constant: % 002 . 0 kW/m 1.37 W/m 0.0247 constant solar 2 2 < = A I 54 •• A copper-bottomed saucepan containing 0.800 L of boiling water boils dry in 10.0 min. Assuming that all the heat is transferred through the flat copper bottom, which has a diameter of 15.0 cm and a thickness of 3.00 mm, calculate the temperature of the outside of the copper bottom while some water is still in the pan. Picture the Problem We can find the temperature of the outside of the copper bottom by finding the temperature difference between the outside of the saucepan and the boiling water. This temperature difference is related to the rate at which the water is evaporating through the thermal-current equation. Express the temperature outside the pan in terms of the temperature inside the pan: T T T T Δ C 100 Δ in out + ° = + = (1) Relate the thermal current through the bottom of the saucepan to its thermal conductivity, area, and the temperature gradient between its surfaces: x T kA t Q Δ Δ = Δ Δ ⇒ x t Q kA T Δ Δ Δ = Δ 1 Because the water is boiling: v mL Q = Δ Chapter 20 1898 Substitute for ΔQ to obtain: t kA x mL T Δ Δ Δ v = Substituting for ΔT in equation (1) yields: t kA x mL T Δ Δ C 100 v out + ° = Substitute numerical values and evaluate T out : ( ) ( ) ( ) ( ) C 101 s 600 m 0.150 4 K m W 401 m 10 3.00 kg MJ 2.26 kg 0.800 C 100 2 3 out ° = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ × ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + ° = − π T 55 •• A cylindrical steel hot-water tank of cylindrical shape has an inside diameter of 0.550 m and inside height of 1.20 m. The tank is enclosed with a 5.00-cm-thick insulating layer of glass wool whose thermal conductivity is 0.0350 W/(m⋅K). The insulation is covered by a thin sheet-metal skin. The steel tank and the sheet-metal skin have thermal conductivities that are much greater than that of the glass wool. How much electrical power must be supplied to this tank in order to maintain the water temperature at 75.0ºC when the external temperature is 1.0ºC? Picture the Problem Thermal energy is lost from this tank through its cylindrical side and its top and bottom. The power required to maintain the temperature of the water in the tank is equal to the rate at which thermal energy is conducted through the insulation. Express the total thermal current as the sum of the thermal currents through the top and bottom and the side of the hot-water tank: side bottom and top I I I + = (1) Express I through the top and bottom surfaces: x T k d x T kA I Δ Δ Δ Δ 2 2 2 1 bottom and top π = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = Substitute numerical values and evaluate : bottom and top I ( ) ( ) W 6 . 24 m 0.0500 C 0 . 1 C 75.0 K m W 0.0350 m 550 . 0 2 2 1 bottom and top = ° − ° ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = π I Thermal Properties and Processes 1899 Letting r represent the inside radius of the tank, express the heat current through the cylindrical side: dr dT kLr dr dT kA I π 2 side − = − = where the minus sign is a consequence of the heat current being opposite the temperature gradient. Separating the variables yields: r dr kL I dT π 2 side − = Integrate from r = r 1 to r = r 2 and T = T 1 to T = T 2 and simplify to obtain: ∫ ∫ − = 2 1 2 1 2 side r r T T r dr kL I dT π and ] ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = − = − 2 1 side 1 2 side side 1 2 ln 2 ln 2 ln 2 2 1 r r kL I r r kL I r kL I T T r r π π π Solving for I side gives: ( ) 1 2 2 1 side ln 2 T T r r kL I − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = π Substitute numerical values and evaluate I side : ( ) ( ) W 117 C 0 . 1 C 0 . 75 m 0.275 m 0.325 ln m 1.20 K m W 0.0350 2 side = ° − ° ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = π I Substitute numerical values in equation (1) and evaluate I: W 142 W 17 1 W 4.6 2 = + = I 56 ••• The diameter d of a tapered rod of length L is given by d = d 0 (1 + ax), where a is a constant and x is the distance from one end. If the thermal conductivity of the material is k what is the thermal resistance of the rod? Chapter 20 1900 Picture the Problem We can use R = ΔT/I and I = −kAdT/dt to express dT in terms of the linearly increasing diameter of the rod. Integrating this expression will allow us to find ΔT and, hence, R. The thermal resistance of the rod is given by: I T R Δ = (1) where I the thermal current in the rod. Relate the thermal current in the rod to its thermal conductivity k, cross- sectional area A, and temperature gradient: dx dT kA I − = where the minus sign is a consequence of the heat current being opposite the temperature gradient. Using the dependence of the diameter of the rod on x, express the area of the rod: ( ) 2 2 0 2 1 4 4 ax d d A + = = π π Substitute for A to obtain: ( ) dx dT ax d k I ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − = 2 2 0 1 4 π Separating variables yields: ( ) ( ) 2 2 0 2 2 0 1 4 1 4 ax dx kd I ax d k Idx dT + − = ⎥ ⎦ ⎤ ⎢ ⎣ ⎡ + − = π π Integrating T from T 1 to T 2 and x from 0 to L gives: ( ) ∫ ∫ + − = L T T ax dx kd I dT 0 2 2 0 1 4 2 1 π and ( ) aL kd IL T T T + = Δ = − 1 4 2 0 1 2 π Substitute for ΔT and I in equation (1) and simplify to obtain: ( ) ( ) aL kd L I aL kd IL R + = + = 1 4 1 4 2 0 2 0 π π 57 ••• A solid disk of radius r and mass m is spinning about a frictionless axis through its center and perpendicular to the disk, with angular velocity ω 1 at temperature T 1 . The temperature of the disk decreases to T 2 . Express the angular velocity ω 2 , rotational kinetic energy K 2 , and angular momentum L 2 in terms of Thermal Properties and Processes 1901 their values at the temperature T 1 and the linear expansion coefficient α of the disk. Picture the Problem Let ΔT = T 2 – T 1 . We can apply Newton’s second law to establish the relationship between L 2 and L 1 and angular momentum conservation to relate ω 2 and ω 1 . We can express both E 2 and E 1 in terms of their angular momenta and rotational inertias and take their ratio to establish their relationship. Apply t L Δ Δ = ∑ τ to the spinning disk: Because 0 = ∑ τ , ΔL = 0 and 1 2 L L = Apply conservation of angular momentum to relate the angular velocity of the disk at T 2 to the angular velocity at T 1 : 1 1 2 2 ω ω I I = ⇒ 1 2 1 2 ω ω I I = (1) Express I 2 : ( ) ( ) ( ) 2 1 2 2 1 2 2 2 Δ Δ 2 1 Δ 1 T T I T mr mr I α α α + + = + = = Because (αΔT) 2 is small compared to αΔT: ( ) T I I Δ 2 1 1 2 α + ≈ Substitute for I 2 in equation (1) to obtain: ( ) 1 1 1 2 Δ 2 1 ω α ω T I I + = (2) Expanding ( binomially yields: ) 1 Δ 2 1 − + T α ( ) s order term higher Δ 2 1 Δ 2 1 1 + − = + − T T α α Because the higher order terms are very small: ( ) T T Δ 2 1 Δ 2 1 1 α α − ≈ + − Substituting in equation (2) yields: ( ) 1 2 2 1 ω α ω T Δ − ≈ Express E 2 in terms of L 2 and I 2 : 2 2 1 2 2 2 2 2 2 I L I L E = = because L 2 = L 1 . Express E 1 in terms of L 1 and I 1 : 1 2 1 1 2I L E = Chapter 20 1902 Express the ratio of E 2 to E 1 and simplify to obtain: 2 1 1 2 1 2 2 1 1 2 2 2 I I I L I L E E = = ⇒ 2 1 1 2 I I E E = Substituting for the ratio of I 1 to I 2 yields: ( ) T E E E Δ 2 1 1 1 2 1 2 α ω ω − = = 58 ••• Write a spreadsheet program to graph the average temperature of the surface of Earth as a function of emissivity, using the results of Problem 22. How much does the emissivity have to change in order for the average temperature to increase by 1 K? This result can be thought of as a model for the effect of increasing concentrations of greenhouse gases like methane and CO 2 in Earth’s atmosphere. Picture the Problem The amount of heat radiated by Earth must equal the solar flux from the Sun, or else the temperature on Earth would continually increase. The emissivity of Earth is related to the rate at which it radiates energy into space by the Stefan-Boltzmann law . 4 r AT e P σ = Using the Stefan-Boltzmann law, express the rate at which Earth radiates energy as a function of its emissivity e and temperature T: 4 r A'T e P σ = where A′ is the surface area of Earth. Use its definition to express the intensity of the radiation P a absorbed by Earth: A P I a = or AI P = a where A is the cross-sectional area of Earth. For 70% absorption of the Sun’s radiation incident on Earth: ( )AI P 70 . 0 a = Equate P r and P a and simplify: ( ) 4 70 . 0 A'T e AI σ = or ( ) ( ) 4 2 2 4 70 . 0 T R e I R σ π π = Solve for T to obtain: ( ) 4 1 4 4 70 . 0 − = = Ce e I T σ (1) Substitute numerical values for I and σ and simplify to obtain: ( )( ) ( ) ( ) 4 1 4 4 2 8 2 K 255 K W/m 10 670 . 5 4 W/m 1370 70 . 0 − − = ⋅ × = e e T Thermal Properties and Processes 1903 A spreadsheet program to evaluate T as a function of e is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content Algebraic Form B1 255 B4 0.4 e B5 B4+0.01 e + 0.1 C4 $B$1/(B4^0.25) ( ) 4 1 K 255 − e A B C D 1 T= 255 K 2 3 e T 4 0.40 321 5 0.41 319 6 0.42 317 7 0.43 315 23 0.59 291 24 0.60 290 25 0.61 289 26 0.62 287 A graph of T as a function of e follows. 285 290 295 300 305 310 315 320 325 0.40 0.45 0.50 0.55 0.60 e T ( K ) Treating e as a variable, differentiate equation (1) to obtain: de Ce de dT 4 5 4 1 − − = (2) Chapter 20 1904 Divide equation (2) by equation (1) to obtain: e de Ce de Ce T dT 4 1 4 1 4 1 4 5 − = − = − − Use a differential approximation to obtain: e e T T Δ − = Δ 4 1 ⇒ T T e e Δ − = Δ 4 Substitute numerical values (e ≈ 0.615 for T Earth = 288 K) and evaluate e e Δ : 014 . 0 K 288 K 1 4 − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − = Δ e e or about a 1.4% change. 59 ••• [SSM] A small pond has a layer of ice 1.00 cm thick floating on its surface. (a) If the air temperature is –10ºC on a day when there is a breeze, find the rate in centimeters per hour at which ice is added to the bottom of the layer. The density of ice is 0.917 g/cm 3 . (b) How long do you and your friends have to wait for a 20.0-cm layer to be built up so you can play hockey? Picture the Problem (a) We can differentiate the expression for the heat that must be removed from water in order to form ice to relate dQ/dt to the rate of ice build-up dm/dt. We can apply the thermal-current equation to express the rate at which heat is removed from the water to the temperature gradient and solve this equation for dm/dt. In Part (b) we can separate the variables in the differential equation relating dm/dt and ΔT and integrate to find out how long it takes for a 20.0-cm layer of ice to be built up. (a) Relate the heat that must be removed from the water to freeze it to its mass and heat of fusion: f mL Q = ⇒ dt dm L dt dQ f = Using the definition of density, relate the mass of the ice added to the bottom of the layer to its density and volume: Ax V m ρ ρ = = Differentiate with respect to time to express the rate of build-up of the ice: dt dx A dt dm ρ = Substitute for dt dm to obtain: dt dx A L dt dQ ρ f = Thermal Properties and Processes 1905 The thermal-current equation is: x T kA dt dQ Δ = Equate these expressions and solve for dt dx : x T kA dt dx A L Δ = ρ f ⇒ x T L k dt dx Δ = ρ f (1) Substitute numerical values and evaluate dt dx : ( ) ( ) ( )( ) cm/h 0.70 cm/h 0.697 h s 3600 s m 94 . 1 m 0.0100 kg/m 917 kg kJ 333.5 C 10 C 0 K m W 0.592 3 = = × = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ° − − ° ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = μ dt dx (b) Separating the variables in equation (1) gives: dt L T k xdx ρ f Δ = Integrate x from x i to x f and t′ from 0 to t: ' dt L T k xdx t x x ∫ ∫ Δ = 0 f f i ρ and ( ) t L T k x x f 2 i 2 f 2 1 ρ Δ = − ⇒ ( ) T k x x L t Δ − = 2 2 i 2 f f ρ Substitute numerical values and evaluate t: ( ) ( ) ( ) ( ) [ ] d 12 h 24 d 1 s 3600 h 1 s 10 03 . 1 m 0.010 m 0.200 C 10 C 0 K m W 0.592 2 kg kJ 333.5 m kg 917 6 2 2 3 = × × × = − ° − − ° ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ = t 60 ••• [SSM] A blackened copper cube that has 1.00-cm-long edges is heated to a temperature of 300ºC, and then is placed in a vacuum chamber whose walls are at a temperature of 0ºC. In the vacuum chamber, the cube cools radiatively. (a) Show that the (absolute) temperature T of the cube follows the differential equation: dT dt = − eσA C T 4 − T 0 4 ( ) where C is the heat capacity of the cube, A is its surface area, e the emissivity, and T o the temperature of the vacuum chamber. (b) Using Euler’s method (Section 5.4 of Chapter 5), numerically solve Chapter 20 1906 the differential equation to find T(t), and graph it. Assume e = 1.00. How long does it take the cube to cool to a temperature of 15ºC? Picture the Problem We can use the Stefan-Boltzmann equation and the definition of heat capacity to obtain the differential equation expressing the rate at which the temperature of the copper block decreases. We can then approximate the differential equation with a difference equation for the purpose of solving for the temperature of the block as a function of time using Euler’s method. (a) The rate at which heat is radiated away from the cube is given by: ( ) 4 0 4 T T A e dt dQ − = σ Using the definition of heat capacity, relate the thermal current to the rate at which the temperature of the cube is changing: dt dT C dt dQ − = Equate these expressions and solve for dt dT to obtain: ( ) 4 0 4 T T C A e dt dT − − = σ Approximating the differential equation by the difference equation gives: ( ) 4 0 4 T T C A e t T − − = Δ Δ σ Solving for ΔT yields: ( ) t T T C A e T Δ − − = Δ 4 0 4 σ or ( ) t T T C A e T T n n n Δ − − = + 4 0 4 1 σ (1) Use the definition of heat capacity to obtain: Vc mc C ρ = = Substitute numerical values (see Figure 13-1 for ρ Cu and Table 19-1 for c Cu ) and evaluate C: ( ) K J 45 . 3 K kg kJ 386 . 0 m 10 00 . 1 m kg 10 93 . 8 3 6 3 3 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⋅ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = − C Thermal Properties and Processes 1907 (b) A spreadsheet program to calculate T as a function of t using equation (1) is shown below. The formulas used to calculate the quantities in the columns are as follows: Cell Formula/Content Algebraic Form B1 5.67E−08 σ B2 6.00E−04 A B3 3.45 C B4 273 T 0 B5 10 Δt A9 A8+$B$5 t+Δt B9 B8-($B$1*$B$2/$B$3) *(B8^4−$B$4^4)*$B$5 ( ) t T T C A e T n n Δ − − 4 0 4 σ A B C 1 σ= 5.67 × 10 −8 W/m 2 ⋅K 4 2 A= 6.00 × 10 −4 m 2 3 C= 3.45 J/K 4 T 0 = 273 K 5 Δt= 10 s 6 7 t (s) T (K) 8 0 573.00 9 10 562.92 10 20 553.56 11 30 544.85 248 2400 288.22 249 2410 288.08 250 2420 287.95 251 2430 287.82 Chapter 20 1908 From the spreadsheet solution, the time to cool to 15°C (288 K) is about 2420 s or . min 5 . 40 A graph of T as a function of t follows. 270 320 370 420 470 520 570 0 500 1000 1500 2000 2500 t (s) T ( K )