Ch 17 HWDue: 11:59pm on Tuesday, September 8, 2015 To understand how points are awarded, read the Grading Policy for this assignment. Heat Radiated by a Person In this problem you will consider the balance of thermal energy radiated and absorbed by a person. Assume that the person is wearing only a skimpy bathing suit of negligible area. As a rough approximation, the area of a human body may be considered to be that of the sides of a cylinder of length L = 2.0 m and circumference C = 0.8 m. For the StefanBoltzmann constant use σ = 5.67 × 10 −8 W/m 2 /K 4 . Part A If the surface temperature of the skin is taken to be T body described in the introduction radiate? Take the emissivity to be e ∘ , how much thermal power P rb does the body = 30 C . = 0.6 Express the power radiated into the room by the body numerically, rounded to the nearest 10 W. Hint 1. Area of person Find the area A of the person, ignoring the ends of the cylinder that represents the person's area. Express the area in terms of L, C , and any constants. ANSWER: A = LC Hint 2. Symbolic expression Find P rb , the total power radiated into the room by the body of the person. Express the total power radiated in terms of the StefanBoltzmann constant σ, the body area A , the body temperature T body , and the emissivity e . ANSWER: P rb = eσAT body ANSWER: P rb = 460 W Correct Typesetting math: 38% 4 Part B The basal metabolism of a human adult is the total rate of energy production when a person is not performing significant physical activity. It has a value around 125 W, most of which is lost by heat conduction to the surrounding air and especially to the exhaled air that was warmed while inside the lungs. Given this energy production rate, it would seem impossible for a human body to radiate 460 W as you calculated in the previous part. Which of the following alternatives seems to best explain this conundrum? ANSWER: The human body is quite reflective in the infrared part of the spectrum (where it radiates) so e is in fact less than 0.1. The surrounding room is near the temperature of the body and radiates nearly the same power into the body. Correct The human body contains significant amounts of water and organic compounds. These typically have many absorption bands in the infrared part of the spectrum where room temperature objects radiate thermal energy. Consequently, human skin has a high emissivity in this part of the spectrum regardless of the emissivity of the skin in the visible part of the spectrum. Part C Now calculate P br , the thermal power absorbed by the person from the thermal radiation field in the room, which is assumed to be at T room = 20∘ C. If you do not understand the role played by the emissivities of room and person, be sure to open the hint on that topic. Express the thermal power numerically, giving your answer to the nearest 10 W. Hint 1. Role of emissivities The emissivity e is the complement of the reflectivity. The power incident on the person is partly absorbed and partly reflected. The fraction absorbed is proportional to the emissivity of the person and thus is less than or equal to the thermal power striking the person by the amount of this power that is reflected. Because the room is closed, no allowance has to be made for its emissivity. If the room has an emissivity of 0.7, then when you look at the wall (with an infrared viewer) 0.7 of what you see is the power radiated by that spot on the wall; the remaining 0.3 of what you see is what is reflected by the wall. But in a closed container like a room, this reflection is simply another part of the wall (since the room is closed). Being at the same temperature, this other part of the wall radiates power at the same rate as the spot you are looking at. When all the emissions and reflections are added together, the net effect is that the radiation from any part of the wall has the same total thermal power as if it were perfectly black. This accounts for the remarkable fact that the thermal radiation field in any closed cavity is independent of the material of the cavity and depends only on the temperature. Such radiation is called cavity radiation or blackbody radiation. The fact that the intensity at each color depended only on the temperature was so striking to physicists that the explanation of this phenomenom was assiduously studied until it unlocked the key to quantum mechanics and revealed a fundamental quantum of nature: Planck's constant h . A good example of this is a kiln. Just after a part is placed in a kiln, the heaters are much hotter than the cool part, and the part is visible through the peephole. As thermal equilibrium is reached everything glows a Hint 2. Area of person Find the area A of the person. becomming invisible. to the nearest 10 W. the net power radiated by the person when in a room with temperature T room Express the net radiated power numerically. ANSWER: P br = eσAT room 4 ANSWER: P br = 400 W Correct Part D Find P net . ANSWER: A = LC Hint 3. and the part disappears into the background color. and the emissivity e . How to set up the problem The net power radiated is simply the power radiated minus the power absorbed. the body area A . Symbolic expression Find P br . Express your answer in terms of the StefanBoltzmann constant σ. and any constants. ANSWER: P net = 60 W ∘ . the total power absorbed by the person. C . Hint 1. the room temperature T room .uniform red or orange. = 20 C . Express the area in terms of L. Under these circumstances a person would feel cool. ANSWER: T = 178 ∘ C Correct Part D K .Correct This net thermal power radiated is less than half of the basal metabolic rate. ANSWER: T = 260 ∘ F Correct Part C The temperature at the tops of the clouds in the atmosphere of Saturn (95 K ). Part A The midday temperature at the surface of the moon (400 K ). but not cold (if we neglect loss of heat by conduction). This explains why lightweight survival blankets have a shiny side: The low emissivity dramatically reduces radiative heat loss (and the air trapped between blanket and body reduces heat loss by conduction). the net radiative loss will become a significant loss mechanism for body heat.6 Convert the following Kelvin temperatures to the Celsius and Fahrenheit scales. if the surrounding temperature is much cooler than this. ANSWER: T = 127 ∘ C Correct Part B The midday temperature at the surface of the moon (400 K ). Exercise 17. However. Relation of Celsius and Kelvin temperature scales A temperature increase of one kelvin corresponds to a temperature increase of one degree also on the Celsius scale. the units of this scale are the same size as those on the Fahrenheit scale ( F ) rather than the Celsius ∘ scale ( C). The Kelvin temperature scale and the Celsius temperature scale differ only in their zero point. However. Give your answer to two significant figures. the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine (0 ∘ ∘ R). and that it boils at 100∘ C. which corresponds to 212∘ F .The temperature at the tops of the clouds in the atmosphere of Saturn (95 K ). which corresponds to 32∘ F . ANSWER: T = 2.55 × 107 K) . ANSWER: T = 289 ∘ F Correct Part E The temperature at the center of the sun (1.79×107 ∘ F Correct Rankine Temperature Scale Like the Kelvin scale. .55 × 107 K) . ANSWER: T = 1.55×107 ∘ C Correct Part F The temperature at the center of the sun (1. Part A Given that water at standard pressure freezes at 0∘ C. Hint 1. calculate the temperature difference ΔT in degrees Fahrenheit that corresponds to a temperature difference of 1 K on the Kelvin scale. and vapor) at 273. Triplepoint temperature On the Kelvin temperature scale. ANSWER: ΔV = 3. ANSWER: T triple = 492 ∘ R Correct Exercise 17.50 m 3 of ethanol when both the tank and the ethanol are at a temperature of 35. water freezes and coexists in three phases (solid. what additional volume of ethanol can be put into the tank? Express your answer using two significant figures. This temperature is known as the triple point.18 A steel tank is completely filled with 2.0 ∘ C .ANSWER: = 1. liquid.16 K at standard pressure.0 ∘ C .4×10−2 m 3 Correct Hot Rods .8 ∘ F ΔT = 1 K Correct Part B What is the numerical value of the triplepoint temperature T triple of water on the Rankine scale? Give your answer to three significant figures. Part A When the tank and its contents have cooled to 16. Hint 1. both of length L and having the same diameter. Correct The length of the combined rod remains the same. the stress on each rod must be the same. Since the rods have the same diameter. Therefore. Part B After the rods have been heated. A tensile stress arises that is different for the two rods. which of the following statements is true? Choose the best answer. Correct Stress is a force per unit area. but because the rods have different expansion coefficients. the force on rod A due to rod B is the same as that on rod B due to rod A.Two circular rods. In other words. A compressive stress arises that is different for the two rods. The temperature of the rods is now raised by ΔT . the lengths of the individual rods change. which of the following statements is true? Choose the best answer. The coefficient of linear expansion and Young's modulus for rod A are α A and Y A respectively; those for rod B are α B and Y B respectively. The length of each rod changes but the combined length of the rods is still 2L. ANSWER: The length of each rod is still L. A compressive stress arises that is the same for both rods. are placed end to end between rigid supports with no initial stress in the rods. Part C What is the stress F /A in the rods after heating? Y Y ΔT . A compressive stress arises in one rod and a tensile stress arises in the other rod. ΔLA + ΔLB = 0 even though ΔLA ≠ 0 and ΔLB ≠ 0 . their crosssectional area is the same. Both rods are "normal" materials with α > 0. Part A After the rods have been heated. A tensile stress arises that is the same for both rods. By Newton's 3rd law. ANSWER: The stress in each rod remains zero. launch the video below. and ΔT . The corresponding formulas are ΔLthermal = αLΔT and ΔL compressive stress The sum gives the net change in length for each rod. Hint 2. close the video window and answer the question on the right. Hint 1. ANSWER: F A = −(α A +α B )ΔT 1 Y A + 1 Y B Correct Another way of thinking about this is that the combination of rods has a net thermal expansion coefficient and a net Young's modulus given by α = α1 + α2 Y = Y1Y2 Y 1 +Y 2 . Then. Y B . α B . . Find the change in length of each rod as a function of the stress on the rods. Video Tutor: Heating Water and Aluminum First. then set the total change in length to zero and solve for the stress.Express the stress in terms of α A . Change in length for each rod The change in length for each rod is due to both thermal as well as compressive stresses. Y A . =L F AY . How to approach the problem You know that the length of the two rods remains equal to 2L during the heating process. You can watch the video again at any point. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Part A Suppose that we replace the aluminum with a mystery metal and repeat the experiment in the video. Room temperature is about 20∘ C before the start of the experiment. the same. ANSWER: half as great. where m is the mass of the substance and c is its heat capacity. Part A If no heat is lost to the surroundings.) Exercise 17. How to approach the problem Recall that the heat Q delivered to a substance can be written Q = mc ΔT .55 kg and containing 1. As in the video. how much heat must be added to raise the temperature from 21. the temperature of the metal increased three times as much as the temperature of the water.0 ∘ C to 80. and the mystery metal heats up to 80∘ C. The water heats up to 40∘ C. (Recall that the heat Q delivered to a substance can be written Q = mc ΔT . the metal has onethird the heat capacity of water. Compared to that of water. where m is the mass of the substance and c is its heat capacity. three times greater.0 ∘ C . the heat capacity of our mystery metal is Hint 1.27 An aluminum tea kettle with mass 1. the mass of the metal is the same as that of the water. Therefore. Correct Given the same input of energy. onethird as great.70 kg of water is placed on a stove. two times greater. Power due to the torque If you've found the torque. Hint 1.03×105 J Answer Requested Thermal Energy from Friction on a Rope A capstan is a rotating drum or cylinder over which a rope or cord slides to provide a great amplification of the rope's tension while keeping both ends free .) Hint 2. Torque applied to the rope In this case. Since the added tension in the rope is due to friction. rounded to the nearest 10 W. with \texttip{t_{\rm 0}}{t_0} as the time for one revolution of the capstan. (Finding a numerical value for this torque can certainly be done as an intermediate calculation. find the rate \texttip{P_{\rm thermal}}{P_thermal} at which thermal energy is being generated. Part A If the difference in tension (T 0 − T ′ ) between the two ends of the rope is 520 N and the capstan has a diameter of 10. the power is given by P=\tau\omega. we can still say that the net torque is the product of the difference between the tensions (T_0T') and the radius. the capstan generates thermal energy. either symbolically or numerically. in watts.) However. the contact force between the capstan and the rope is not localized at a single point but is distributed along the rope. but it doesn't help much in reaching the answer. Give a numerical answer.∘ C ? ANSWER: Q = 5. Power from force . (This is more realistic than assuming that forces act at a single point.0 cm and turns once in 0. where \omega=2\pi/t_0. Hint 3.90 \;{\rm s}. Use the chain rule In Part A. you found the rate at which heat is added to the capstan. but once this difference has been achieved. you've found that P_{\rm thermal}= Fv=\left(T_0T'\right){2\pi r\over t_0}. The net force on the rope is T_0T' and the relative speed is the circumference of the capstan divided by the period of the capstan's rotation. and of course we can. only on the difference in tensions! A larger number of turns might well create a larger tension difference.064 ^\circ {\rm C/s} Correct . and this rate of temperature rise will be proportional to \texttip{P_{\rm thermal}}{P_thermal}. ANSWER: {d\Theta\over dt} = 0. Part B If the capstan is made of iron (with a specific heat capacity C_{\rm iron}=470\;{\rm J/(kg}\cdot ^\circ{\rm C}) and has a mass of 6. Hint 1. P_{\rm thermal}={dQ\over dt}={dQ\over d\Theta}{d\Theta\over dt}=(mC){d\Theta\over dt}. If you've done this problem symbolically using forces. rounded to two significant figures. This added heat will of course cause a temperature rise. Note that \texttip{\Theta }{Theta} is a temperature.We should be able to find the power as the product of a force and a relative speed. Give a numerical answer. so that {d\Theta\over dt}={P_{\rm thermal}\over mC}. you've found that P_{\rm thermal}=\tau\omega=(Fr)\left({2\pi\over t_0}\right)={2\pi r\left(T_0T'\right)\over t_0}\;. Hint 4. in degrees Celsius per second. the work done per rotation of the capstan will not depend on the number of turns.00 \;{\rm kg}. How both methods work equally well If you've done this problem symbolically using torques. at what rate d\Theta/ dt does its temperature rise? Assume that the temperature in the capstan is uniform and that all the thermal energy generated flows into it. ANSWER: \texttip{P_{\rm thermal}}{P_thermal} = 180 W Correct The answer does not depend on the number of turns. Symbolically. The fact that these expressions are the same should not be surprising. the proportionality being the reciprocal of the heat capacity. where \texttip{C}{C} is the specific heat capacity. launch the video below. If you make the rubber skin of a water balloon thicker. Use a thicker balloon. the water balloon did not pop because most of the candle's thermal energy went into heating the water. the less the substance's temperature will rise in response to a given input of heat. How to approach the problem. Part A Which of the following changes would make the water balloon more likely to pop? (Ignore effects of convection within the fluid.) Contact with this relatively cool water kept the balloon's thin rubber skin from getting hot enough to melt. Then. Recall that in the video. You will be asked to use your knowledge of physics to predict the outcome of an experiment. close the video window and answer the question at right.Video Tutor: Water Balloon Held over Candle Flame First. Rubber conducts heat more slowly than water. Hint 1. (Recall that the higher a substance's heat capacity. You can watch the video again at any point. .) Select all that apply. ANSWER: Use a liquid that has a lower heat capacity than water. Use a liquid that has a higher heat capacity than water. Let's think about the process in more detail. will the skin be more likely or less likely to melt? Think about the conduction of heat across the skin between the hot side facing the candle and the cooler side in contact with water. Use a thinner balloon. The high heat capacity of water means that the water in the balloon warmed only slowly. which of the following statements is true? ANSWER: Steam burns the skin worse than hot water because the thermal conductivity of steam is much higher than that of liquid water. 34^\circ{\rm C}.0 {\rm ^\circ C} . are cooled down to skin temperature. Part A Find the final temperature of the system.Correct Using a liquid with a lower heat capacity than water means that the candle flame will raise the liquid's temperature more quickly than for water. assuming no heat loss to the surroundings. initially at 100^\circ{\rm C}. Hot water and steam both burn skin about equally badly. Assume that the steam condenses extremely fast. .41 A copper pot with a mass of 0. HotWater Burns Just about everyone at one time or another has been burned by hot water or steam. Hot water burns the skin worse than steam because the thermal conductivity of hot water is much higher than that of steam. This problem compares the heat input to your skin from steam as opposed to hot water at the same temperature. when they come in contact with your skin. Steam burns the skin worse than hot water because the latent heat of vaporization is released as well. so that the outer part of the balloon wall will heat up more quickly to its melting point.0 {\rm ^\circ C} is dropped into the pot. Part A Under these conditions.170 {\rm {\rm kg}} of water. Exercise 17. A thicker balloon will conduct heat through to the liquid more slowly.500 {\rm {\rm kg}} contains 0.250 {\rm {\rm kg}} block of iron at 85. We will further assume a constant specific heat capacity c = 4190\; {\rm J/(kg \cdot K)} for both liquid water and steam. A 0.5 {\rm ^\circ C} Correct Steam vs. ANSWER: T = 27. and both are at a temperature of 20. Assume that water and steam. and the temperature difference \texttip{\Delta T}{DeltaT} between the initial temperature of the water and the skin temperature. Express the heat transferred. Actually the local temperature in the area where the steam condenses can be raised quite significantly. the latent heat of vaporization \texttip{L}{L}. Express your answer in terms of \texttip{c}{c}. Determine the heat transferred from steam to skin Find an expression for the heat \texttip{H_{\rm 1}}{H_1} transferred from the vapor to the skin. Hint 1.0 {\rm g} of steam onto the skin? The latent heat of vaporization for steam is L = 2. Determine the heat transferred from water to skin Find an expression for the heat \texttip{H_{\rm 2}}{H_2} transferred from the hot water to the skin.0 {\rm g} of water onto the skin? To compare this to the result in the previous part. and the temperature difference \texttip{\Delta T}{DeltaT} between the initial temperature of the steam and the skin temperature. Part C How much heat \texttip{H_{\rm 2}}{H_2} is transferred by 25. Express the heat transferred. to three significant figures. Hint 1. in joules.33×104 {\rm J} Correct Here we assumed that the skin continues to remain at 34^{\circ}{\rm C}. ANSWER: .Correct The key point is that the latent heat of vaporization has to be taken into account for the steam. Part B How much heat \texttip{H_{\rm 1}}{H_1} is transferred to the skin by 25. Express your answer in terms of \texttip{c}{c}. the mass of water \texttip{m}{m}. to three significant figures.256 \times 10^6 \; {\rm J/kg}. the mass of steam \texttip{m}{m}. in joules. ANSWER: \texttip{H_{\rm 1}}{H_1} = m \left(c {\Delta{T}}+L\right) ANSWER: \texttip{H_{\rm 1}}{H_1} = 6. continue to assume that the skin temperature does not change. what is the temperature reading \texttip{T_{\rm 2}}{T_2} of the outlet thermometer? Express the outlet temperature in terms of \texttip{T_{\rm 1}}{T_1}. Water is flowing at the rate \texttip{F}{F}.\texttip{H_{\rm 2}}{H_2} = m c {\Delta{T}} ANSWER: \texttip{H_{\rm 2}}{H_2} = 6910 {\rm J} Correct The amount of heat transferred to your skin is almost 10 times greater when you are burned by steam versus hot water. For these reasons.. \texttip{C}{C}. steam burns are often far more severe than hotwater burns. the voltmeter reads \texttip{V}{V}. One trial design is shown in the figure. the inlet thermometer registers \texttip{T_{\rm 1}}{T_1}. Then the power (i. \texttip{F}{F}. The temperature of steam can also potentially be much greater than 100^\circ{\rm C}. Hint 1.e. Combine these two calculations to determine the increase in temperature of the water as it flows through the heater. Then use the definition of heat capacity to compute how much the temperature of a unit mass of water will increase for a given quantity of heat. Assume that the heat capacity of water is \texttip{C}{C} and that the heat capacity of the heater apparatus is \texttip{C_{\rm h}}{C_h}. . and the ammeter reads current \texttip{I}{I}. the heat generated per unit time by the heating element) is VI. An Electric Water Heater An engineer is developing an electric water heater to provide a continuous ("on demand") supply of hot water. and any other given quantities. How to approach the problem Determine the amount of heat input per unit mass of water that flows through the heater. Part A When a steady state is finally reached. Hint 2. ANSWER: . ANSWER: \texttip{\Delta T}{DeltaT} = \large{\frac{Q}{m} \frac{1}{C}} ANSWER: \texttip{T_{\rm 2}}{T_2} = \large{\frac{VI}{FC}+T_{1}} Correct Part B Why is it unnecessary to take into account the heat capacity of the apparatus itself? Choose the best explanation. ANSWER: Q/m = \large{\frac{V I}{F}} Hint 3.Hint 2. then \large{\frac{P}{F}=\frac{\frac{\mbox{heat}}{\mbox{time}}}{\frac{\mbox{mass}} {\mbox{time}}}=\frac{\mbox{heat}}{\mbox{mass}}}. Hint 1. and \texttip{C}{C}. Find the change in water temperature If an amount of heat \texttip{Q}{Q} is added uniformly to a substance of mass \texttip{m}{m} and heat capacity \texttip{C}{C}. by how much will the temperature of that substance increase? Express the temperature increase \texttip{\Delta T}{DeltaT} in terms of \texttip{Q}{Q}. Electrical power The electrical power disspated in a device is given by the product of the voltage and the current: P = V I. What relationships are involved The power \texttip{P}{P} dissipated in the resistor is equal to the energy dissipated per unit time. The flow rate \texttip{F}{F} is equal to the mass of the water that flows through the heater per unit time. Thus. Find heat input per unit mass of water How much heat \texttip{Q}{Q} is delivered by the resistor per unit mass \texttip{m}{m} of water flowing through the heater? Express the heat per unit mass in terms of quanitities given in the problem introduction. if all of the energy dissipated by the resistor goes into heating the water. \texttip{m}{m}. the voltmeter reads V = 120\; {\rm V}. ANSWER: \texttip{T_{\rm 2}}{T_2} = 69 ^\circC Correct Part E . The heat capacity of any of the materials used in a water heater is much smaller than that of water. Part D Calculate the temperature \texttip{T_{\rm 2}}{T_2} of the water leaving the heater.In steady state the temperature of the apparatus doesn't change. Hint 1. Electrical power The electrical power disspated in a device is given by the product of the voltage and the current: P = V I.0 \;{\rm A}. but reasonable. Since the actual heating unit is immersed in the flow all the heat goes directly into the water. Express your answer numerically. Hence its heat capacity is irrelevant.500\; {\rm kg/min}. ANSWER: \texttip{P}{P} = 1800 W Correct This is a large. to the nearest integer. Correct Imagine that the input temperature of the water is T_1 = 18^\circ {\rm C}. in watts. power requirement for a household appliance. the ammeter reads I = 15. in degrees Celsius. whereas the power is measured in watts (joules/second). Part C What is the power \texttip{P}{P} at which the heater operates? Express your answer numerically. The heat capacity of water C = 4200\; {\rm J/(kg \cdot K)}. and the flow rate is F= 0. Note that \texttip{F}{F} is defined as the number of kilograms of water flowing through the heater per minute. Exercise 17.5 {\rm {\rm ^\circ C}} . rather than using electricity. Most homes in the United States are wired for a maximum of 200 A for the entire house.0 {\rm {\rm ^\circ C}} . If the heater is to operate on a US standard V = 120 \;{\rm V} wall plug.00^\circ {\rm C} and the output temperature should be at least T_2=40. and only 20 A on any particular outlet.71 A picture window has dimensions of 1.0^\circ {\rm C} for a moderately warm shower. ANSWER: H = 1. For the electric heater to meet the modest design requirements given. to the nearest integer. in amperes. the outside temperature is 17. it would consume almost 20 kW of power! For this reason.60 {\rm {\rm mm}} thick. most "on demand" water heaters generate heat by burning a fuel such as propane. The input water temperature (during the winter) can be as low as T_1=5. ANSWER: . Assume a conservative flow rate of F=8. On a winter day. Part A At what rate is heat being lost through the window by conduction? Express your answer using three significant figures. how much current \texttip{I}{I} would it need to draw to meet the design requirements? Express the current numerically.40 {\rm m} \times2.750 {\rm mm}thick layer of paper (thermal conductivity 0.5 gallons per minute). ANSWER: \texttip{I}{I} = 163 A Correct 163 A is a tremendous amount of current for a home appliance.00\; {\rm kg/min} (corresponding to about 2. while the inside temperature is a comfortable 20.Consider using this heater to generate "on demand" hot water for a home shower.88×104 {\rm W} Correct Part B At what rate would heat be lost through the window if you covered it with a 0.0500 {\rm W/m \cdot K})? Express your answer using three significant figures.50 {\rm m} and is made of glass 5. What is the relationship between the heat current in rod 1 ( \texttip{H_{\rm 1}}{H_1}) to the heat currents in rods 2 and 3 ( \texttip{H_{\rm 2}}{H_2} and \texttip{H_{\rm 3}}{H_3} respectively)? Express \texttip{H_{\rm 1}}{H_1} in terms of \texttip{H_{\rm 2}}{H_2} and \texttip{H_{\rm 3}}{H_3}. breaking into two flows at the junction point. Similarly. The crosssectional area of each rod is \texttip{A} {A}. Use an analogy to electric currents If you are familiar with electrical current. ANSWER: . Hint 1. Hint 1. the algebraic sum of all heat currents into a junction is zero (any heat current leaving the junction must be subtracted from the total). Find a relationship among the heat currents Heat is flowing from the free end of rod 1 to the free ends of rods 2 and 3. which states that the algebraic sum of the currents into any junction is zero (any current leaving the junction must be subtracted from the total). You may assume that there is no heat loss from the surfaces of the rods. and they have length \texttip{L}{L} and thermal conductivity \texttip{k}{k}.H = 5970 {\rm W} Correct Heat Flow through Three Rods Three identical rods are welded together to form a Yshaped figure. The free end of rod 1 is maintained at \texttip{T_{\rm 1}}{T_1} and the free ends of rods 2 and 3 are maintained at a lower temperature \texttip{T_{\rm 0}}{T_0}. the temperature of the junction point? Express your answer in terms of \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}. you can solve this problem by using the analog of Kirchoff's junction rule. Part A What is \texttip{T_{\rm j}}{T_j}. in steady state. Also. ANSWER: \texttip{H}{H} = \large{\frac{k A {\Delta{T}}}{L}} Hint 2.\texttip{H_{\rm 1}}{H_1} = H_{2} + H_{3} Hint 2. Find \texttip{H_{\rm 1}}{H_1} in terms of the junction temperature Now. you can substitute these into the formula H_1=H_2+H_3 and solve algebraically for \texttip{T_{\rm j}}{T_j}. ANSWER: \texttip{H_{\rm 1}}{H_1} = \large{\frac{k A \left(T_{1}T_{j}\right)}{L}} ANSWER: \texttip{T_{\rm j}}{T_j} = \large{\frac{T_{1}+2T_{0}}{3}} Correct This is a weighted average of the temperatures at the two ends of the system and so always lies between them as you would expect. and other quantities given in the problem introduction. \texttip{A}{A}. Determine the general formula for heat current Given the difference in temperature between the ends of the rod ( \texttip{\Delta T}{DeltaT}). This is because the heat current in rods 2 and 3 separately must be less than the heat current in rod 1 (since H_2 + H_3 = H_1). it is closer to \texttip{T_{\rm 0}}{T_0} than to \texttip{T_{\rm 1}}{T_1}. the rod's thermal conductivity \texttip{k}{k}. and the temperatures \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}. Hint 1. \texttip{A}{A}. Find the temperature difference What is the temperature difference \texttip{\Delta T}{DeltaT} between the ends of rod 1? . what is the general formula for the heat current \texttip{H}{H} in the rod? Express your answer in terms of \texttip{k}{k}. and \texttip{\Delta T} {DeltaT} (typed as DeltaT). Part B What is the heat current \texttip{H_{\rm 1}}{H_1} in rod 1? Express the heat current in terms of any or all of \texttip{k}{k}. Derive a general expression for the heat current through rod 1 in terms of the temperature of the junction. and its length \texttip{L}{L} and crosssectional area \texttip{A}{A}. if you can find expressions for the heat currents in the rods in terms of their physical properties and temperature difference across them. \texttip{L}{L}. \texttip{T_{\rm j}}{T_j}. \texttip{L}{L}. ANSWER: H_2 = H_3 = \large{kA{\frac{T_{1}T_{0}}{3L}}} Correct Problem 17. you must increase the crosssectional area of the remaining two rods by a factor of 1. Find this heat current. use the relationship you found in Part A of this problem. ANSWER: \texttip{\Delta T}{DeltaT} = \large{\frac{2}{3} \left(T_{1}T_{0}\right)} ANSWER: \texttip{H_{\rm 1}}{H_1} = \large{{\frac{2}{3}}{\frac{kA}{L}}\left(T_{1}T_{0}\right)} Correct This is greater than the heat current due to only 2 rods. H_2 = H_3. Use your answer to Part A The temperature difference between the ends of rod 1 is \Delta T = T_1 T_{\rm j}.33.Express your answer in terms of \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}. Hint 1. Express your answer in terms of \texttip{k}{k}. by symmetry. How to approach this problem Recall that H_1 = H_2 + H_3. Also. This information is sufficient to find \texttip{H_{\rm 2}}{H_2} and \texttip{H_{\rm 3}}{H_3}. Your answer should be positive.92 . Hint 1. Part C By symmetry. \texttip{L}{L}. You would expect this to be the case since the effective surface area is increased by having a third rod. \texttip{A}{A}. To express this in terms of \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}. You can check that if you wanted to remove the third rod while keeping the same heat current. and the temperatures \texttip{T_{\rm 1}}{T_1} and \texttip{T_{\rm 0}}{T_0}. the heat current in rods 2 and 3 will be the same. 5 {\rm {\rm W}} ? ANSWER: T = 29. but we are interested in hibernation. ANSWER: T_{\rm final} = 100 {\rm ^\circ C} Correct Part B What is the final mass of the iron and the remaining water? Express your answer using three significant figures. Assume the surface area of each layer is constant and given by the surface area of the spherical model constructed for the black bear. Part A What is the final temperature of the water? Express your answer using three significant figures.90 {\rm ^\circ C} during hibernation. the thickness varies with the season.9 {\rm ^\circ C}.0 {\rm g} of water at 20.60 {\rm {\rm m}} in diameter having a layer of fat 4.6 {\rm ^\circ C} . (Actually. Assume no heat exchange with the surroundings.104 Animals in cold climates often depend on two layers of insulation: a layer of body fat [of thermal conductivity 0.) In studies of bear hibernation. We can model a black bear (Ursus americanus) as a sphere 1.0 {\rm g} slug of redhot iron (temperature 745 {\rm ^\circ C}) by dropping it into an insulated cup of negligible mass containing 75.0 {\rm ^\circ C}. ANSWER: m_{\rm final} = 181 {\rm g} Correct Problem 17.00 {\rm {\rm cm}} thick. it was found that the outer surface layer of the fur is at 2.You cool a 110. when the fat layer is thickest.200 {\rm W/(m \cdot K)} ] surrounded by a layer of air trapped inside fur or down. Part A What should the temperature at the fatinner fur boundary be so that the bear loses heat at a rate of 51. while the inner surface of the fat layer is at 30. 0 {\rm cm} Correct Score Summary: Your score on this assignment is 101%. Your instructor adjusted your score by awarding 1 additional point.22 out of a possible total of 16 points. . Part B How thick should the air layer (contained within the fur) be so that the bear loses heat at a rate of 51. only rounding as a final step before submitting your answer. use your full precision answer.Correct Should you be required to use your answer to this part in a subsequent part. This adjustment is reflected in your score above. You received 15.5 {\rm {\rm W}} ? ANSWER: L_{\rm air} = 10.