Chapter 20F - CHAPTER 20 The Second Law of Thermodynamics 1...

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Unformatted text preview: CHAPTER 20 The Second Law of Thermodynamics 1* Where does the energy come from in an internal-combustion engine? In a steam engine? Internal combustion engine: From the heat of combustion (see Problems 19-106 to 19-109). Steam engine: From the burning of fuel to evaporate water and to raise the temperature and pressure of the steam. 2 How does friction in an engine affect its efficiency? Friction reduces the efficiency of the engine. 3 John is house-sitting for a friend who keeps delicate plants in her kitchen. She warns John not to let the room get too warm or the plants will wilt, but John forgets and leaves the oven on all day after his brownies are baked. As the plants begin to droop, John turns off the oven and opens the refrigerator door, intending to use the refrigerator to cool the kitchen. Explain why this doesn't work. Since a refrigerator exhausts more heat to the room than it extracts from the interior of the refrigerator, the temperature of the room will increase rather than decrease. 4 Why do power-plant designers try to increase the temperature of the steam fed to engines as much as possible? Increasing the temperature of the steam increases the Carnot efficiency, and generally increases the efficiency of any heat engine. 5* An engine with 20% efficiency does 100 J of work in each cycle. ( a ) How much heat is absorbed in each cycle? ( b ) How much heat is rejected in each cycle? ( a ) From Equ. 20-2, Q h = W / ( b ) Q c = Q h (1 - ) Q h = 100/0.2 J = 500 J Q c = 500 0.8 J = 400 J 6 An engine absorbs 400 J of heat and does 120 J of work in each cycle. ( a ) What is its efficiency? ( b ) How much heat is rejected in each cycle? ( a ) = W / Q h ( b ) Q c = Q h (1 - ) = 120/400 = 0.3 = 30% Q c = 400 0.7 J = 280 J 7 An engine absorbs 100 J and rejects 60 J in each cycle. ( a ) What is its efficiency? ( b ) If each cycle takes 0.5 s, find the power output of this engine in watts. ( a ) Use Equ. 20-2 = 1 - 60/100 = 0.4 = 40% Chapter 20 The Second Law of Thermodynamics ( b ) P = W / t = Q h / t P = 0.4 100/0.5 W = 80 W 8 A refrigerator absorbs 5 kJ of energy from a cold reservoir and rejects 8 kJ to a hot reservoir. ( a ) Find the coefficient of performance of the refrigerator. ( b ) The refrigerator is reversible and is run backward as a heat engine between the same two reservoirs. What is its efficiency? ( a ) W = Q h - Q c ; COP = Q c / W ( b ) = W / Q h W = 3 kJ; COP = 5/3 = 1.67 = 3/8 = 0.375 = 37.5% 9* An engine operates with 1 mol of an ideal gas for which R C 2 3 v = and R C 2 5 p = as its working substance. The cycle begins at P 1 = 1 atm and V 1 = 24.6 L. The gas is heated at constant volume to P 2 = 2 atm. It then expands at constant pressure until V 2 = 49.2 L. During these two steps, heat is absorbed by the gas. The gas is then cooled at constant volume until its pressure is again 1 atm. It is then compressed at constant pressure to its original state. constant volume until its pressure is again 1 atm....
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This homework help was uploaded on 04/09/2008 for the course PHYS 161,260,27 taught by Professor Staff during the Spring '08 term at Maryland.

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Chapter 20F - CHAPTER 20 The Second Law of Thermodynamics 1...

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