1 Note added to Homework set 7: The solution to Problem 16 has an error in it. The specific heat of water is listed as c 1 J/g K but should be c J/g K The final numerical answer given is correct but the math shown does not give that answer.
2 Answer, Key Homework 7 David McIntyre 1 This print-out should have 22 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. Chapter 22 problems. 001 (part 1 of 2) 0 points A heat engine absorbs 362 J of thermal energy and performs 27.2 J of work in each cycle. Find the efficiency of the engine. Correct answer: Given : Q h 362 J and W 27.2 J The thermal efficiency of a heat engine is e W Q h 27.2 J 362 J (part 2 of 2) 0 points Find the thermal energy expelled in each cycle. Correct answer: J. The work done by a heat engine through a cyclic process ( U 0) is W Q h Q c Q c Q h W J. 003 (part 1 of 3) 4 points of a system that absorbs 555 cal of thermal energy while doing 580 J of external work. Correct answer: J. Given : Q 555 cal and W 580 J. According to the first law of thermodynamics, we have U Q W where Q is the thermal energy transferred into the system and W is the work done by the system. Then we have U 555 cal J cal J. 580 J 004 (part 2 of 3) 3 points of a system that absorbs 769 cal of thermal energy while 626 J of external work is done on the system. Correct answer: J. U Q ( W ) 769 cal J cal J J 005 (part 3 of 3) 3 points of a system that is maintained at a constant volume while 1270 cal is removed from the system. Correct answer: J. Since the volume is maintained constant W P V 0 U Q 1270 cal J cal J. 006 (part 1 of 2) 0 points An ideal gas is compressed to half its original volume while its temperature is held constant.
3 Answer, Key Homework 7 David McIntyre 2 If 750 J of energy is removed from the gas during the compression, how much work is done on the gas? Correct answer: 750 J. Given : Q 750 J. According to the first law of thermodynamics, U Q W, where Q is the thermal energy transferred into the system and W is the work done by the system. Since U 0, then Thus Q h W eff P t 1 T c (126 kw) (3600 s) K 928 K J. W Q 750 J. 007 (part 2 of 2) 0 points What is the change in the internal energy of the gas during the compression? Correct answer: 0 J. If the temperature remains constant, U 0 J. 008 (part 1 of 2) 5 points A Carnot engine has a power output of 126 kw. The engine operates between two reservoirs at 16 C and 655 C. How much thermal energy is absorbed per hour? Correct answer: J. Given : P 126 kw, 655 C 928 K, T c 16 C 289 K. The efficiency of heat engine is eff W Q h 1 T c and the work done by the system is W P t. and 009 (part 2 of 2) 5 points How much thermal energy is lost per hour? Correct answer: J. The work done by a heat engine through a cyclic process ( U 0) is Then W Q h Q c. Q c Q h W Q h P t ( J) (126 kw) (3600 s) J. W KW 010 (part 1 of 2) 0 points A steam engine is operated in a cold climate where the exhaust temperature is 26 C. Calculate the theoretical maximum efficiency of the engine using an intake steam temperature of 114 C. Correct answer: Given : 114 C 387 K and T c 26 C 247 K.
4 Answer, Key Homework 7 David McIntyre 3 According to Carnot s theorem, the theoretical maximum efficiency is e 1 T c K 387 K (part 2 of 2) 0 points If, instead, superheated steam at 286 C is used, find the maximum possible efficiency. Correct answer: C 559 K The maximum efficiency is e K 559 K (part 1 of 1) 0 points The efficiency of a 840 MW nuclear power plant is 27.2%. If a river of flow rate kg/s were used to transport the excess thermal energy away, what would be the average temperature increase of the river? Correct answer: C. Given : P output 840 MW 10 6 W and e 27.2% The excess thermal energy transported per second by the river is P excess P input (1 e) ( ) Poutput (1 e) e ( ) 840 MW ( ) MW where e is efficiency and P output is power output of the plant. Then the temperature of the river is increased (per second) by dm dt c T dq dt P excess where c is heat capacity of water and dm is dt flow rate of the water. Thus T P excess kg/s C. 013 (part 1 of 2) 0 points A house loses thermal energy through the exterior walls and roof at a rate of 4860 W when the interior temperature is 20.1 C and the outside temperature is 0.2 C. Calculate the electric power required to maintain the interior temperature at T i for the following two cases: The electric power is used in electric resistance heaters (which convert all of the electricity supplied to thermal energy). Correct answer: 4860 W. Given : Q/ t 4860 W, T i 20.1 C, T o 0.2 C. and Since all the electricity supplied is converted to thermal energy, we have Thus Q t E t P El P El 4860 W. 014 (part 2 of 2) 0 points The electric power is used to operate the compressor of a heat pump (which has a coefficient of performance equal to ν 0.7 of the Carnot cycle value). Correct answer: W.
5 Answer, Key Homework 7 David McIntyre 4 For a heat pump we have T i (COP ) Carnot T i T o 20.1 C K 20.1 C ( 0.2 C) Hence to bring 4860 W of heat in the house requires only P h Q/ t (COP ) actual W 0.6 (COP ) carnot 4860 W (0.7) ( ) W. 015 (part 1 of 1) 0 points An ice tray contains 375 g of water at 0 C. Calculate the change in entropy of the water as it freezes completely and slowly at 0 C. Correct answer: J/K. Given : m 375 g kg, L J/kg, and T 0 C 273 K. In the freezing process T is constant, so Q m L where m is mass of water, and l is latent heat of fusion. Thus S Q T m L T (0.375 kg)( J/kg) 273 K J/K. Given : T i 15.8 C K, T f 80.5 C K, m 210 g, and c 1 J/g K. The heat absorbed in the process is dq r m c dt. The change in entropy in an arbitrary reversible process between an initial state and final state is S f i f i ds m c dt T m c log T f T i f i dq r T (210 g) (1 J/g K) log J/K. ( ) K K 017 (part 1 of 6) 2 points One mole of an ideal monatomic gas is taken through the cycle abca shown schematically in the diagram. State a has volume V a m 3 and pressure P a Pa, and state c has volume V c m 3. Process ca lies along the T 231±1 K isotherm. The molar heat capacities for the gas are c p 20.8 J/mol K and c v 12.5 J/mol K. p ( 10 5 Pa) K a b 016 (part 1 of 1) 10 points Calculate the change in entropy of 210 g of water heated slowly from 15.8 C to 80.5 C. Correct answer: J/K c 250 K V ( 10-3 m 3 )
6 Answer, Key Homework 7 David McIntyre 5 This schematic plot is intended to give an example of a P V diagram (not to scale). Use the values of P, V, and T given above. Determine the temperature T b of state b. Correct answer: K. Given : P b Pa, V b m 3, and T b J/mol K. We use the ideal gas equation T P V n R, where P is the pressure, V is the volume (both evaluated at b ), R is the molar gas constant, and n is the number of moles. T b P V R ( Pa) ( m3 ) J/mol K K. 018 (part 2 of 6) 2 points Determine the heat Q ab added to the gas during process ab. Correct answer: J. For state a Given : P a Pa, V a m 3, and T a J/mol K. T a P V R ( Pa) ( m3 ) J/mol K K Thus Q n c p T (1 mol)(20.8 J/mol K) ( K K) J, where Q is the heat transferred, n is the number of moles, c p is the the molar heat capacity for a constant pressure process (such as process ab ), and T is the change in temperature from a to b. 019 (part 3 of 6) 2 points U ab U b U a. Correct answer: J. In an isobaric process the change in internal energy is given by U ab Q ab W Q ab P V Q ab P [V b V a ] J ( Pa) ( m m 3 ) J, 020 (part 4 of 6) 2 points Determine the work W bc done by the gas on its surroundings during process bc. Correct answer: 0. W P V and V 0, so W (part 5 of 6) 1 points The net heat added to the gas for the entire cycle is 2140 J. Determine the net work done by the gas on its surroundings for the entire cycle. Correct answer: 2140 J. Given : Q 2140 J. For a complete cycle the change in internal energy U is zero, so W Q 2140 J.
7 The work is simply the net heat added to the gas. 022 (part 6 of 6) 1 points Determine the efficiency Eff of a Carnot engine that operates between the maximum and minimum temperatures in this cycle. Correct answer: The Carnot efficiency Eff is given by Eff 1 T c. The maximum temperature is clearly that of state b, determined to be K in question 1. The minimum temperature will be that of the isotherm, K. Therefore Eff 1 T a T b K K Answer, Key Homework 7 David McIntyre 6
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