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OpenStax College Physics Instructor Solutions Manual Chapter 14 CHAPTER 14: HEAT AND HEAT TRANSFER METHODS 14.2 TEMPERATURE CHANGE AND HEAT CAPACITY 1. Solution On a hot day, the temperature of an 80,000-L swimming pool increases by 1.50C . What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation. 1 m3 m V (1.00 10 kg/m )(80,000 L) 8.00 10 4 kg . 1000 L 3 3 Therefore, Q mcT (8.00 104 kg)(4186 J/kg C)(1.50C) 5.02 108 J 2. Show that 1 cal/g C 1 kcal/kg C . Solution 1 cal 1 kcal 1000 g 1 kcal/kg C g C 1000 cal 1 kg 3. To sterilize a 50.0-g glass baby bottle, we must raise its temperature from 22.0C to 95.0C . How much heat transfer is required? Solution Q mcT (50.0 10 -3 kg)( 840 J/kg C)(73.0C) 3066 J 3.07 10 3 J 4. The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at 20.0C : (a) water; (b) concrete; (c) steel; and (d) mercury. Solution Q mcT T (a) T Q mc 1.00 kcal 1.00 C T 20.0 C 1.00 C 21.0 C (1.00 kg)(1.00 kcal/kg C) OpenStax College Physics 5. Instructor Solutions Manual Chapter 14 (b) T 1.00 kcal 5.0C T 20.0C 5.0C 25.0C (1.00 kg)(0.20 kcal/kg C) (c) T 1.00 kcal 9.26C T 20.0C 9.26C 29.3C (1.00 kg)(0.108 kcal/kg C) (d) T 1.00 kcal 30.0C T 20.0C 30.0C 50.0C (1.00 kg)(0.0333 kcal/kg C) Rubbing your hands together warms them by converting work into thermal energy. If a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm per rub, and with an average frictional force of 40.0 N, what is the temperature increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and fingers. Solution Let N be the number of hand rubs and F be the average frictional force of a hand NFd 20(40.0 N)(7.50 10 2 m) rub: Q NFd mcT T 0.171C mc (0.100 kg)(3500 J/kg C) 6. A 0.250-kg block of a pure material is heated from 20.0C to 65.0C by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed. Solution Q mcT c Q 1.04 kcal 0.0924 kcal/kg C mT (0.250 kg)( 45.0C) It is copper. 7. Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water? Solution mw cw T Q mc cc T mc cw 1 kcal/kg C 10.8 mw cc 0.0924 kcal/kg C OpenStax College Physics 8. Instructor Solutions Manual Chapter 14 (a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00-g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a 54.9C temperature increase? (b) Compare your answer to labeling information found on a package of peanuts and comment on whether the values are consistent. Solution (a) Q mw cw T mAlcAl T mw cw mAlcAl T (0.500 kg)(1.00 kcal/kg C) Q (54.9C) 28.63 kcal (0.100 kg)( 0.215 kcal/kg C) Q 28.63 kcal 5.73 kcal/g mp 5.00 g (b) A label for unsalted dry roasted peanuts says that 33 g contains 200 calories (kcal), Q 200 kcal which is 6 kcal/g , which is consistent with our results to part (a), mp 33 g to one significant figure. 9. Solution Following vigorous exercise, the body temperature of an 80.0-kg person is 40.0C . At what rate in watts must the person transfer thermal energy to reduce the body temperature to 37.0C in 30.0 min, assuming the body continues to produce energy at the rate of 150 W? 1 watt = 1 joule/seco nd or 1 W = 1 J/s Q mchuman bodyT (80.0 kg)(3500 J/kg C)(40C - 37C) 8.40 10 5 J Pcooling Q 8.40 10 5 J 4.67 10 2 W t (30 min)(60 s/1 min) Thus, Prequired Pcooling Pbody 467 W 150 W 617 W . OpenStax College Physics Instructor Solutions Manual Chapter 14 10. Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails (1 watt = 1 joule/seco nd or 1 W = 1 J/s and 1 MW = 1 megawatt) . (a) Calculate the rate of temperature increase in degrees Celsius per second (C/s ) if the mass of the reactor core is 1.60 105 kg and it has an average specific heat of 0.3349 kJ/kg C . (b) How long would it take to obtain a temperature increase of 2000C , which could cause some metals holding the radioactive materials to melt? (The initial rate of temperature increase would be greater than that calculated here because the heat transfer is concentrated in a smaller mass. Later, however, the temperature increase would slow down because the 5 105 - kg steel containment vessel would also begin to heat up.) Solution (a) Q mcT T Q mc Recall that 1 W = 1 J/s. Thus T for 1 s is given by T (b) t (150 10 6 J)(1 kcal/4186J ) 2.80C Rate 2.80C/s (1.60 10 5 kg)(0.0800 kcal/kg C) 2000C 714 s 11.9 min 2.80C/s 14.3 PHASE CHANGE AND LATENT HEAT 11. How much heat transfer (in kilocalories) is required to thaw a 0.450-kg package of frozen vegetables originally at 0C if their heat of fusion is the same as that of water? Solution Q mLf (0.450 kg)(79.8 kcal/kg) 35.9 kcal 12. A bag containing 0C ice is much more effective in absorbing energy than one containing the same amount of 0C water. (a) How much heat transfer is necessary to raise the temperature of 0.800 kg of water from 0C to 30.0C ? (b) How much heat transfer is required to first melt 0.800 kg of 0C ice and then raise its temperature? (c) Explain how your answer supports the contention that the ice is more effective. OpenStax College Physics Solution Instructor Solutions Manual Chapter 14 (a) Q mcT (0.800 kg)(4186 J/kg C)(30.0C) 1.00 10 5 J (b) Q mLf mcT (0.800 kg)(334 103 J/kg) 1.005 105 J 3.68 105 J (c) The ice is much more effective in absorbing heat because it first must be melted, which requires a lot of energy, then it gains the same amount of heat as the bag that started with water. The first 2.67 10 5 J of heat is used to melt the ice, then it absorbs the 1.00 10 5 J of heat as water. 13. (a) How much heat transfer is required to raise the temperature of a 0.750-kg aluminum pot containing 2.50 kg of water from 30.0C to the boiling point and then boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is 500 W [ 1 watt = 1 joule/seco nd (1 W = 1 J/s) ]? Solution ' (a) Q mw cw T mAlcAl T mw Lv Q (2.50 kg)(1.00 kcal/kg C)(70.0C) (0.750 kg)(0.215 kcal/kg C)(70.0C) (0.750 kg)(539 kcal/kg) Q 590.5 kcal 591 kcal Q , where P power and t time . P 4186 J/kcal 3 t (590.5 kcal) 4.94 10 s 500 W (b) Q Pt t 14. Solution The formation of condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.00 g of condensation forms on a glass containing both water and 200 g of ice, how many grams of the ice will melt as a result? Assume no other heat transfer occurs. mw Lv mice Lf mice mw Lv 580 kcal/kg 58.1 g (8.00 g) Lf 79.8 kcal/kg (Note that Lv for water at 37C is used here as a better approximation than Lv for 100C water.) OpenStax College Physics Instructor Solutions Manual Chapter 14 15. On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at 0C and completely melts to 0C water in exactly one day [ 1 watt = 1 joule/seco nd (1 W = 1 J/s) ]? Solution P 16. On a certain dry sunny day, a swimming pool’s temperature would rise by 1.50C if not for evaporation. What fraction of the water must evaporate to carry away precisely enough energy to keep the temperature constant? Solution Let M be the mass of pool water and m be the mass of pool water that evaporates. (3.50 kg)(334 kJ/kg C) Q mLf 13.53 W 13.5 W t t 86400 s McT mLv(37C) m cT (1.00 kcal/kg C)(1.50C) 2.59 10 3 M Lv(37C) 580 kcal/kg (Note that Lv for water at 37C is used here as a better approximation than Lv for 100C water.) 17. (a) How much heat transfer is necessary to raise the temperature of a 0.200-kg piece of ice from 20.0C to 130C , including the energy needed for phase changes? (b) How much time is required for each stage, assuming a constant 20.0 kJ/s rate of heat transfer? (c) Make a graph of temperature versus time for this process. Solution (a) (i) Heat needed to warm ice to 0C Q1 mi ci T (0.200 kg)(2.090 kJ/kg C)(20C) 8.36 kJ (ii) Heat needed to melt ice at 0C Q2 mi Lf (0.200 kg)(334 kJ/kg) 66.8 kJ (iii) Heat required to warm 0C water to 100C Q3 mi cw T (0.200 kg)(4.186 kJ/kg o C)(100 o C) 83.73 kJ 83.7 kJ (iv) Heat required to vaporize water at 100C Q4 mi Lv (0.200 kg)(2256 kJ/kg) 451.2 kJ 451 kJ (v) Heat required to warm 100C vapor to 130C OpenStax College Physics Instructor Solutions Manual Chapter 14 Q5 mi c v T (0.200 kg)(1.520 kJ/kg C)(30C) 9.12 kJ Total heat required Q Q1 Q2 Q3 Q4 Q5 619.0 kJ 147.9 kcal 148 kcal . (b) P Q Q t t P (i) t1 Q1 8.36 kJ 0.418 s P 20 kJ/s (ii) t 2 Q2 66.8 kJ 3.34 s P 20 kJ/s (iii) t 3 Q3 83.7 kJ 4.185 s 4.19 s P 20 kJ/s (iv) t 4 Q4 451 kJ 22.6 s P 20 kJ/s (v) t5 Q5 9.12 kJ 0.456 s P 20 kJ/s Total time t t1 t 2 t 3 t 4 t 5 31.00 s 31 s (c) 18. In 1986, a gargantuan iceberg broke away from the Ross Ice Shelf in Antarctica. It was approximately a rectangle 160 km long, 40.0 km wide, and 250 m thick. (a) What is the mass of this iceberg, given that the density of ice is 917 kg/m 3 ? (b) How much heat transfer (in joules) is needed to melt it? (c) How many years would it take sunlight alone to melt ice this thick, if the ice absorbs an average of 100 W/m 2 , 12.00 h per day? OpenStax College Physics Solution Instructor Solutions Manual Chapter 14 3 3 3 (a) m V lwh (917 kg/m )(160 10 m)(40.0 10 m)(250 m) 1.467 1015 kg 1.47 1015 kg (b) Q mLf (1.467 1015 kg)(79.8 kcal/kg)(4 186 J/kcal) 4.90 1020 J 3600 s 16 (c) Qday (100 W/m 2 )(160 103 m)(40.0 103 m)(12 h) 2.765 10 J 1h 17 Q (1.171 10 kcal )(4186 J/kcal) 1y n 1.773 10 4 d 48.5 y 16 Qday 2.765 10 J 365.25 d 19. How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to cool the coffee from 95.0C to 45.0C ? You may assume the coffee has the same thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560 cal/g). (You may neglect the change in mass of the coffee as it cools, which will give you an answer that is slightly larger than correct.) Solution The heat gained in evaporating the coffee equals the heat leaving the coffee and glass to lower its temperature, so that MLv mc cc ΔT mg cg ΔT , where M is the mass of coffee that evaporates. Solving for the evaporated coffee gives: M T (mc cc mg cg ) Lv (95.0C 45.0C) (350 g)(1.00 cal/g C) (100 g)(0.20 cal/g C) 33.0 g 560 cal/g 20. (a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 10 7 J of energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water has its temperature raised from 20.0C to 100C , it boils, and the resulting steam is raised to 300 C . (b) Discuss additional complications caused by the fact that crude oil has a smaller density than water. Solution (a) Q mcw Tw mLv mcs Ts OpenStax College Physics m Instructor Solutions Manual Chapter 14 Q c w Tw Lv cs Ts 2.80 10 7 J 9.67 kg (4186 J/kg C)(80.0C) 2256 10 3 J/kg (1520 J/kg C)(200C) V 9.67 kg 1 m3 1L 3 3 9.67 L 3 1.00 10 kg 10 m (b) Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less able to absorb the heat generated by the oil. 21. The energy released from condensation in thunderstorms can be very large. Calculate the energy released into the atmosphere for a small storm of radius 1 km, assuming that 1.0 cm of rain is precipitated uniformly over this area. Solution We have a phase change Q mLv . We need to find mass of rain in a cloud of radius 1 km. m V (1000 kg/m 3 )(0.01 m)( 106 m 2 ) 107 kg . With Q mLv and Lv 2256 kJ/kg , we find Q 7 1013 J – about the energy released in the first atomic bomb explosion. 22. To help prevent frost damage, 4.00 kg of 0C water is sprayed onto a fruit tree. (a) How much heat transfer occurs as the water freezes? (b) How much would the temperature of the 200-kg tree decrease if this amount of heat transferred from the tree? Take the specific heat to be 3.35 kJ/kg C , and assume that no phase change occurs. Solution (a) Q mLf (4.00 kg)(79.8 kcal/kg C) 319.2 kcal 319 kcal (b) Q mcT T Q 319.2 kcal 2.00C mc (200 kg)(0.800 kcal/kg C) 23. A 0.250-kg aluminum bowl holding 0.800 kg of soup at 25.0C is placed in a freezer. What is the final temperature if 377 kJ of energy is transferred from the bowl and soup, assuming the soup’s thermal properties are the same as that of water? Explicitly show how you follow the steps in the Problem-Solving Strategies for the Effects of Heat Transfer. Solution To bring the system to 0C requires heat, Q , of: OpenStax College Physics Instructor Solutions Manual Chapter 14 Q mAl cAl T ms cs T (0.250 kg)(0.215 kcal/kg C) (0.800 kg)(1.00 kcal/kg C)(25.0C) 21.34 kcal This leaves (90.0 21.34) kcal 68.66 kcal to freeze all the soup, leaving Q' ' (68.66 63.84) kcal 4.82 kcal to be removed. So, we can now determine the final temperature of the frozen soup: Q" (mAlcAl ms cs )T (mAlcAl ms cs )( 0C-Tf ). Tf Q" mAlcAl ms cs 4.82 kcal 10.6C (0.250 kg)(0.215 kcal/kg C) (0.800 kg)(0.500 kcal/kg C) 24. A 0.0500-kg ice cube at 30.0C is placed in 0.400 kg of 35.0C water in a very well-insulated container. What is the final temperature? Solution First bring the ice up to 0C and melt it with heat Q1 : Q1 mcT1 mLf (0.0500 kg) 0.500 kcal/kg C(30.0C) (79.8 kcal/kg) 4.74 kcal This lowers the temperature of water by T2 : Q1 4.74 kcal 11.85C mc (0.400 kg)(1.00 kcal/kg C) New Tw 35.0C 11.85C 23.15C Q1 mcT2 T2 Now, the heat lost by the hot water equals that gained by the cold water ( Tf is the final temperature): mc c w (Tf Tc ) mh c w (Th Tf ) Tf mh Th mcTc (0.400 kg)(296.3 K) (0.0500 kg)(273.15 K) 293.7 K 20.6C mc m h 0.450 kg 25. If you pour 0.0100 kg of 20.0C water onto a 1.20-kg block of ice (which is initially at 15.0C ), what is the final temperature? You may assume that the water cools so rapidly that effects of the surroundings are negligible. Solution First, we need to calculate how much heat would be required to raise the OpenStax College Physics Instructor Solutions Manual Chapter 14 temperature of the ice to 0C : Qice mcT (1.20 kg)(2090 J/kg C)(15C) 3.762 10 4 J Now, we need to calculate how much heat is given off to lower the water to 0C : Q1 mcT1 (0.0100 kg)(4186 J/kg C)(20.0C) 837.2 J Since this is less than the heat required to heat the ice, we need to calculate how much heat is given off to convert the water to ice: Q2 mLf (0.0100 kg)(334 103 J/kg) 3.340 103 J Thus, the total amount of heat given off to turn the water to ice at 0C : Qwater 4.177 10 3 J . Since Qice Qwater , we have determined that the final state of the water/ice is ice at some temperature below 0C . Now, we need to calculate the final temperature. We set the heat lost from the water equal to the heat gained by the ice, where we now know that the final state is ice at Tf 0C : Qlost by water Qgained by ice, or mwater c water T200 mwater Lf mwater cice T0? mice cice T15? Substituting for the change in temperatures (being careful that T is always positive) and simplifying gives: mwater [cwater (20C) Lf (cice )(0 Tf )] micecice[Tf (15C)]. Solving for the final temperature gives Tf mwater [cwater (20C) Lf ] mice cice (15C) (mwater mice )cice and so finally, (0.0100 kg)[(4186 J/kg C)(20C) 334 103 J/kg] Tf (0.0100 kg 1.20 kg)(2090 J/kg C) (1.20 kg)(2090 J/kg C)(15C) (0.0100 kg 1.20 kg)(2090 J/kg C) 13.2C 26. Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500C rock must be placed in 4.00 kg of 15.0C water to bring its temperature to 100C , if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings and take the average specific heat of the rocks to be that of granite. OpenStax College Physics Solution Instructor Solutions Manual Chapter 14 Let the subscripts r, e, v, and w represent rock, equilibrium, vapor, and water, respectively. mr cr (T1 Te ) mv Lv mw cw (Te T2 ) mr mv Lv mw cw (Te T2 ) cr (T1 Te ) (0.0250 kg)(2256 10 3 J/kg) (3.975 kg)(4186 J/kg C)(100C 15C) (840 J/kg C)(500C 100C) 4.38 kg 27. What would be the final temperature of the pan and water in Calculating the Final Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a Hot Pan if 0.260 kg of water was placed in the pan and 0.0100 kg of the water evaporated immediately, leaving the remainder to come to a common temperature with the pan? Solution Let the subscripts Al, e, v, and w represent aluminum pan, equilibrium, vapor, and water, respectively. mAlcAl (T1 Te ) mv Lv mw cw (Te T2 ) Te mAl c AlT1 m w c w T2 m v Lv m w c w mAl c Al (0.500 kg)(900 J/kg C)(150C) (0.250 kg)(4186 J/kg C)(20.0C) (0.0100 kg)(2256 10 3 J/kg) Te (0.250 kg)(4186 J/kg C) (0.500 kg)(900 J/kg C) 44.0C 28. In some countries, liquid nitrogen is used on dairy trucks instead of mechanical refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a density of 808 kg/m 3 . (a) Calculate the heat transfer necessary to evaporate this amount of liquid nitrogen and raise its temperature to 3.00C . (Use cp and assume it is constant over the temperature range.) This value is the amount of cooling the liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt-hours? (c) Compare the amount of cooling obtained from melting an identical mass of 0C ice with that from evaporating the liquid nitrogen. OpenStax College Physics Solution Instructor Solutions Manual Chapter 14 (a) Q mLv mcp T mLv c p T 48.0 kcal/kg (0.248 kcal/kg C) Q (200 10 3 m 3 )(808 kg/m 3 ) 3.00C (195.8C) 4 1.57 10 kcal 1 kW h (b) (1.572 10 4 kcal)(4186 J/kcal) 6 3.60 10 18.28 kW h 18.3 kW h J (c) Qice mLf (161.6 kg)(79.8 kcal/kg) 12,895 kcal 1.29 10 4 kcal 29. Solution Some gun fanciers make their own bullets, which involves melting and casting the lead slugs. How much heat transfer is needed to raise the temperature and melt 0.500 kg of lead, starting from 25.0C ? Q mcT mLf mcT Lf Q (0.500 kg) 0.0305 kcal/kg C 327C 25.0C 5.85 kcal/kg 7.53 kcal 14.5 CONDUCTION 30. (a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The surface area of the walls is 120 m 2 and their inside surface is at 18.0C , while their outside surface is at 5.00C . (b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction? Solution (a) Q kAT2 T1 2(0.042 J/s m C)(120 m 2 )(18.0 C 5.00C) t d 0.130 m 3 3 1.008 10 W 1.01 10 W (b) 1 one-kilowatt room heater is needed. 31. The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, 2 calculate the rate of conduction in watts through a 3.00 - m window that is 0.635 cm thick (1/4 in) if the temperatures of the inner and outer surfaces are 5.00C and 10.0C , respectively. This rapid rate will not be maintained—the inner surface will cool, and even result in frost formation. OpenStax College Physics Solution Instructor Solutions Manual Chapter 14 Q kAT2 T1 t d (0.84 J/s m C)(3.00 m 2 )5.00C (10.0C) 5953 W 6.0 103 W -2 0.635 10 m 32. Calculate the rate of heat conduction out of the human body, assuming that the core internal temperature is 37.0C , the skin temperature is 34.0C , the thickness of the tissues between averages 1.00 cm , and the surface area is 1.40 m 2 . Solution Q kAT2 T1 (0.2 J/s m C)(1.40 m 2 )(37.0C 34.0C) 84.0 W t d 0.0100 m 33. Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet, making contact over an area of 80.0 cm 2 with each foot. Both the ceramic and the carpet are 2.00 cm thick and are 10.0C on their bottom sides. At what rate must heat transfer occur from each foot to keep the top of the ceramic and carpet at 33.0C ? Solution Q kAT2 T1 t d For the wool carpet: Qw (0.04 J/s m C)(80.0 10 -4 m 2 )(33.0C - 10.0C) 0.368 W t 0.0200 m For the ceramic tile: Qc (0.84 J/s m C)(80.0 10 -4 m 2 )( 23C) 7.73 W t 0.0200 m 34. A man consumes 3000 kcal of food in one day, converting most of it to maintain body temperature. If he loses half this energy by evaporating water (through breathing and sweating), how many kilograms of water evaporate? Solution Q mLv(37C) m 35. Q Lv(37C) 1500 kcal 2.59 kg 580 kcal/kg (a) A firewalker runs across a bed of hot coals without sustaining burns. Calculate the OpenStax College Physics Instructor Solutions Manual Chapter 14 heat transferred by conduction into the sole of one foot of a firewalker given that the bottom of the foot is a 3.00-mm-thick callus with a conductivity at the low end of the range for wood and its density is 300 kg/m 3 . The area of contact is 25.0 cm 2 , the temperature of the coals is 700C , and the time in contact is 1.00 s. (b) What temperature increase is produced in the 25.0 cm3 of tissue affected? (c) What effect do you think this will have on the tissue, keeping in mind that a callus is made of dead cells? Solution (a) kAT2 T1 t Q d (0.0800 J/s m C)(25.0 10 -4 m 2 )(700C - 37.0C)(1.00s) 44.2 J 0.00300 m (b) Taking the density of the callus to be 300 kg/m 3 , the change in temperature can be found from: Q mcT VcT T Q 41.7 J 1.68C 3 Vc (300 kg/m )(25.0 10 6 m 3 )(3500 J/kg C) (c) At a temperature change of 2C , the heat probably won’t do much damage, since a callus is made of dead cells. 36. Solution (a) What is the rate of heat conduction through the 3.00-cm-thick fur of a large 2 animal having a 1.40 - m surface area? Assume that the animal’s skin temperature is 32.0C , that the air temperature is 5.00C , and that fur has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace this heat transfer? (a) Q kAT2 T1 (0.023 J/s m C)(1.40 m 2 )(37.0C) 39.71 W 39.7 W t d 0.0300 m 1 kcal (b) W Pt (39.71 J/s)(8.64 10 4 s) 819.7 kcal 820 kcal 4186 J 37. A walrus transfers energy by conduction through its blubber at the rate of 150 W when immersed in 1.00C water. The walrus’s internal core temperature is 37.0C , and it has a surface area of 2.00 m 2 . What is the average thickness of its blubber, which has the conductivity of fatty tissues without blood? OpenStax College Physics Solution Instructor Solutions Manual Chapter 14 Q kAT2 T1 t d kAT2 T1 (0.2 J/s m C)(2.00 m 2 )(38.0 C) d 0.101 m 10.1 cm Q/t 150 W 38. Compare the rate of heat conduction through a 13.0-cm-thick wall that has an area of 10.0 m 2 and a thermal conductivity twice that of glass wool with the rate of heat conduction through a window that is 0.750 cm thick and that has an area of 2.00 m 2 , assuming the same temperature difference across each. Solution Q kA(T2 T1 ) t d , so that (Q / t ) wall k A d (2 0.042 J/s m C)(10.0 m 2 )(0.750 10 2 m) wall wall window (Q / t ) window k window Awindowd wall (0.84 J/s m C)(2.00 m 2 )(13.0 10 2 m) 0.0288 wall : window, or 35 :1 window : wall 39. Suppose a person is covered head to foot by wool clothing with average thickness of 2.00 cm and is transferring energy by conduction through the clothing at the rate of 50.0 W. What is the temperature difference across the clothing, given the surface area is 1.40 m 2 ? Solution Q kA(T2 T1 ) kAT t d d d (Q / t ) (2.00 10 -2 m)(50.0 W) T 17.86C 17.9C kA (0.04 J/s m C)(1.40 m 2 ) 40. Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.600 cm thick and heat conduction occurs through the same area and at the same rate as computed in Example 14.6, what is the temperature difference across it? Ceramic has the same thermal conductivity as glass and brick. Solution Q kA(T2 T1 ) kAT t d d d (Q / t ) (6.00 10 -3 m)(2256 W) T 1046 C 1.05 10 3 K -2 2 kA (0.84 J/s m C)(1.54 10 m ) OpenStax College Physics 41. Solution Instructor Solutions Manual Chapter 14 One easy way to reduce heating (and cooling) costs is to add extra insulation in the attic of a house. Suppose the house already had 15 cm of fiberglass insulation in the attic and in all the exterior surfaces. If you added an extra 8.0 cm of fiberglass to the attic, then by what percentage would the heating cost of the house drop? Take the single story house to be of dimensions 10 m by 15 m by 3.0 m. Ignore air infiltration and heat loss through windows and doors. Q kAT . We need to consider all t d 6 sides that contribute to the heat loss. We will put the loss through the attic in a separate part. The original heat loss by conduction is given by Q (0.042 J/s/m/ o C)(10 m 3 m 2) (15 m 3 m 2) (15 m 10 m) T t 0.15 m o 0.042 J/s/m/ C(10 m 15 m) T 0.15 m (84 J/s/ C 42 J/s/ o C)T (126 J/s/ o C)T If we add 8 cm to the attic, the new addition is [84 J/s/ o C (0.042 J/s/m/ o C 150 m 2 )/0.23 m] T (84 J/s/ o C 27 J/s/ o C)T 111 J/s/ o C T So the percentage of savings in heat transfer = (126 111) / 126 12% . 42. (a) Calculate the rate of heat conduction through a double-paned window that has a 1.50 - m 2 area and is made of two panes of 0.800-cm-thick glass separated by a 1.00cm air gap. The inside surface temperature is 15.0C , while that on the outside is 10.0C . (Hint: There are identical temperature drops across the two glass panes. First find these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.) (b) Calculate the rate of heat conduction through a 1.60-cm-thick window of the same area and with the same temperatures. Compare your answer with that for part (a). OpenStax College Physics Solution Q Q1 Instructor Solutions Manual Q3 Q2 TL T1 d1 T2 TR d2 d1 In equilibrium, the heat flows across each “slab” are equal. Q1 Q2 Q3 and T1 TL TR T2 t t t (a) a1 K1 , a 2 K 2 , a3 K 3 a1 d1 d2 d3 Q1 Q2 per unit time a1 A(T1 TL ) a 2 A(T2 T1 ) Q2 Q3 per unit time a 2 A(T2 T1 ) a3 A(TR T2 ) a1 A(TR T2 ) (since a1 a3 ) a1T1 a1TL a 2T2 a 2T1 a 2T2 a 2T1 a1TR a1T2 Adding two equations, we obtain: a1 (TR TL ) a1 (TR TL ) a1 (T1 T2 ) 2a 2 (T2 T1 ) T2 T1 2 a a 2 1 Q2 a1 a 2 A(TR TL ) a 2 A(T2 T1 ) t 2a 2 a1 Now, we have K 1 0.84 J/s m C 105 J/s m 2 C d1 0.800 10 -2 m K 0.023 J/s m C a2 2 2.3 J/s m 2 C -2 d2 1.00 10 m TR 15.0C; TL 10.0C . Thus a1 Chapter 14 OpenStax College Physics Instructor Solutions Manual Chapter 14 Q2 a1a2 A(TR TL ) 105 J/s m 2 C 2.3 J/s m 2 C (1.50 m 2 )25.0C t 2a2 a1 2(2.3 J/s m 2 C) 105 J/s m 2 C 82.6 W Since (b) Q2 Q1 Q3 Q Q 82.6 W 83 W t t t t t Q kA(TR TL ) (0.84 J/s m C)(1.50 m 2 )(25.0C) 1969 W 1.97 10 3 W -2 t d 1.60 10 m The single-pane window has a rate of heat conduction equal to 1969/83, or 24 times that of a double pane window. 43. Solution Many decisions are made on the basis of the payback period: the time it will take through savings to equal the capital cost of an investment. Acceptable payback times depend upon the business or philosophy one has. (For some industries, a payback period is as small as two years.) Suppose you wish to install the extra insulation in Problem 14.41. If energy cost $1.00 per million joules and the insulation was $4.00 per square meter, then calculate the simple payback time. Take the average T for the 120 day heating season to be 15.0C . Q 126T J/s C as baseline energy use. So the t total heat loss during this period is Q (126J/s C)(15.0C)(120 days) (86.4 103 s/day) 1960 106 J . We found in Problem 14.41 that At the cost of $1/MJ, the cost is $1960. From Problem 14.41, the savings is 12% or $235 / y . We need 150 m 2 of insulation in the attic. At $4 / m 2 this is a $600 cost. So the payback period is $600 / $235 / year 2.6 years (excluding labor cost) . 44. For the human body, what is the rate of heat transfer by conduction through the body’s tissue with the following conditions: the tissue thickness is 3.00 cm, the change in temperature is 2.00C , and the skin area is 1.50 m 2 . How does this compare with the average heat transfer rate to the body resulting from an energy intake of about 2400 kcal per day? (No exercise is included.) Solution The rate of heat transfer by conduction is OpenStax College Physics Instructor Solutions Manual Chapter 14 Q kAT (0.2 J/s m o C) 1.50 m 2 2.00 C 20.0 W. t d 3.00 10 2 m On a daily basis, this is 1,728 kJ/day. Daily food intake is 2400 kcal/d 4186 J/kcal 10,050 kJ/day. So only 17.2% of energy intake goes as heat transfer by conduction to the environment at this T . 14.6 CONVECTION 45. At what wind speed does 10C air cause the same chill factor as still air at 29C ? Solution 10 m/s (from Table 14.4) 46. At what temperature does still air cause the same chill factor as 5C air moving at 15 m/s? Solution 26C (from Table 14.4) 47. The “steam” above a freshly made cup of instant coffee is really water vapor droplets condensing after evaporating from the hot coffee. What is the final temperature of 250 g of hot coffee initially at 90.0°C if 2.00 g evaporates from it? The coffee is in a Styrofoam cup, so other methods of heat transfer can be neglected. Solution Let M be the mass of coffee that is left after evaporation and m be the mass of coffee that evaporates. mLv (2.00 g)(539 kcal/kg) 4.35C Mc (248 g)(1.00 kcal/kg C) Tf Ti T 90.0C 4.35C 85.7C McT mLv T 48. (a) How many kilograms of water must evaporate from a 60.0-kg woman to lower her body temperature by 0.750°C ? (b) Is this a reasonable amount of water to evaporate in the form of perspiration, assuming the relative humidity of the surrounding air is low? OpenStax College Physics Instructor Solutions Manual Chapter 14 Solution (a) M is the mass of the woman and m is the mass of water that evaporates: McT mLv(37C) m McT (60.0 kg)(0.83 kcal/kg C)(0.750C) 6.44 10 2 kg Lv(37C) 580 kcal/kg (b) Yes, 64.4 g of water is reasonable. If the air is very dry, the sweat may evaporate without even being noticed. 49. Solution On a hot dry day, evaporation from a lake has just enough heat transfer to balance the 1.00 kW/m 2 of incoming heat from the Sun. What mass of water evaporates in 1.00 h from each square meter? Explicitly show how you follow the steps in the Problem-Solving Strategies for the Effects of Heat Transfer. A 1 m 2 , P 1.00 kW 1.00 103 W , so Pt Q mLv(37C) m Pt Lv(37C) (1.00 103 W)(3600 s) 1.48 kg 2430 103 J/kg (Note that we can use the Lv value at 37C as a closer approximation of the temperature on a hot day than 100C .) 50. Solution 51. One winter day, the climate control system of a large university classroom building malfunctions. As a result, 500 m 3 of excess cold air is brought in each minute. At what rate in kilowatts must heat transfer occur to warm this air by 10.0°C (that is, to bring the air to room temperature)? Q mcT VcT 1.29 kg/m 3 500 m 3 721 J/kg C 10.0C t t t 60.0 s 4 7.75 10 W 77.5 kW. P The Kilauea volcano in Hawaii is the world’s most active, disgorging about 5 105 m3 of 1200°C lava per day. What is the rate of heat transfer out of Earth by convection if this lava has a density of 2700 kg/m 3 and eventually cools to 30°C ? Assume that the specific heat of lava is the same as that of granite. OpenStax College Physics Solution Instructor Solutions Manual Chapter 14 Q mcT VcT t t t For c, we use the specific heat of granite, which is formerly molten rock. Q 2700 kg/m 3 5 10 5 m 3 840 J/kg C 1170C t 8.64 10 4 s 1.54 1010 W 2 1010 W 2 10 4 MW 52. During heavy exercise, the body pumps 2.00 L of blood per minute to the surface, where it is cooled by 2.00°C . What is the rate of heat transfer from this forced convection alone, assuming blood has the same specific heat as water and its density is 1050 kg/m 3 ? Solution Q mcT VcT t t t 3 1050 kg/m 2.00 10 -3 m 3 4186 J/kg C2.00C 293 W 60 s 53. A person inhales and exhales 2.00 L of 37.0°C air, evaporating 4.00 102 g of water from the lungs and breathing passages with each breath. (a) How much heat transfer occurs due to evaporation in each breath? (b) What is the rate of heat transfer in watts if the person is breathing at a moderate rate of 18.0 breaths per minute? (c) If the inhaled air had a temperature of 20.0C , what is the rate of heat transfer for warming the air? (d) Discuss the total rate of heat transfer as it relates to typical metabolic rates. Will this breathing be a major form of heat transfer for this person? Solution (a) Q mLv(37C) (4.00 10-5 kg)(2430 103 J/kg) 97.2 J (b) P NQ 18(97.2 J) 29.2 W t 60.0 s (c) Q mcT VcT (1.29 kg/m 3 )(2.00 10 3 m 3 )(721 J/kg C)(37.0C 20.0C) 31.6 J/breath , so the rate of heat loss is : P NQ 18(31.6 J) 9.49 W t 60.0 s OpenStax College Physics Instructor Solutions Manual Chapter 14 (d) The total rate of heat loss would be 29.2 W + 9.49 W = 38.7 W. While sleeping, our body consumes 83 W of power, while sitting it ranges 120-210 W. Therefore, the total rate of heat loss from breathing will not be a major form of heat loss for this person. 54. A glass coffee pot has a circular bottom with a 9.00-cm diameter in contact with a heating element that keeps the coffee warm with a continuous heat transfer rate of 50.0 W. (a) What is the temperature of the bottom of the pot, if it is 3.00 mm thick and the inside temperature is 60.0°C ? (b) If the temperature of the coffee remains constant and all of the heat transfer is removed by evaporation, how many grams per minute evaporate? Take the heat of vaporization to be 2340 kJ/kg. Solution (a) Q kAT (Q / t )d (50.0 W)(3.00 10 3 m) T 28.07C t d kr 2 (0.84 J/s m C) (4.50 10 2 m) 2 T Ti T T 88.07C 88C (b) Pt mLv m Pt (50.0 W)(60.0 s) 1.28 g Lv 2340J/g 14.7 RADIATION 55. At what net rate does heat radiate from a 275 - m 2 black roof on a night when the roof’s temperature is 30.0C and the surrounding temperature is 15.0C ? The emissivity of the roof is 0.900. Solution Q 4 4 eA(T24 T14 ) (5.67 10 8 J/s m 2 K 4 )(0.900)(2 75 m 2 ) 288 K 303 K t 21.7 kW Note that the negative answer implies heat loss to the surroundings. 56. (a) Cherry-red embers in a fireplace are at 850C and have an exposed area of 0.200 m 2 and an emissivity of 0.980. The surrounding room has a temperature of 18.0C . If 50% of the radiant energy enters the room, what is the net rate of radiant heat transfer in kilowatts? (b) Does your answer support the contention that most of the heat transfer into a room by a fireplace comes from infrared radiation? OpenStax College Physics Instructor Solutions Manual Chapter 14 Solution (a) Q 1 eA(T24 T14 ) t 2 1 4 4 (5.67 10 8 J/s m 2 K 4 )(0.980)(0 .200 m 2 ) 291 K 1123 K 2 8.80 kW Note that the negative answer implies heat loss to the surroundings. (b) This answer is quite large, so it does indeed suggest that the heat put into a room by a fireplace comes mainly from infrared radiation (which is hotter than red embers). 57. Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of heat transfer by radiation from 1.00 m 2 of 1200C fresh lava into 30.0C surroundings, assuming lava’s emissivity is 1.00. Solution Q 4 4 eA(T24 T14 ) (5.67 10 8 J/s m 2 K 4 )(1.00)(1. 00 m 2 ) 303 K 1473 K t 266 kW 58. (a) Calculate the rate of heat transfer by radiation from a car radiator at 110°C into a 2 50.0°C environment, if the radiator has an emissivity of 0.750 and a 1.20 - m surface area. (b) Is this a significant fraction of the heat transfer by an automobile engine? To answer this, assume a horsepower of 200 hp (1.5 kW) and the efficiency of automobile engines as 25% . Solution (a) Q eA(T24 T14 ) t 4 4 (5.67 10 8 J/s m 2 K 4 )(0.750)(1 .20 m 2 ) 323 K 383 K 543 W (b) Assuming an automobile engine is 200 horsepower and the efficiency of a gasoline 200 horsepower 800 horsepower engine is 25%, the engine consumes 25% Therefore, 600 horsepower is lost due to heating. The radiator transfers 1 hp 543 W 0.728 hp from radiation, which is not a significant fraction 746 W because the heat is primarily transferred from the radiator by other means. OpenStax College Physics Instructor Solutions Manual Chapter 14 59. Find the net rate of heat transfer by radiation from a skier standing in the shade, given the following. She is completely clothed in white (head to foot, including a ski mask), the clothes have an emissivity of 0.200 and a surface temperature of 10.0C , the surroundings are at 15.0°C , and her surface area is 1.60 m 2 . Solution Q eA(T24 T14 ) t 4 4 (5.67 10 8 J/s m 2 K 4 )(0.200)(1 .60 m 2 ) 258 K 283 K 36.0 W 60. Solution Suppose you walk into a sauna that has an ambient temperature of 50.0°C . (a) Calculate the rate of heat transfer to you by radiation given your skin temperature is 37.0°C , the emissivity of skin is 0.98, and the surface area of your body is 1.50 m 2 . (b) If all other forms of heat transfer are balanced (the net heat transfer is zero), at what rate will your body temperature increase if your mass is 75.0 kg? (a) Q eA(T24 T14 ) t 4 4 (5.67 10 8 J/s m 2 K 4 )(0.98)(1. 50 m 2 ) 323 K 310 K 137 W (b) Q mcT T Q 1 (137 W) 5.24 10 4 C/s 0.0314 C/min t t mc (75.0 kg)(3500 J/kg C) 61. Thermography is a technique for measuring radiant heat and detecting variations in surface temperatures that may be medically, environmentally, or militarily meaningful.(a) What is the percent increase in the rate of heat transfer by radiation from a given area at a temperature of 34.0°C compared with that at 33.0°C , such as on a person’s skin? (b) What is the percent increase in the rate of heat transfer by radiation from a given area at a temperature of 34.0°C compared with that at 20.0°C , such as for warm and cool automobile hoods? Solution 307 K 4 (a) 1 100% 1.31% 306 K 307 K 4 (b) 1 100% 20.5% 293 K OpenStax College Physics Instructor Solutions Manual Chapter 14 62. The Sun radiates like a perfect black body with an emissivity of exactly 1. (a) Calculate the surface temperature of the Sun, given that it is a sphere with a 7.00 108 - m radius that radiates 3.80 10 26 W into 3-K space. (b) How much power does the Sun radiate per square meter of its surface? (c) How much power in watts per square meter is that value at the distance of Earth, 1.50 1011 m away? (This number is called the solar constant.) Solution Q/t Q eAT 4 T (a) 2 t e( 4r ) 1/ 4 (the surface area of a sphere is 4r 2 ) 3.80 10 26 W T 8 4 8 2 (5.67 10 J/s m K )(4 )(7.00 10 m) 1/4 5.74 10 3 K P P 3.80 10 26 W 6.17 10 7 W/m 2 (b) 2 8 2 A 4r 4 (7.00 10 m) (c) Let r be the radius of a sphere with the sun at the center and the earth at a point on the surface of the sphere. P 3.80 10 26 W 1.34 10 3 W/m 2 2 11 2 4r 4 (1.50 10 m) 63. Solution A large body of lava from a volcano has stopped flowing and is slowly cooling. The interior of the lava is at 1200°C , its surface is at 450°C , and the surroundings are at 27.0°C . (a) Calculate the rate at which energy is transferred by radiation from 1.00 m 2 of surface lava into the surroundings, assuming the emissivity is 1.00. (b) Suppose heat conduction to the surface occurs at the same rate. What is the thickness of the lava between the 450°C surface and the 1200°C interior, assuming that the lava’s conductivity is the same as that of brick? (a) Q eA(T24 T14 ) t Q 4 4 (5.67 10 8 J/s m 2 K 4 )(1.00)(1. 00 m 2 ) 300 K 723 K t 1.503 10 4 W 1.50 10 4 W 15.0 kW Note the negative answer implies heat lost to the surroundings. OpenStax College Physics Instructor Solutions Manual Chapter 14 (b) Q kAT t d kAT (0.84 J/s m C)(1.00 m 2 )(750 C) d 4.19 10 2 m 4.2 cm 4 Q/t 1.503 10 W 64. Calculate the temperature the entire sky would have to be in order to transfer energy by radiation at 1000 W/m 2 —about the rate at which the Sun radiates when it is directly overhead on a clear day. This value is the effective temperature of the sky, a kind of average that takes account of the fact that the Sun occupies only a small part of the sky but is much hotter than the rest. Assume that the body receiving the energy has a temperature of 27.0°C . Solution Q Q/t eA(T24 T14 ) T1 T24 t eA Q/t 1000 W/m 2 A 1/ 4 1000 W/m 2 4 T1 (300 K) 8 2 4 (5.67 10 J/s m K )(1.00) 1/4 401 K 65. (a) A shirtless rider under a circus tent feels the heat radiating from the sunlit portion of the tent. Calculate the temperature of the tent canvas based on the following information: The shirtless rider’s skin temperature is 34.0°C and has an emissivity of 0.970. The exposed area of skin is 0.400 m 2 . He receives radiation at the rate of 20.0 W—half what you would calculate if the entire region behind him was hot. The rest of the surroundings are at 34.0°C . (b) Discuss how this situation would change if the sunlit side of the tent was nearly pure white and if the rider was covered by a white tunic. Solution Q A 4 4 (a) t e 2 (T2 T1 ) , Q/t T2 T14 e( A / 2) 1/ 4 2(Q / t ) T14 eA 1/ 4 2(20.0 W) 4 (307 K) 8 2 4 2 (5.67 10 J/s m K )(0.970)(0 .400 m ) 321.63 K 48.5C 1/4 OpenStax College Physics Instructor Solutions Manual Chapter 14 (b) A pure white object reflects more of the radiant energy that hits it, so the white tent would prevent more of the sunlight from heating up the inside of the tent, and the white tunic would prevent that radiant energy inside the tent from heating the rider. Therefore, with a white tent, the temperature would be lower than 48.5C , and the rate of radiant heat transferred to the rider would be less than 20.0 W. 66. Integrated Concepts One 30.0°C day the relative humidity is 75.0% , and that evening the temperature drops to 20.0°C , well below the dew point. (a) How many grams of water condense from each cubic meter of air? (b) How much heat transfer occurs by this condensation? (c) What temperature increase could this cause in dry air? Solution (a) Let x be the vapor density during the day. Percent relative humidity is equal to the vapor density divided by the saturation vapor density. Using the values for relative humidity and saturation vapor density, we have x 0.750 x 22.8 g/m 3 30.4 g/m 3 22.8 g/m 3 17.2 g/m 3 5.60 g/m 3 (b) Q mLv (5.60 10 3 kg)(539 kcal/kg) 3.02 kcal (c) Q mcT T Q 3.02 kcal 13.6C mc (1.29 kg)(0.172 kcal/kg C) 67. Integrated Concepts Large meteors sometimes strike the Earth, converting most of their kinetic energy into thermal energy. (a) What is the kinetic energy of a 10 9 kg meteor moving at 25.0 km/s? (b) If this meteor lands in a deep ocean and 80% of its kinetic energy goes into heating water, how many kilograms of water could it raise by 5.0°C? (c) Discuss how the energy of the meteor is more likely to be deposited in the ocean and the likely effects of that energy. Solution (a) KE 1 mv 2 0.5(10 9 kg)(25.0 10 3 m/s) 2 3.125 1017 J 3 1017 J 2 (b) Q mcT m Q (0.80)(3.1 25 1017 J) 1.19 1013 kg 1 1013 kg cT (4186 J/kg C)(5.0C) (c) When a large meteor hits the ocean, it causes great tidal waves, dissipating a large OpenStax College Physics Instructor Solutions Manual Chapter 14 amount of its energy in the form of kinetic energy of the water. 68. Integrated Concepts Frozen waste from airplane toilets has sometimes been accidentally ejected at high altitude. Ordinarily it breaks up and disperses over a large area, but sometimes it holds together and strikes the ground. Calculate the mass of 0°C ice that can be melted by the conversion of kinetic and gravitational potential energy when a 20.0 kg piece of frozen waste is released at 12.0 km altitude while moving at 250 m/s and strikes the ground at 100 m/s (since less than 20.0 kg melts, a significant mess results). Solution Let M be the mass of the ice block and m be the mass that melts before hitting the ground. KEi PE Q KEf 1 1 Mv 2i Mgh mLf Mv 2f 2 2 M gh 0.5(v 2i v 2f ) m Lf (20.0 kg) (9.80 m/s 2 )(12.0 10 3 m) 0.5(250 m/s) 2 0.5(100 m/s) 2 8.61 kg 334 10 3 J/kg 69. Solution Integrated Concepts (a) A large electrical power facility produces 1600 MW of “waste heat,” which is dissipated to the environment in cooling towers by warming air flowing through the towers by 5.00°C . What is the necessary flow rate of air in m 3 /s ? (b) Is your result consistent with the large cooling towers used by many large electrical power plants? (a) Q mcT m V m Q 1600 10 6 J/s 4.438 10 5 kg/s cT (721 J/kg C)(5.00C) 4.438 10 5 kg/s V 3.44 10 5 m 3 /s F 3.44 10 5 m 3 /s 3 t 1.29 kg/m (b) This is equivalent to 12 million cubic feet of air per second. That is tremendous. This is too large to be dissipated by heating the air by only 5C . Many of these cooling towers use the circulation of cooler air over warmer water to increase the rate of evaporation. This would allow there to be much smaller amounts of air necessary to remove such a large amount of heat, because evaporation removes larger quantities of heat than was considered in part (a). OpenStax College Physics 70. Instructor Solutions Manual Chapter 14 Integrated Concepts (a) Suppose you start a workout on a Stairmaster, producing power at the same rate as climbing 116 stairs per minute. Assuming your mass is 76.0 kg and your efficiency is 20.0% , how long will it take for your body temperature to rise 1.00º C if all other forms of heat transfer in and out of your body are balanced? (b) Is this consistent with your experience in getting warm while exercising? Solution (a) You produce power at a rate of 685 W, and since you are 20% efficient, you must P 685 W have generated: Pgenerated produced 3425 W . efficiency 0.20 If only 685 W of power was useful, the power available to heat the body is Pwasted 3425 W 685 W 2.740 103 W . Q mcT , so that t t mcT (76.0 kg)(3500 J/kg C)(1.00C) t 97.1 s Pwasted 2.74 10 3 W Now, Pwasted (b) This says that it takes about a minute and a half to generate enough heat to raise the temperature of your body by 1.00C , which seems quite reasonable. Generally, within five minutes of working out on a Stairmaster, you definitely feel warm and probably are sweating to keep your body from overheating. 71. Integrated Concepts A 76.0-kg person suffering from hypothermia comes indoors and shivers vigorously. How long does it take the heat transfer to increase the person’s body temperature by 2.00º C if all other forms of heat transfer are balanced? Solution Q Pt mcT t 72. mcT (76.0 kg)(3500 J/kg C)(2.00C) 1.25 10 3 s 20.9 min P 425 W Integrated Concepts In certain large geographic regions, the underlying rock is hot. Wells can be drilled and water circulated through the rock for heat transfer for the generation of electricity. (a) Calculate the heat transfer that can be extracted by cooling 1.00 km3 of granite by 100°C . (b) How long will it take for heat transfer at the rate of 300 MW, assuming no heat transfers back into the 1.00 km3 of rock by its surroundings? OpenStax College Physics Instructor Solutions Manual Chapter 14 Solution (a) Q mcT VcT (2700 kg/m 3 )(1.00 10 3 m) 3 (840 J/kg C)(100C) 2.27 1017 J (b) Pt Q t 73. Q 2.27 1017 J 7.57 10 8 s 24.0 y 6 P 300 10 W Integrated Concepts Heat transfers from your lungs and breathing passages by evaporating water. (a) Calculate the maximum number of grams of water that can be evaporated when you inhale 1.50 L of 37.0°C air with an original relative humidity of 40.0%. (Assume that body temperature is also 37.0°C .) (b) How many joules of energy are required to evaporate this amount? (c) What is the rate of heat transfer in watts from this method, if you breathe at a normal resting rate of 10.0 breaths per minute? Solution (a) To solve this, we calculate the mass of water initially in the breath and subtract this value from the mass of the water in an exhaled breath at 100% humidity. Using the saturation vapor density of water at 37C , 1.50 10 (44.0 g/m min 1.50 10 3 m 3 (44.0 g/m 3 )(0.400) 2.64 10 2 g mex 3 m 3 3 ) 6.60 10 2 g m mex min 6.60 10 2 g 2.64 10 2 g 3.96 10 2 g (b) Q mLv(37C) (3.96 102 g)(2430 J/g) 96.23 J 96.2 J (Note that Lv for water at 37C is used here as a better approximation than Lv for water at 100C. ) (c) P 74. NQ 10(96.23 J) 16.0 W t 60.0 s Integrated Concepts (a) What is the temperature increase of water falling 55.0 m over Niagara Falls? (b) What fraction must evaporate to keep the temperature constant? OpenStax College Physics Solution Instructor Solutions Manual Chapter 14 gh (9.80 m/s 2 )(50.0 m) (a) mgh mcT T 0.117C c 4186 J/kg C (b) Let M be the mass of water that evaporates. M gh (9.80 m/s 2 )(50.0 m) mgh MLv 2.17 10 4 3 m Lv 2430 10 J/kg C (Note that Lv for water at 37C is used here as a better approximation than Lv for water at 100C. ) 75. Integrated Concepts Hot air rises because it has expanded. It then displaces a greater volume of cold air, which increases the buoyant force on it. (a) Calculate the ratio of the buoyant force to the weight of 50.0°C air surrounded by 20.0°C air. (b) What energy is needed to cause 1.00 m3 of air to go from 20.0°C to 50.0°C? (c) What gravitational potential energy is gained by this volume of air if it rises 1.00 m? Will this cause a significant cooling of the air? Solution (a) PV nRT The density of a given volume of air will be proportional to h ~ 1 Th c ~ 1 . T 1 Tc The buoyant force is equal to the weight of the displaced cold air (Archimedes’ principle.) Thus, FB wc cVg and wh hVg FB cVg Th 323 K 1.102 wh hVg Tc 293 K (b) Q mcp T Vcp T (1.29 kg/m 3 )(1.00 m 3 )(721 J/kg C)(30.0C) 2.79 10 4 J (c) PE mgh (1.29 kg/m 3 )(1.00 m 3 )(9.80 m/s 2 )(1.00 m) 12.6 J This will not cause a significant cooling of the air because it is much less than the energy found in part (b), which is the energy required to warm the air from 20.0C to 50.0C . OpenStax College Physics Instructor Solutions Manual Chapter 14 76. Unreasonable Results (a) What is the temperature increase of an 80.0 kg person who consumes 2500 kcal of food in one day with 95.0% of the energy transferred as heat to the body? (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? Solution (a) Q mcT , so that T Q (0.950)(25 00 kcal) 36C. mc (80.0 kg)(0.83 kcal/kg C) This says that the temperature of the person is 37C 36C 73C ! (b) Any temperature increase greater than about 3C would be unreasonably large. In this case the final temperature of the person would rise to 73C 163F . (c) The assumption that the person retains 95% of the energy as body heat is unreasonable. Most of the food consumed on a day is converted to body heat, losing energy by sweating and breathing, etc. 77. Unreasonable Results A slightly deranged Arctic inventor surrounded by ice thinks it would be much less mechanically complex to cool a car engine by melting ice on it than by having a water-cooled system with a radiator, water pump, antifreeze, and so on. (a) If 80.0% of the energy in 1.00 gal of gasoline is converted into “waste heat” in a car engine, how many kilograms of 0°C ice could it melt? (b) Is this a reasonable amount of ice to carry around to cool the engine for 1.00 gal of gasoline consumption? (c) What premises or assumptions are unreasonable? Solution Qgas (a) Qgas MLf M L f Qgas 0.800 1.3 10 8 J. Thus, M 0.800 1.3 10 8 J 311.4 kg 3.1 10 2 kg 3 334 10 J/kg (b) No, the mass of ice is greater than 1/4 of a ton. (c) Not all waste heat goes into the engine. 78. Unreasonable Results (a) Calculate the rate of heat transfer by conduction through a window with an area of 1.00 m 2 that is 0.750 cm thick, if its inner surface is at 22.0°C and its outer surface is at 35.0°C . (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? OpenStax College Physics Solution (a) Instructor Solutions Manual Chapter 14 Q kAT2 T1 (0.84 J/s m C)(1.00 m 2 )(35.0C - 22.0C) t d 0.750 10 -2 m 1456 W 1.46 10 3 W 1.46 kW (b) This is very high power loss through a window. An electric heater of this power can keep an entire room warm. (c) The surface temperatures of the window do not differ by as great an amount as assumed. The inner surface will be warmer, and the outer surface will be cooler. 79. Solution Unreasonable Results A meteorite 1.20 cm in diameter is so hot immediately after penetrating the atmosphere that it radiates 20.0 kW of power. (a) What is its temperature, if the surroundings are at 20.0°C and it has an emissivity of 0.800? (b) What is unreasonable about this result? (c) Which premise or assumption is responsible? Q 20.0 kW 20.0 10 3 W and R 0.600 cm. (Note that the negative t sign indicates that the meteorite radiates heat to the surroundings.) (a) Given Q/t T1 T24 eA 1/ 4 20.0 10 3 W (293 K) 4 8 2 4 2 2 (5.67 10 J/s m K )(0.800)4 (0.600 10 m) 5.59 10 3 K (b) The meteorite has too high a temperature. It would completely melt. (c) The rate of radiation is probably too high. This file is copyright 2016, Rice University. All Rights Reserved. 1/4