HEAT ENGINES AND REFRIGERATORS

Transcription

1 EA ENGINES AND REFRIGERAORS 9 onceptual uestions 9.. s =. (a) < 0, > 0. ork is done by the system; the area under the curve is positive. s (b) > 0, < 0. ork is done on the system to compress it to a smaller volume. s (c) > 0, < 0. More work is done on the system than by the system. s 9.. > = >. he amount of work done by the gas is the area side the closed cycle loop (traversed a clockwise direction). 9.. No. out what you get h = =. You cannot get out more than you put. what you had to pay 9.. (a) (b) (c) (d) In stage the volume is fixed (so no work is done) but the temperature creases, so heat was added. In stage on the isotherm work is done by the gas. In stage work is done on the gas and heat is removed. 9.. out 0 6 η > η= η > η. η = η = η = η = he thermal efficiency is larger for enge ; the same amount of heat is added per cycle both enges, but the cycle for enge has a larger out due to the larger enclosed area: out η = It is an isothermal process with equal amounts of heat added to the system and work done by the system. η = (a) No; cannot have Jout > 0 J. (b) Yes, this is a heat enge with out η = = = 0., which is less than η arnot = (c) No, it isn t possible to have η > η arnot : 6 η = = = 0. 6, but η arnot = opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist. 9-

2 9- hapter (a) No, the purpose of a refrigerator is to remove heat from the cold reservoir, and this diagram dumps heat to the cold reservoir. (b) Yes, 0 K = = = 0., which is less than K arnot =. 0 (c) No, 0 K = = =, which is less than K arnot = No. he first law of thermodynamics (energy conservation) for a refrigerator or air conditioner requires = + here are no perfect refrigerators (second law), so > 0 (work must be done by the compressor) and. thus >. Because the air conditioner exhausts more heat to the room than it extracts from the room, the net effect is to crease the room temperature, not decrease it. 9.. Yes, the first law says that energy is conserved, so we will never get more work out of the heat enge than heat energy is transferred to the system. In fact, the second law (formal statement #) says that there are no perfect heat enges with η =, so there is always some waste heat exhausted to the cold reservoir. Exercises and Problems Section 9. urng eat to ork Section 9. eat Enges and Refrigerators 9.. Solve: (a) he enge has a thermal efficiency of η = 0% = 0. 0 and a work output of 00 J per cycle. he heat put is calculated as follows: η = 00 J J. = = (b) Because out =, the heat exhausted is = = 0 J 00 J = 0 J 9.. Solve: Durg each cycle, the work done by the enge is out = 00 J and the enge exhausts = 00 J of heat energy. By conservation of energy, = + = 00 J + 00 J = 600 J hus, the efficiency of the enge is 00 J η = = = J 9.. Solve: (a) Durg each cycle, the heat transferred to the enge is = kj, and the heat exhausted is = 0 kj. he thermal efficiency of the heat enge is 0 kj η = = = 0. 7 = 7% kj (b) he work done by the enge per cycle is = = kj 0 kj = kj 9.. Solve: he coefficient of performance of the refrigerator is 600 J 00 J K = = = = J opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

3 eat Enges and Refrigerators Solve: (a) he heat extracted from the cold reservoir is calculated as follows: (b) he heat exhausted to the hot reservoir is K =. 0 = = 00 J 0 J = + = 00 J + 0 J = 0 J 9.6. Model: Assume that the car enge follows a closed cycle. Solve: (a) Sce 00 rpm is 0 cycles per second, the work output of the car enge per cycle is (b) he heat put per cycle is calculated as follows: he heat exhausted per cycle is out kj s kj kj = 00 =. s 0 cycles cycle cycle. kj η = = = 6. kj 0. 0 = = 6. kj. kj = 0 kj 9.7. Solve: he amount of heat discharged per second is calculated as follows: 9 η = = = = (900 M) = η 0. out hat is, each second the electric power plant discharges J of energy to the ocean. Sce a typical American house needs. 0 0 J of energy per second for heatg, the number of houses that could be heated with the 9 waste heat is (. 9 0 J)/(. 0 0 J) = 96, Solve: he amount of heat removed from the water coolg it down hour is = mwatercwater. he mass of the water is mwater = rwatervwater = (000 kg/m )( L) = (00 kg/m )(0 m ) =. 0 kg he rate of heat removal from the refrigerator is = (. 0 kg)(90 J/kg K)(0 ) = J J = = 7. 6 J/s 600 s he refrigerator does work = 8.0 = 8.0 J/s to remove this heat. hus the performance coefficient of the refrigerator is Section 9. Ideal-Gas eat Enges Section 9. Ideal-Gas Refrigerators 7. 6 J/s K = = J/s 9.9. Model: Process A is isochoric, process B is isothermal, process is adiabatic, and process D is isobaric. Solve: Process A is isochoric, so the crease pressure creases the temperature and hence the thermal energy. Because Eth = s and s = 0 J, creases for process A. Process B is adiabatic, so = 0 J. s is positive because of the crease volume. Sce = 0 J = s + Eth, Eth is negative for process B. Process is isothermal, opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

4 9- hapter 9 so is constant and hence Eth = 0 J. he work done s is positive because the gas expands. Because = + E is positive for process B. Process D is isobaric, so the decrease volume leads to a decrease s th, temperature and hence a decrease the thermal energy. Due to the decrease volume, s is negative. Because = + E also decreases for process D. s th, E th S A B D 9.0. Model: Process A is adiabatic, process B is isochoric, and process is isothermal. Solve: Process A is adiabatic, so = 0 J. ork s is positive as the gas expands. Sce = s + Eth = 0 J, Eth must be negative. he temperature falls durg an adiabatic expansion. Process B is isochoric. No work is done ( s = 0 J), and is positive as heat energy is added to raise the temperature ( Eth positive). Process is isothermal so = 0 and Eth = 0 J. he gas is compressed, so s is negative. = s for an isothermal process, so is negative. eat energy is withdrawn durg the compression to keep the temperature constant. E th s A + 0 B Solve: he work done by the gas per cycle is the area side the closed p-versus-v curve. he area side the triangle is out 6 6 = ( atm atm)(600 0 m 00 0 m ). 0 0 Pa 6 = atm (00 0 m ) 0 J = atm 9.. Solve: he work done by the gas per cycle is the area enclosed with the pv curve. e have (60 J) 60 J = ( p max 00 kpa)(800 cm 00 cm ) = p 6 max. 0 0 Pa m p max =. 0 0 Pa = 00 kpa 9.. Model: he heat enge follows a closed cycle, which consists of four dividual processes. Solve: (a) he work done by the heat enge per cycle is the area enclosed by the p-versus-v graph. e get 6 out = (00 kpa 00 kpa)(00 0 m ) = 0 J he heat energy leavg the enge is = 90 J + J = J. he heat put is calculated as follows: (b) he thermal efficiency of the enge is = = + = J + 0 J = J 0. kj 0 J η = = = 0. J Assess: Practical enges have thermal efficiencies the range η opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

5 eat Enges and Refrigerators Model: he heat enge follows a closed cycle, startg and endg the origal state. he cycle consists of three dividual processes. Solve: (a) he work done by the heat enge per cycle is the area enclosed by the p-versus-v graph. e get out 6 = (00 kpa)(00 0 m ) = 0 J he heat energy transferred to the enge is = 0 J + 8 J = J. Because out =, the heat energy exhausted is (b) he thermal efficiency of the enge is = = J 0 J = 0 J 0. 0 kj 0 J η = = = J Assess: Practical enges have thermal efficiencies the range η Model: he heat enge follows a closed cycle. Solve: he work done by the gas per cycle is the area side the closed p-versus-v curve. e get 6 out = (00 kpa 00 kpa)(600 cm 00 cm ) = (00 0 Pa)(00 0 m ) = 0 J Because out =, the heat exhausted is = = ( J + 90 J) 0 J = J 0 J = 8 J 9.6. Model: he heat enge follows a closed cycle. Solve: (a) he work done by the gas per cycle is the area side the closed p-versus-v curve. e get 6 out = (00 kpa 00 kpa)(600 cm 00 cm ) = (00 0 Pa)(00 0 m ) = 0 J he heat exhausted is = 80 J + 00 J = 80 J. he thermal efficiency of the enge is 0 J η = = = = J + 0 J out (b) he heat extracted from the hot reservoir is = + = 0 J Model: he Brayton cycle volves two adiabatic processes and two isobaric processes. he adiabatic processes volve compression and expansion through the turbe. ( γ)/ γ Solve: he thermal efficiency for the Brayton cycle is ηb = rp, where γ = P/ V and r p is the pressure ratio. For a diatomic gas γ =.. For an adiabatic process, Because the volume is halved, V = V so he efficiency is γ γ γ = = pv p V p / p ( V/ V ) p γ. r = p / p = () = =. 69 B 0./. η = (.69) = Model: he refrigerator follows a closed cycle. Solve: (a) he net work done on the refrigerator one cycle is = s = 78 J + 9 J = J opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

6 9-6 hapter 9 his is the work needed to push the heat from the cold reservoir up to the hot reservoir. he heat exhausted to the hot reservoir is = 0 J. From the first law of thermodynamics, the heat extracted from the cold reservoir is (b) he coefficient of performance is + = = = 0 J J = 6 J 6 J K = = =.6 J Assess: his is a reasonable value for the coefficient of performance for a refrigerator. Section 9. he Limits of Efficiency Section 9.6 he arnot ycle 9.9. Model: he efficiency of a arnot enge ( η arnot ) depends only on the temperatures of the hot and cold reservoirs. On the other hand, the thermal efficiency ( η ) of a heat enge depends on the heats and. Solve: (a) Accordg to the first law of thermodynamics, = out +. For enge (a), = 00 J, = 00 J and out = 00 J, so the first law of thermodynamics is obeyed. For enge (b), = 00 J, = 00 J and out = 00 J, so the first law is violated. For enge (c) = 00 J, = 00 J and out = 00 J, so the first law of thermodynamics is obeyed. (b) For the three heat enges, the maximum or arnot efficiency is Enge (a) has 00 K η arnot = = = K η = out 00 J 0 60 = = 00 J =. his is larger than η arnot, thus violatg the second law of thermodynamics. For enge (b), 00 J η = = = 0. 0 < η 00 J so the second law is obeyed. Enge (c) has a thermal efficiency of so the second law of thermodynamics is obeyed. 00 J η = = 0. < η 00 J 9.0. Model: For a refrigerator = +, and the coefficient of performance and the arnot coefficient of performance are arnot arnot K = =, Karnot Solve: (a) For refrigerator (a) = + (60 J = 0 J + 0 J), so the first law of thermodynamics is obeyed. For refrigerator (b) 0 J = 0 J + 0 J, so the first law of thermodynamics is obeyed. For the refrigerator (c) 0 J 0 J + 0 J, so the first law of thermodynamics is violated. (b) For the three refrigerators, the maximum coefficient of performance is K arnot = 00 K = 00 K 00 K = opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

7 eat Enges and Refrigerators 9-7 For refrigerator (a), 0 J K = = = < K 0 J so the second law of thermodynamics is obeyed. For refrigerator (b), 0 J K = = = > K 0 J so the second law of thermodynamics is violated. For refrigerator (c), so the second law is obeyed. 0 J K = =. < K 0 J 9.. Model: he efficiency of a arnot enge depends only on the absolute temperatures of the hot and cold reservoirs. Solve: he efficiency of a arnot enge is arnot arnot arnot η arnot = = = 80 K = 7 (7 + 7) K Assess: A real enge would need a lower temperature than 7 to provide 60% efficiency because no real enge can match the arnot efficiency. 9.. Model: he efficiency of an ideal enge (or arnot enge) depends only on the temperatures of the hot and cold reservoirs. Solve: (a) he enge s thermal efficiency is 0 J η = = = = 0. 0 = 0% + J + 0 J out (b) he efficiency of a arnot enge is η arnot = /. he mimum temperature the hot reservoir is found as follows: 9 K 0. 0 = = 88 K = his is the mimum possible temperature. In a real enge, the hot-reservoir temperature would be higher than because no real enge can match the arnot efficiency. 9.. Model: Assume that the heat enge follows a closed cycle. Solve: (a) he enge s thermal efficiency is 00 J η = = = = 0. = % J + 00 J out (b) he thermal efficiency of a arnot enge is η arnot = /. For this to be %, 9.. Solve: (a) he efficiency of the arnot enge is 0. = = 0. 8 K = (00 + 7) K 00 K η arnot = = = 0. 0 = 0% 00 K (b) An enge with power output of 000 does out = 000 J of work durg each t = s. A arnot enge has a heat put that is 000 J = = = 00 J η 0. 0 arnot opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

8 9-8 hapter 9 durg each t = s. he rate of heat put is 00 J/s = 00. (c) = out, so the heat output durg t = s is out = = 00 J. he rate of heat output is thus 00 J/s = Model: e will use Equation 9.7 for the efficiency of a arnot enge. η arnot = e are given = 67 K and the origal efficiency η arnot = Solve: First solve for. Solve for aga with η arnot = = ( η ) = (67 K)( 0. 0) = 0 K arnot arnot = ( η ) = (67 K)( 0. 60) = 69 K he difference of these values is K, so the temperature of the cold reservoir should be decreased by to raise the efficiency from 0% to 60%. Assess: e expected to have to lower by quite a bit to get the better efficiency Model: he maximum possible efficiency for a heat enge is provided by the arnot enge. Solve: he maximum efficiency is η max (7 + 0) K = ηarnot = = = ( ) K Because the heat enge is runng at only 0% of the maximum efficiency, η = (0. 0) η = he amount of heat that must be extracted is 000 J = = =. 0 kj η Model: e are given = 77 K and = 7 K, therefore (by Equation 9.7) the arnot efficiency is 7 K n arnot = = e are also given η =. η 77 K arnot Solve: Rearrange Equation 9.: = out ( η ). out is the same for both enges, so it cancels. ( η) 0. 60( ηarnot ) 0. 60(0. 67) = = = = 7. ( ) ( η ) η arnot out arnot arnot Assess: his enge requires.7 times as much heat energy durg each cycle as a arnot enge to do the same amount of work Model: he coefficient of performance of a arnot refrigerator depends only on the temperatures of the cold and hot reservoirs. Solve: (a) he arnot performance coefficient of a refrigerator is K arnot ( 0 + 7) K = = = (0 + 7) K ( 0 + 7) K (b) he rate at which work is done on the refrigerator is found as follows: (c) he heat exhausted to the hot side per second is 00 J/s K = J/s = K = 6. = = = + = 00 J/s + J/s = J/s 0. k max opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

9 eat Enges and Refrigerators Model: he mimum possible value of occurs with a arnot refrigerator. Solve: (a) For the refrigerator, the coefficient of performance is he heat energy exhausted per cycle is K = = K = (. 0)(0 J) = 0 J = + = 0 J + 0 J = 60 J (b) If the hot-reservoir temperature is 7 = 00 K, the lowest possible temperature of the cold reservoir can be obtaed as follows: K arnot = 0 0 K. = 00 K = = 9.0. Model: Equation 9.7 gives arnot η =. e are given η arnot = /. Solve: η = = = = arnot Equation 9.8 gives the coefficient of performance for the arnot refrigerator. K = = = = arnot Assess: his result is the ballpark for coefficients of performance. 9.. Solve: he work done by the enge is equal to the change the gravitational potential energy. hus, he efficiency of this enge is = Ugrav = mgh = (000 kg)(9. 8 m/s )(0 m) = 88,000 J (7 + 0) K η = 0. 0 ηarnot = 0. 0 = 0. 0 = 0. 8 ( ) K he amount of heat energy transferred is calculated as follows: 9.. Solve: he mass of the water is 88,000 J 6 η = = = =. 7 0 J η m 000 kg (00 0 L) = kg L m he heat energy is removed from the water three steps: () coolg from + to 0, () freezg at 0, and () coolg from 0 to. he three heat energies are = mc = (0. 00 kg)(86 J/kg K)( K) = 679 J = mlf = (0. 00 kg)(. 0 J/kg) =,00 J = mc = (0. 00 kg)(090 J/kg K)( K) = J = + + =,7 J Usg the performance coefficient,,7 J,7 J K = 0 0,679 J. = = 0. = opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

10 9-0 hapter 9 he heat exhausted to the room is thus = + =,7 J + 0,679 J =. 0 J 9.. Solve: An adiabatic process has = 0 and thus, from the first law, s = E th. For any ideal-gas process, Eth = nv, so s = nv. e can use the ideal-gas law to fd onsequently, the work is pv ( pv ) ( pv ) f ( pv ) i pfvf pivi = = = = nr nr nr nr pfvf pv i i V s = nv = nv = ( pfvf pivi) nr R Because P = V + R, we can use the specific heat ratio γ to fd ith this, the work done an adiabatic process is P V + R V/ R+ V γ = = = = / R R γ V V V p V p V pv p V pv pv R γ γ V f f i i s = ( f f i i) = ( f f i i) = 9.. Model: e are given = K and = K. See Figure 9.. Solve: Every second, the refrigerator must draw enough heat from the cold reservoir to compensate for the heat lost through the staless-steel panel. herefore, the heat transferred from the cold reservoir to the system is A (0.0 m)(0.0 m) = k t = ( /m K) [ ( 0 )](.0 s) = 0,080 J L 0.00 m Usg the coefficient of performance for a arnot refrigerator, we can fd the energy required to operate for one second: ( ) (0,080 J)(70 K) Karnot = = = = =.8 kj K he power required is therefore P = t = (.8 kj)(.0 s) =.8 k. Assess: his is much more power than is required for the Brayton-cycle refrigerator of Example 9, which shows why refrigerators are sulated with more than simple steel doors. 9.. Solve: For any heat enge, η = /. For a arnot heat enge, η arnot = /. hus a property of the arnot cycle is that / = /. onsequently, the coefficient of performance of a arnot refrigerator is K arnot / / = = = = = / / 9.6. Model: e are given = 98 K and = 7 K. See Figure Solve: = mlf = (0 kg)(. 0 J/kg) =. 0 J. (a) For a arnot cycle η arnot = but that must also equal, η = so =. 98 K ( 0 J) 6 0 J 6 6 = =. =. 7 K (b) = =. 6 0 J. 0 J = J =. 0 0 J Assess: his is a reasonable amount of work to freeze 0 kg of water. opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

11 eat Enges and Refrigerators Model: e will use the arnot enge to fd the maximum possible efficiency of a floatg power plant. Solve: he efficiency of a arnot enge is η max (7 + ) K = ηarnot = = = % (7 + 0) K 9.8. Model: he ideal gas the arnot enge follows a closed cycle four steps. Durg the isothermal expansion at temperature, heat is transferred from the hot reservoir to the gas. Durg the isothermal compression at, heat is removed from the gas. No heat is transferred durg the remag two adiabatic steps. Solve: he thermal efficiency of the arnot enge is Usg = + out, we obta η out K out arnot = 6 J = 7 K = 000 J = isothermal = = = 000 J 6 J 0. 6 kj 9.9. Solve: Substitutg to the formula for the efficiency of a arnot enge, η arnot = 0 0 K. = + 80 K = = he hot-reservoir temperature is = + 80 K = 0 K = Solve: From the thermal efficiency of the arnot enge, we can fd the work done each cycle: η arnot 7 K = = = ( J) = 0 J K he work required to lift a 0 kg mass 0 m is = Fd = (0 kg)(9.8 m/s )(0 m) = 980 J. At 0 J/cycle, the arnot enge will have to cycle 98 times to do this work. 9.. Model: Assume the soda is essentially made of water. e are given = 8 K, = K, and = 0 J. See Figure 9.. Solve: he total amount of heat to transfer from the soda is otal ( ) = Mc = Vρc = (.00 0 m )(000 kg/m )(90 J/kg K)(0 K) =,900 J For a arnot cycle η arnot = but also, η = so =. herefore, the heat extracted from the soda each cycle is K = = (0 J) = 9.8 J 8 K o remove,900 J will therefore take ( ) otal,900 J = = 8 cycles 9.8 J 9.. Solve: (a) is given as 000 J. Usg the energy transfer equation for the heat enge, he thermal efficiency of a arnot enge is = + = + = 00 K η = = = 0. 0 = 600 K out = η = ( η ) = (000 J)( 0. 0) = 00 J opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

12 9- hapter 9 o determe and, we turn our attention to the arnot refrigerator, which is driven by the output of the heat enge with = out. he coefficient of performance is K 00 K = = = 0. = = = 00 K 00 K η out = Kη = (. 0)(0. 0)(000 J) = 000 J Usg now the energy transfer equation + =, we have = + = η + = (0. 0)(000 J) J = 00 J (b) From part (a) = 00 J and = 000 J, so >. (c) Although = 000 J and = 00 J, the two devices together do not violate the second law of thermodynamics. his is because the hot and cold reservoirs are different for the heat enge and the refrigerator. 9.. Solve: he work done by the arnot enge powers the refrigerator, so ( ) arnot Eng = ( ) Refrigerator. e are given that = 0 K and = 0 K for both the arnot enge and the refrigerator and ( ) arnot Eng = 0.0 J for the arnot enge. he work done by the arnot enge is ( out ) arnot Eng 0 K η = = ( ) arnot Eng = (0.0 J) =.87 J ( ) 0 K arnot Eng he heat extracted from the cold reservoir by the refrigerator may be found from it coefficient of performance: K = (.00)(.87 J).7 J ( ) = ( ) = = Refrigerator out arnot Eng he heat exhausted by the refrigerator to the hot reservoir may be found from the first law of thermodynamics: ( ) + = =.87 J+.7 J = 8.7 J Refrigerator Assess: he work done on the refrigerator is less than the heat exhausted to the hot reservoir, as expected. 9.. Model: A heat pump is a refrigerator that is coolg the already cold outdoors and warmg the doors with its exhaust heat. Solve: (a) he coefficient of performance for this heat pump is K = 0. = /, where is the amount of heat removed from the cold reservoir. is the amount of heat exhausted to the hot reservoir. = +, where is the amount of work done on the heat pump. e have = 0. = 0. + = 60. If the heat pump is to deliver kj of heat per second to the house, then kj = kj = 6. 0 = =. kj 60. In other words,. k of electric power is used by the heat pump to deliver kj/s of heat energy to the house. (b) he monthly heatg cost the house usg an electric heater is kj 600 s $ (00 h) = $70 s h 0 MJ he monthly heatg cost the house usg a heat pump is. kj 600 s $ (00 h) = $ s h 0 MJ opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

13 eat Enges and Refrigerators Visualize: e are given = 7 K, = 9 K. e are also given that one second = 00 J and = ( s)(00 kj/m)( m/60 s) = 667 J. Solve: he coefficient of performance of a refrigerator is given Equation J K = = = J owever the coefficient of performance of a arnot refrigerator is given Equation 9.8. K arnot = 7 K 7 = 0 K =. owever, formal statement #8 of the second law says that the coefficient of performance cannot exceed the arnot coefficient of performance, so the salesman is makg false claims. You should not buy the DreamFridge. Assess: he second law imposes real-world restrictions Solve: he maximum possible efficiency of the heat enge is 00 K η max = = = K he efficiency of the enge designed by the first student is 0 J η = = = 0. 0 J Because η > η max, the first student has proposed an enge that would violate the second law of thermodynamics. is or her design will not work. he efficiency of the enge designed by the second student is η = 90 J / 0 J = 0. 6 < ηmax agreement with the second law of thermodynamics. Applyg the first law of thermodynamics, = + 0 J = 70 J + 90 J we see the first law is violated. his design will not work as claimed. he design by the third student satisfies the first law of thermodynamics because = + = 0 J. he thermal efficiency of this enge is η = 0. 6 < η max, which satisfies the second law of thermodynamics. he data presented by students and are faulty. Only student has an acceptable design Model: he power plant is to be treated as a heat enge. Solve: (a) Every hour 00 metric tons or 0 kg of coal is burnt. he volume of coal is he height of the room will be 8 m. (b) he thermal efficiency of the power plant is 0 kg m ( h) = 800 m h 00 kg J/s J h = = = = 0. = % kg 8 0 J h. 0 J h kg 600 s Assess: An efficiency of % is typical of power plants Model: he power plant is treated as a heat enge. Solve: e are given that = 00 = 7 K and = = 98 K. (a) he maximum possible thermal efficiency of the power plant is 98 K η max = = = 0. 8 = 8% 7 K opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

14 9- hapter 9 6 (b) In one second, the plant generates = J of work and = J of heat energy to replace out 6 the energy taken from the hot reservoir to heat the water. he plant s actual efficiency is J η = = 6 0. = % J (c) Because = + out, = =. 0 0 J/s. 0 0 J/s =. 0 0 J/s he mass of water that flows per second through the condenser is 8 L hr 0 m 000 kg m =. 0 (s) =. 0 kg h 600 s L m 9 he change the temperature as =. 0 0 J of heat is transferred to m =. 0 kg of water is 9 = mc. 0 0 J = (. 0 kg)(86 J/kg K) = he exit temperature is 8 + = Model: he power plant is treated as a heat enge. Solve: he mass of water per second that flows through the plant every second is 8 L hr 0 m 000 kg m =. 0 0 = kg/s h 600 s L m he amount of heat transferred per second to the coolg water is thus 6 9 = mc = ( kg/s)(86 J/kg K)(7 6 ) = J/s he amount of heat per second put to the power plant is Fally, the power plant s thermal efficiency is = + = J/s J/s = J/s 9.0. Solve: (a) he energy supplied one day is (b) he volume of water is out J/s η = = = 0. 7 = 7% J/s 9 9 J 600 s h =. 0 0 = J s h d 9 V = km = 0 m. he amount of energy is 000 kg (0 m ) (90 J/kg K)( K) 0 J m 9 = mc = = (c) hile it s true that the ocean contas vast amounts of thermal energy, that energy can be extracted to do useful work only if there is a cold reservoir at a lower temperature. hat is, the ocean has to be the hot reservoir of a heat enge. But there s no readily available cold reservoir, so the ocean s energy cannot readily be tapped. here have been proposals for usg the colder water near the bottom of the ocean as a cold reservoir, pumpg it up to the surface where the heat enge is. Although possible, the very small temperature difference between the surface and the ocean depths implies that the maximum possible efficiency (the arnot efficiency) is only a few percent, and the efficiency of any real ocean-driven heat enge would likely be less than % perhaps much less. hus the second law of thermodynamics prevents us from usg the thermal energy of the ocean. Save your money. Don t vest. opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

15 eat Enges and Refrigerators Visualize: If we do this problem on a per-second basis then one second = ( s)(. 0 0 J/m) ( m/60 s) = 8. 0 J. = ( s)( J/m)( m/ 60 s) =. 0 J. Solve: (a) Aga, one second = =. 0 J 8. 0 J =. 0 0 J Sce this is per second, the power required by the compressor is (b) he coefficient of performance is Assess: he result is typical for air conditioners.. K P =.0 k. 8 0 J = = = J 9.. Model: he heat enge follows a closed cycle with process and process beg isochoric and process and process beg isobaric. For a monatomic gas, = R and = R. Visualize: Please refer to Figure P9.. Solve: (a) he first law of thermodynamics is = E th + S. For the isochoric process, S = 0 J. hus, o fd volume V, V P = 70 J = Eth = nv 70 J 70 J 70 J = = = = 0 K n (. 0 mol) (. 0 mol) (8. J/mol K) ( R ) V = K = K + 00 K = 60 K nr (. 0 mol)(8. J/mol K)(00 K) V = = = = 8. 0 m 0 0 Pa V p. he pressure p can be obtaed from the isochoric condition as follows: p p = = =. =. 00 K p p 60 K ( 00 0 Pa) Pa ith the above values of p, V, and, we can now obta p, V, and. e have For the isobaric process, V = V = m p = p = Pa V = = = = 0 K V V V P S th S = n = (. 0 mol) R ( ) = (. 0 mol) (8. J/mol K)(60 K) =,80 J = p ( V V ) = ( Pa)(8. 0 m ) = 990 J E = =,80 J 990 J = 790 J e are now able to obta p, V, and. e have V = V = m p = p = Pa p Pa = = = (0 K) = 600 K p p p Pa opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

16 9-6 hapter 9 For isochoric process, S th S = nv = (. 0 mol) R ( ) = (. 0 mol) (8. J/mol K)( 60) = 700 J = 0 J E = = 700 J For isobaric process, P = n = (. 0 mol) (8. J/mol K)(00 K 600 K) = 60 J S th S = p ( V V ) = ( Pa) (8. 0 m m ) = 90 J E = = 60 J ( 90 J) = 70 J S (J) (kj) E th (kj) , Net (b) he thermal efficiency of this heat enge is 00 J η = = = = 0. = % + 70 J +,80 J Assess: Note that more than two significant figures are retaed part (a) because the results are termediate. For a closed cycle, as expected, ( s ) net = net and ( Eth ) net = 0 J Model: he heat enge follows a closed cycle. For a diatomic gas, = R and = R. Visualize: Please refer to Figure P9.. Solve: (a) Sce = 9 K, the number of moles of the gas is 6 V pv ( Pa)(0 0 m ) n = = = mol R (8. J/mol K)(9 K) At pot, V = V and p = p. he temperature is calculated as follows: pv p V p V = = = ()()(9 K) = 6 K p V At pot, V = V = V and p = p. he temperature is calculated as before: p V = = ()()(9 K) = 7 K p V For process, the work done is the area under the p-versus-v curve. hat is, he change the thermal energy is s =. +.. (0 atm)(0 cm 0 cm ) ( atm 0 atm)(0 cm 0 cm ) Pa = (0 0 m )( atm) =. 0 J atm th V P E = n = ( mol) (8. J/mol K)(6 K 9 K) =. 9 J he heat is = s + Eth = J. For process, the work done is s = 0 J and = E = n = n R ( ) th V = ( mol) (8. J/mol K)(7 K 6 K) = 0. J opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

17 eat Enges and Refrigerators 9-7 For process, 6 s th V = (0. atm)(0 cm 0 cm ) = ( Pa)( 0 0 m ) =. J E = n = ( mol) (8. J/mol K)(9 K 7 K) =. 80 J he heat is = Eth + s =. J. (b) he efficiency of the enge is (c) he power output of the enge is s (J) (J) Eth Net.. 0 net. J η = = = = 9. 0% J revolutions m net = (. J/s) = m 60 s revolution 60 Assess: Note that more than two significant figures are retaed part (a) because the results are termediate. For a closed cycle, as expected, ( s ) net = net and ( Eth ) net = 0 J. 9.. Model: For the closed cycle, process is isothermal, process is isobaric, and process is isochoric. Visualize: Please refer to Figure P9.. Solve: (a) e first need to fd the conditions at pots,, and. e can then use that formation to fd S and for each of the three processes that make up this cycle. Usg the ideal-gas equation the number of moles of the gas is 6 pv (. 0 0 Pa)(600 0 m ) n = = = 0. 0 mol R (8. J/mol K)(00 K) e are given that γ =., which means this is not a monatomic or a diatomic gas. he specific heats are R V = = R P = V + R= R γ At pot, process is isothermal, so we can fd the pressure p as follows: V m V m pv = p V p = p = p = p = atm = Pa At pot, process is isobaric, so we can fd the temperature as follows: m m V V V = = = = = 900 K V Pot P (Pa).0 atm = atm = atm =.09 0 V (m ) (K) opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

18 9-8 hapter 9 Process is isothermal: Process is isobaric: Process is isochoric: e fd that ( S ) = pv ln( V/ V ) = J = ( S ) = J ( S ) = p V = p( V V ) =. 6 J = np = np( ) = 608. J ( S) = 0 J = nv = nv( ) = J ( S) cycle = J +. 6 J + 0 J =. 8 J cycle = J J J =. 8 J hese are equal, as they should be. Knowg that the work done is = ( S ) cycle =.8 J/cycle, an enge operatg at 0 cycles/s has a power output of P out. 8 J 0 cycle J = = 096 = k cycle s s (b) Only is positive, so = = 608 J. hus, the thermal efficiency is. 8 J η = = = = 9. 0% 608. J 9.. Model: For the closed cycle of the heat enge, process is isobaric, process is isochoric, and process is adiabatic. = R and = R for a monatomic gas, so γ = /. V P Visualize: Please refer to Figure P9.. Solve: (a) e can use the adiabat to calculate p as follows: γ / γ γ V 600 cm V 00 cm pv = pv p = p = (00 kpa) = 98 kpa can be determed by takg the ratio of the ideal-gas equation applied to pots and. his gives pv pv = V 00 cm V 600 cm = = (600 K) = 00 K where we have used the fact that p = p. Applyg the same strategy at pot gives pv pv = p 00 kpa = = (600 K) = 0. K p 98 kpa where we have used the fact that V = V. Before we calculate the work and heat exchanged for each cycle, we need to know the number of moles. his may be calculated by applyg the ideal gas law at any pot on the cycle: Now we can calculate s,, and th 6 pv (98 kpa)(00 0 m ) n = = = 0.8 mol R (8. J/mol K)(600 K) E for the three processes volved the cycle. For process, ( ) E = n ( ) = n R ( ) =. 86 kj th, V = n ( ) = n R ( ) =. 76 kj P opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

19 eat Enges and Refrigerators 9-9 he work done s is the area under the p-versus-v graph. e have For process, s = 0 J and For process, = 0 J and 6 6 s = (98 kpa)(600 0 m 00 0 m ) = kj E = = n ( ) = n R ( ) =. 69 kj th, V E = n ( ) = n R ( ) = kj th, V Because Eth = + and = s, s = Eth, = kj for process. s (kj) (kj) E th (kj) Net (b) he thermal efficiency of the enge is 78 J η = = = 0. = % 76 J Assess: Note that more than two significant figures are retaed part (a) because the results are termediate. As expected for a closed cycle, ( s ) net = net and ( Eth ) net = 0 J Model: For the closed cycle of the heat enge, process is isochoric, process is adiabatic, and 7 process is isothermal. For a diatomic gas = R and γ = V Solve: (a) From the graph V = 000 cm. he pressure p lies on the adiabat from. e can fd the pressure as follows: 7/ γ γ V 000 cm 000 cm γ pv = pv p = p = ( Pa) = Pa 696 kpa V he temperature can be obtaed from the ideal-gas equation relatg pots and : pv pv pv Pa = = = (00 K) () =. K K p V Pa (b) he number of moles of the gas is For isochoric process, = 0 J and For adiabatic process, = 0 J and. pv ( Pa)( m ) n = = = mol R (8. J/mol K)(00 K) s th V = E = n = n R = 7. J E = n = n R ( ) = 7. J th V Usg the first law of thermodynamics, Eth = s +, which means s = Eth = + 7. J. s can also be determed from p V p V nr( ) J/K (00 K. K) = = = = 7. J s γ γ opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

20 9-0 hapter 9 For isothermal process, Eth = 0 J and V s = nrln =. J V Usg the first law of thermodynamics, Eth = s +, = s =. J. E th (J) s (J) (J) Net (c) he work per cycle is 87 J and the thermal efficiency is s J η = = = 0. = % 7. J 9.7. Model: For this heat enge, process is adiabatic, process is isothermal, and process is isobaric. For a diatomic gas, 7 = R, = R, and γ =. 0. V P Solve: (a) From the graph, p = 00 kpa. Pot is connected by an adiabatic process to pot, where p = 00 kpa γ γ and V = 000 cm. So pv = pv and thus / γ /. 0 p 00 kpa = V = (000 cm ) = 69 cm 690 cm p 00 kpa V Pot is connected by an isochoric process to pot, where V = 000 cm and = 00 K. hus V 69 cm = = (00 K) = 69 K 000 cm V Altogether, p = 00 kpa, V = 690 cm, and = 69 K. (b) For the adiabatic process, = 0 and s = E th. For any process, Eth = nv. he number of moles can be found from pot : hus pv (. 0 0 Pa)( m ) n = = = 0. 0 mol R (8. J/mol K)(00 K) th ( R) E = (0. 0 mol) (00 K 69 K) = 7 J = For the isothermal process, E th = 0 and = s. he work done is For the isobaric process, V s = nrln = J = V s = p V = (. 0 0 Pa)( m m ) = J he heat exhausted to the cold reservoir is P Fally, from the first law, Eth = s = 7 J. 7 = n = (0. 0 mol) R (69 K 00 K) = 8 J E th s Net s opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

21 eat Enges and Refrigerators 9- (c) he work per cycle is out = 9 J. eat is put only durg process, so = J and the enge s thermal efficiency is 9 J η = = = 0. 7 = 7, J Assess: As expected for a closed cycle, ( E th ) net = 0 and ( s ) net = net Model: For the closed cycle of the refrigerator, process is isochoric, process is adiabatic, process is isochoric, and process is adiabatic. For a monatomic gas = R and γ = Visualize: Please refer to the figure below. V. Solve: (a) he number of moles of gas may be found by applyg the ideal-gas equation to pot. he result is pv ( 0 kpa )(.00 0 m ) n = = = 7. 0 mol R (8. J/mol K)(0 K) he temperatures at pots and may be found usg able 9.: / γ V 00 cm = = (0 K) = 6 K V 0 cm / γ V 00 cm = = (00 K) = 68 K V 0 cm Only the adiabatic segments do work, so the total work done by the system is = + = n ( + ) = n R ( + ) S S, S, v = (7. 0 mol) (8. J/molK)(6 K 0 K + 00 K 69 K) =.97 J hus, the work done on the system is = S =.79 J. Durg the adiabatic segments, no heat is exchanged with the heat reservoirs, so heat is exchanged only durg the isochoric segments. For a refrigerator, the heat exchanged with the cold reservoir is the heat that is put to the system (i.e., > 0), which occurs segment. ith the help of able 9., this is = n = n R ( ) = (7. 0 mol) (8. J/molK)(0 K 00 K) =.0 J hus, the coefficient of performance is V K.0 J = = =.7 J.9 (b) he power needed to run the refrigerator is cycles P= ( cycle ) 60 = 7 s opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

22 9- hapter Model: Process of the cycle is isochoric, process is isothermal, and process is isobaric. For a monatomic gas, = R and =. V P R Visualize: Please refer to Figure P Solve: (a) At pot, the pressure p = atm =.0 0 Pa and the volume V = m = 0 m. he number of moles is Usg the ideal-gas law, 0. 0 g n = = 0. 0 mol g/mol pv (. 0 0 Pa)(. 0 0 m ) = = = 06 K 0. kk nr (0. 00 mol)(8. J/mol K) At pot, the pressure p = atm = Pa and V = 0 m. he temperature is pv ( Pa)(. 0 0 m ) = = = 00 K kk nr (0. 00 mol)(8. J/mol K) At pot, the pressure is p = atm =. 0 0 Pa and the temperature is = = 00 K. he volume is p atm = = ( 0 m ) = 0 m atm V V p (b) For the isochoric process, = 0 J and V For the isothermal process, Eth = 0 J and = n = (0.00 mol) R (00 K 06 K) = 607 J V. 0 0 m = = nrln = (0. 00 mol)(8. J/mol K)(00 K) ln = 8 J V. 0 0 m For the isobaric process, = p V = (. 0 0 Pa)(. 0 0 m. 0 0 m ) = 0 J = n = (0.00 mol) (8. J/mol K)(06 K 00 K) = 0 J P he total work done is net = + + = 0 J. he total heat put is = + = J. he thermal efficiency of the enge is net 0 J η = = = 9, J (c) he maximum possible efficiency of a heat enge that operates between max and m is m 06 K η max = = = 80, 00 K max Assess: he actual efficiency of an enge is less than the maximum possible efficiency Model: he process of the heat enge cycle is isochoric and the process is isobaric. For a monatomic gas = R and = R. V P Solve: (a) he three temperatures are pv (.0 0 Pa)(0.0 m ) = = = 60.7 K 0.60 kk nr (.0 mol)(8. J/mol K) opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

23 eat Enges and Refrigerators 9- pv (6.0 0 Pa)(0.00 m ) = = = 80. K.8 kk nr (.0 mol)(8. J/mol K) pv (.0 0 Pa)(0.00 m ) = = = 0. K. kk nr (.0 mol)(8. J/mol K) (b) For process, the work done is the area under the p-versus-v graph. he work and the change ternal energy are s =. 0 J = (6.0 0 Pa.0 0 Pa)(0.00 m 0.0 m ) + (.0 0 Pa)(0.00 m 0.0 m ) E = n = (.0 mol) R ( ) th V = (.0 mol) (8. J/mol K)(80. K 60.7 K) =.00 0 J he heat put is = s + Eth =. 0 J. For isochoric process, s = 0J and th V = E = n = (.0 mol) (8. J/mol K)(0. K 80. K) =.0 0 J For isobaric process, the work done is the area under the p-versus-v curve. ence, s = (.0 0 Pa)(0.0 m 0.00 m ) =.0 0 J E = n = n R ( ) = (.0 mol) (8. J/mol K)(60.7 K 0. K) =. 0 J th V he heat put is = S + Eth =.0 0 J E th (J) S (J) (J) Net 0. 0 (c) he thermal efficiency is net. 0 J η = = =.9,. 0 J 9.6. Model: he closed cycle this heat enge cludes adiabatic process, isobaric process, and 7 7 isochoric process. For a diatomic gas, = R, = R, and γ = =.. V P Visualize: Please refer to Figure P9.6. Solve: (a) e can fd the temperature from the ideal-gas equation as follows: pv (.0 0 Pa)(.0 0 m ) = = = 07 K. kk nr (0.00 mol)(8. J/mol K) γ γ e can use the equation pv = pv to fd V, /γ /. p.0 0 Pa V= V = (.0 0 m ) =.69 0 m p.0 0 Pa he ideal-gas equation can now be used to fd, pv (.0 0 Pa)(.69 0 m ) = = = 60 K.6 kk nr (0.00 mol)(8. J/mol K) opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

24 9- hapter 9 At pot, V = V so we have pv ( 0 Pa)(.69 0 m ) = = = 679 K 6. kk nr (0.00 mol)(8. J/mol K) (b) For adiabatic process, = 0 J, E th = s, and pv p V nr( ) (0.00 mol)(8. J/mol K)(07 K 60 K) s = = = = 7.0 J γ γ (.) For isobaric process, 7 7 = n = n R ( ) = (0.00 mol) (8. J/mol K)(679 K 07 K) = 69 J P th V E = n = n R = 69 J he work done is the area under the p-versus-v graph. ence, For isochoric process, s = 0J and s = (.0 0 Pa)(.69 0 m.0 0 m ) = 677 J E = = n = (0.00 mol) (8. J/mol K)(60 K 679 K) = 09 J th V E th (J) S (J) (J) Net (c) he enge s thermal efficiency is net 0 J η = = = 0. =, 69 J 9.6. Model: For the closed cycle of the heat enge, process is isothermal, process is isobaric, and 7 process is isochoric. For a diatomic gas = R and γ = Visualize: Please refer to the figure below. V. Solve: (a) Beg by expressg the pressure, volume, and temperature terms of the pressure, volume, and temperature at pot. In the isothermal expansion, the volume is halved so the pressure must double (ideal gas equation). herefore p = p. Because is isobaric, p = p = p. e are given that V = V / and that V = V. Fally, we know that = because they are on the same isotherm, and the ideal gas equation gives pv nr pv nr = = = opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

25 eat Enges and Refrigerators 9- he table below summarizes: p V p = p V V / p = = = p V = V = ith the help of able 9., we can fd expressions for the work and heat for each segment the cycle. he results are given the table below. S nrln( V/ V) = nrln nr ln p( V V) = ( p)( V/) = p V = nr 7 n ( ) = nr P 0 n ( ) = nr V he heat transferred from the hot reservoir to the heat enge (> 0) is done the isochoric segment. he total work done by the system is = nr ln + nr = nr ( ln ) he thermal efficiency is therefore nr ( ln ) η = = = = 8.8, 7 nr (b) he thermal efficiency of a arnot enge operatg between and is η arnot = 0. 0 = = =, Assess: he efficiency is much less than the arnot efficiency Model: he closed cycle of the heat enge volves the followg four processes: isothermal expansion, isochoric coolg, isothermal compression, and isochoric heatg. For a monatomic gas V = R. Visualize: Solve: Usg the ideal-gas law, nr (0.0 mol)(8. J/mol K)(600 K) p = = = Pa V.0 0 m At pot, because of the isothermal conditions, = = 600 K and V.0 0 m p = p = ( Pa) =.9 0 Pa V.0 0 m opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

26 9-6 hapter 9 At pot, because it is an isochoric process, V = V = 000 cm and Likewise at pot, = = 00 K and 00 K p = p = = 600 K (.9 0 Pa).7 0 Pa V.0 0 m p = p = (.7 0 Pa) =.9 0 Pa V.0 0 m Let us now calculate net = For the isothermal processes, ln V = nr V = (0.0 mol)(8. J/mol K)(600 K) ln() = 69. J ln V = = (0.0 mol)(8. J/mol K)(00 K) ln =.6 J nr V For the isochoric processes, = = 0 J. hus, the work done per cycle is net =.6 J 0 J. Because = S + E th, = + ( Eth) = 69. J+ 0 J = 69. J For the first isochoric process, For the second isothermal process For the second isochoric process, = n = (0.0 mol) R ( ) V = (0.0 mol) (8. J/mol K)(00 K 600 K) = 77.9 K = + ( E ) =.6 J+ 0 J =.6 J th ( ) = n = n R ( ) V = (0.0 mol) (8. J/mol K)(600 K 00 K) = 77.9 K hus, = + = 9. J. he thermal efficiency of the enge is net.6 J η = = = 0. =, 9. J 9.6. Model: Processes and are isobaric. Processes and are isochoric. Visualize: Solve: (a) Except an adiabatic process, heat must be transferred to the gas to raise its temperature. hus heat is transferred durg processes and. his is the reverse of the heat enge Example 9.. opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

27 eat Enges and Refrigerators 9-7 (b) eat flows from hot to cold. Sce heat energy is transferred to the gas durg processes and, which end with the gas at temperature 700 K, the reservoir temperature must be > 700 K. his is the hot reservoir, so the heat transferred is. Similarly, heat energy is transferred out of the gas durg processes and. his requires that the reservoir temperature be < 00 K. his is the cold reservoir, and the energy transferred durg these two processes is. (c) he heat energies were calculated Example 9., but now they have the opposite signs. = + = J J =.8 0 J = + =.7 0 J J = 6. 0 J (d) For a counterclockwise cycle the pv diagram, the work is. Its value is the area side the curve, which is = ( p) ( V) = ( 0,00 Pa )( m ) =.0 0 J. Note that = Η, as expected from energy conservation. (e) No. A refrigerator uses work put to transfer heat energy from the cold reservoir to the hot reservoir. his device uses work put to transfer heat energy from the hot reservoir to the cold reservoir Solve: (a) If you wish to build a arnot enge that is 80% efficient and exhausts heat to a cold reservoir at 0, what temperature ( ) must the hot reservoir be? (b) (0 + 7) = = 0.0 =. 0 ( + 7) Solve: (a) A refrigerator with a coefficient of performance of.0 exhausts 00 J of heat each cycle. hat work is required each cycle and how much heat is removed each cycle from the cold reservoir? (b) e have.0 = / =. his means 00 J = + = + = = = = 0 J ence, = = 00 J 0 J = 80 J Solve: (a) A heat enge operates at 0% efficiency and produces 0 J of work each cycle. hat is the net heat extracted from the hot reservoir and the net heat exhausted each cycle? (b) e have 0.0 = /. Usg the first law of thermodynamics, Substitutg to the defition of efficiency, = = 0 J = 0 J out 0 J 0 J 0 J 0 J 0.0 = = + = = = 00 J 0.0 he heat exhausted is = 0 J = 00 J 0 J = 80 J Solve: (a) opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

28 9-8 hapter 9 In this heat enge, 00 kj of work is done each cycle. hat is the maximum pressure? (b) (.0 0 Pa)(.0 m ).0 0 J.0 0 Pa 00 kpa max max p = p = = Model: he heat enge follows a closed cycle, startg and endg the origal state. Visualize: he figure dicates the followg seven steps. First, the p is serted when the heat enge has the itial conditions. Second, heat is turned on and the pressure creases at constant volume from to atm. hird, the p is removed. he flame contues to heat the gas and the volume creases at constant pressure from 0 cm to 00 cm. Fourth, the p is serted and some of the weights are removed. Fifth, the contaer is placed on ice and the gas cools at constant volume to a pressure of atm. Sixth, with the contaer still on ice, the p is removed. he gas contues to cool at constant pressure to a volume of 0 cm. Seventh, with no ice or flame, the p is serted back and the weights returned brgg the enge back to the itial conditions and ready to start over. Solve: (a) (b) he work done per cycle is the area side the curve: 6 = ( p)( V) = ( 0,00 Pa)(0 0 m ) = 0 J (c) eat energy is put durg processes and, so = +. his is a diatomic gas, with V = R and = R. he number of moles of gas is P 6 pv (0,00 Pa)(0 0 m ) n = = = mol R (8. J/mol K)(9 K) Process is isochoric, so = ( p/ p) = = 879 K. Process is isobaric, so = ( V / V ) = = 78 K. hus Similarly, V = n = nr( ) = ( mol)(8. J/mol K)(86 K) =. J 7 7 P = n = nr( ) = ( mol)(8. J/mol K)(879 K) =.8 J hus =. J +.8 J = 78.0 J and the enge s efficiency is 0. J η = = = 0. =, 78.0 J Model: System undergoes an isochoric process and system undergoes an isobaric process. Solve: (a) eat will flow from system to system because system is hotter. Because there is no heat put from (or loss to) the outside world, we have + = 0 J. eat, which is negative, will change the temperature of system. eat will both change the temperature of system and do work by liftg the piston. But these consequences of heat flow don t change the fact that + = 0 J. System undergoes constant volume coolg from opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.

29 eat Enges and Refrigerators 9-9 i = 600 K to. System, whose pressure is controlled by the weight of the piston, undergoes constant pressure f heatg from i = 00 K to f. hus, + = 0J = n ( ) + n ( ) = n R ( ) + n R ( ) Solvg this equation for f gives V f i P f i f i f i n i + ni (0.060 mol)(600 K) + (0. 00 mol)(00 K) f = = = 6 K n + n ( mol) + (0. 00 mol) (b) Knowg f, we can compute the heat transferred from system to system : (c) he change of thermal energy system is = n ( ) = n R ( ) = 0 J P f i f i th V f i E = n = n R ( ) = = 6. J Accordg to the first law of thermodynamics, = S + E th. hus, the work done by system is S = Eth = 0.0 J 6. J = 0.8 J. he work is done to lift the weight of the cylder and the air above it by a height y. he weight of the air is wair = pa = ppr = (0. 0 N/m ) p(0.00 m) = 79.6 N. herefore, 0.8 J = ( w + w ) y y = 0.00 m ( + ) = (. 0 kg)(9.8 m/s ) N = s s cyl air w cyl wair (d) he fraction of heat converted to work is s 0.8 J = 0.0 J = =, 9.7. Model: Process and process are adiabatic, and process and process are isochoric. Visualize: Please refer to Figure P9.7. Solve: (a) For adiabatic process, = 0J and pv p V nr( ) = = γ γ For isochoric process, = 0 J and = nv( ). For adiabatic process, = 0J and pv pv nr( ) = = γ γ For isochoric process, = 0 J and = nv( ). he work done per cycle is nr( ) nr( ) nr net = = + 0J+ + 0J = ( + ) γ γ γ (b) he thermal efficiency of the heat enge is out V( ) n η = = = = n ( ) V he last step follows from the fact that > and >. e will now simplify this expression further as follows: Similarly, γ γ γ γ γ γ V γ = = = = = V pv pvv nrv nr V nr V r = r γ he equation for thermal efficiency now becomes. η = = r r r γ γ γ opyright 0 Pearson Education, Inc. All rights reserved. his material is protected under all copyright laws as they currently exist.