College Physics Student s Manual Chapter CHAPTER : HEAT AND HEAT TRANSFER METHODS. TEMPERATURE CHANGE AND HEAT CAPACITY. On a hot day, the temperature of an 80,000- L swimming pool increases by.0 C. What is the net heat transfer during this heating? Ignore any complications, such as loss of by evaporation. The heat input is given by mcδt, where the specific heat of is c 86 /kg C. The mass is given by m ρ V (.00 and the temperature change is ΔT mcδt (8.00 m kg/m (80,000L 8.00 00 L.0 C. Therefore, kg(86/kg C(.0 C.0 kg, 8 9. Following vigorous exercise, the body temperature of an 80.0- kg person is 0.0 C. At what rate in watts must the person transfer thermal energy to reduce the body temperature to 7.0 C in 0.0 min, assuming the body continues to produce energy watt joule/second orw /s at the rate of W? ( First, calculate how much heat must be dissipated: mchuman bodyδt (80.0kg(00 /kg C(0 C - 7 C 8.0 Then, since power is heat divided by time, we can get the power required to produce the calculated amount of heat in 0.0 minutes:
College Physics Student s Manual Chapter P 8.0.67 t (0 min(60 s/min cooling W. Now, since the body continues to produce heat at a rate of W, we need to add that to the required cooling power: Prequired Pcooling + Pbody 67 W + W 67 W.. PHASE CHANGE AND LATENT HEAT. A bag containing 0 C is much more effective in absorbing energy than one containing the same amount of 0 C. (a How much heat transfer is necessary to raise the temperature of 0.800 kg of from 0 C to 0.0 C? (b How much heat transfer is required to first melt 0.800 kg of 0 C and then raise its temperature? (c Explain how your answer supports the contention that the is more effective. (a Use mcδt, since there is no phase change involved in heating the : mcδt (0.800 kg(86 /kg C(0.0 C.00 (b To determine the heat required, we must melt the, using ml, and then add the heat required to raise the temperature of melted using mcδt, so that mlf + mcδt (0.800 kg( /kg +.00.68 (c The is much more effective in absorbing heat because it first must be melted, which requires a lot of energy, then it gains the heat that the also would. The first.67 of heat is used to melt the, then it absorbs the.00 of heat that the absorbs. f 6
College Physics Student s Manual Chapter 9. How many grams of coffee must evaporate from 0 g of coffee in a 0- g glass cup to cool the coffee from 9.0 C to.0 C? You may assume the coffee has the same thermal properties as and that the average heat of vaporization is 0 k/kg (60 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. The heat gained in evaporating the coffee equals the heat leaving the coffee and glass to lower its temperature, so that ML m c ΔT + m cgδt, where M is the mass of coffee that evaporates. Solving for the evaporated coffee gives: v c c g M ΔT ( mccc + mgc L (9.0 C.0 C 60 cal/g v g [(0g(.00 cal/g C + (0g(0.0 cal/g C ].0 g Not that we did the problem in calories and grams, since the latent heat was given in those units, and the result we wanted was in grams. We could have done the problem in standard units, and then converted back to grams to get the same answer.. If you pour 0.00 kg of 0.0 C onto a.0- kg block of (which is initially at.0 C, what is the final temperature? You may assume that the cools so rapidly that effects of the surroundings are negligible. The heat gained by the equals the heat lost by the. Since we do not know the final state of the / combination, we first need to compare the heat needed to raise the to 0 C and the heat available from the. First, we need to calculate how much heat would be required to raise the temperature of the to 0 C : mcδt (.0 kg(090 /kg C( C.76 Now, we need to calculate how much heat is given off to lower the to mcδt (0.00kg(86/kg C(0.0 C 87. 0 C : Since this is less than the heat required to heat the, we need to calculate how 7
College Physics Student s Manual Chapter much heat is given off to convert the to : mlf (0.00 kg( /kg.0 Thus, the total amount of heat given off to turn the to at.77. 0 C : Since >, we have determined that the final state of the / is at some temperature below 0 C. Now, we need to calculate the final temperature. We set the heat lost from the equal to the heat gained by the, where we now know that the final state is at T < 0 C: f m c ΔT + m L + m c ΔT m c ΔT lost by gained by, or 0 0 f 0?? Substituting for the change in temperatures (being careful that Δ T is always positive and simplifying gives m c (0 C + L + ( c (0 T ] m c [ T ( C]. [ f f f Solving for the final temperature gives T f m [ c (0 C + Lf ] m ( m + m c c ( C and so finally, T f (0.00 kg[(86 /kg C(0 C + (0.00 kg +.0 kg(090 /kg C (.0kg(090 /kg C( C (0.00 kg +.0 kg(090 /kg C. C /kg]. CONDUCTION. A man consumes 000 kcal of food in one day, converting most of it to maintain body temperature. If he loses half this energy by evaporating (through breathing and sweating, how many kilograms of evaporate? 8
College Physics Student s Manual Chapter Use mlv, where 000 kcal and Lv(7 C 80 kcal/kg, ml 0 kcal 80 kcal/kg v(7 C m Lv(7 C.9 kg 8. Compare the rate of heat conduction through a.0- cm- thick wall that has an area of.0 m and a thermal conductivity tw that of glass wool with the rate of heat conduction through a window that is 0.70 cm thick and that has an area of assuming the same temperature difference across each..00 m, ka( T T Use the rate of heat transfer by conduction,, and take the ratio of t d the wall to the window. The temperature difference for the wall and the window will be the same: ( / t ( / t wall window k k wall window A A wall d window window d wall ( 0.0 /s m C(.0 m (0.70 m (0.8 /s m C(.00 m (.0 m 0.088 wall : window, or :window : wall So windows conduct more heat than walls. This should seem reasonable, since in the winter the windows feel colder than the walls..6 CONVECTION 0. One winter day, the climate control system of a large university classroom building malfunctions. As a result, 00 m of excess cold air is brought in each minute. At what rate in kilowatts must heat transfer occur to warm this air by.0 C (that is, to bring the air to room temperature? 9
College Physics Student s Manual Chapter Use mcδt in combination with the equations mcδt ρvcδt P t t t 7.7 W 77. kw. P and m ρv to get: t 7/kg C.0 C (.9 kg/m ( 00 m ( ( 60.0 s.7 RADIATION 8. (a Calculate the rate of heat transfer by radiation from a car radiator at C into a 0.0 C environment, if the radiator has an emissivity of 0.70 and a.0 - m surface area. (b Is this a significant fraction of the heat transfer by an automobile engine? To answer this, assume a horsepower of 00 hp (. kw and the efficiency of automobile engines as %. (a Using the Stefan- Boltzmann law of radiation, making sure to convert the temperatures into units of Kelvin, the rate of heat transfer is: t σ ea( T T (.67 8 /s m K (0.70(.0 m [( K ( 8 K ] W (b Assuming an automobile engine is 00 horsepower and the efficiency of a gasoline 00 horsepower engine is %, the engine consumes 800 horsepower in order % to generate the 00 horsepower. Therefore, 800 horsepower is lost due to hp heating. Since hp 76 W, the radiator transfers W 0.78 hp 76 W from radiation, which is not a significant fraction because the heat is primarily transferred from the radiator by other means.
College Physics Student s Manual Chapter 6. (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.0 C and has an emissivity of 0.970. The exposed area of skin is 0.00 m. He receives radiation at the rate of 0.0 W half what you would calculate if the entire region behind him was hot. The rest of the surroundings are at.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. (a Use the Stefan- Boltzmann law of radiation: t T A σe ( T T / t + T σe( A/, / ( / t + T σea / (0.0 W 8 (.67 /s m K (0.970(0.00 m.6 K 8. C + (07 K (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 8. C, and the rate of radiant heat transferred to the rider would be less than 0.0 W. / 70. Integrated Concepts (a Suppose you start a workout on a Stairmaster, producing power at the same rate as climbing 6 stairs per minute. Assuming your mass is 76.0 kg and your efficiency is 0.0%, how long will it take for your body temperature to rise.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?
College Physics Student s Manual Chapter (a You produce power at a rate of 68 W, and since you are 0% efficient, you must produced 68 W have generated: P generated P W. efficiency 0.0 If only 68 W of power was useful, the power available to heat the body is P W 68 W.70 W. wasted mcδt Now, Pwasted, so that t t mcδt (76.0 kg(00 /kg C(.00 C t P.7 W wasted 97.s (b This says that it takes about a minute and a half to generate enough heat to raise the temperature of your body by.00 C, 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. 76. Unreasonable Results (a What is the temperature increase of an 80.0 kg person who consumes 00 kcal of food in one day with 9.0% of the energy transferred as heat to the body? (b What is unreasonable about this result? (c Which premise or assumption is responsible? (0.90(00 kcal (a mcδt, so thatδt 6 C. mc (80.0 kg(0.8 kcal/kg C This says that the temperature of the person is 7 C + 6 C 7 C! (b Any temperature increase greater than about C would be unreasonably large. 7 C 6 F. In this case the final temperature of the person would rise to ( (c The assumption that the person retains 9% 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.