Chapter 19

Chapter 19 - 19.1. Solve: (a) The engine has a thermal...

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19.1. Solve: (a) The engine has a thermal efficiency of 40% 0.40 η = = and a work output of 100 J per cycle. The heat input is calculated as follows: out H HH 100 J 0.40 250 J W Q QQ =⇒= = (b) Because out H C WQ Q =− , the heat exhausted is C H out 250 J 100 J 150 J QQW = =
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19.2. Solve: During each cycle, the work done by the engine is out 20 J W = and the engine exhausts C 30 J Q = of heat energy. Because out H C WQ Q =− , H out C 20 J 30 J 50 J QW Q =+ = + = Thus, the efficiency of the engine is C H 30 J 1 1 0.40 50 J Q Q η =
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19.3. Solve: (a) During each cycle, the heat transferred into the engine is H 55 kJ Q = , and the heat exhausted is C 40 kJ Q = . The thermal efficiency of the heat engine is C H 40 kJ 1 1 0.27 27% 55 kJ Q Q η =− = = (b) The work done by the engine per cycle is out H C 55 kJ 40 kJ 15 kJ WQ Q =−= =
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19.4. Solve: The coefficient of performance of the refrigerator is CH i n in in 50 J 20 J 1.5 20 J QQ W K WW −− == = =
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19.5. Solve: (a) The heat extracted from the cold reservoir is calculated as follows: CC C in 4.0 200 J 50 J QQ KQ W =⇒= = (b) The heat exhausted to the hot reservoir is HCi n 200 J 50 J 250 J QQW =+= + =
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19.6. Model: Assume that the car engine follows a closed cycle. Solve: (a) Since 2400 rpm is 40 cycles per second, the work output of the car engine per cycle is out kJ 1 s kJ 500 12.5 s 40 cycles cycle W = (b) The heat input per cycle is calculated as follows: out H H 12.5 kJ 62.5 kJ 0.20 W Q Q η =⇒ = = The heat exhausted per cycle is CHi n 62.5 kJ 12.5 kJ 50 kJ QQW =−= =
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19.7. Solve: The amount of heat discharged per second is calculated as follows: () 9 out out C out H C out 11 1 900 MW 1 1.913 10 W 0.32 WW QW QQ W η ⎛⎞ = = ⇒ = −= × ⎜⎟ + ⎝⎠ That is, each second the electric power plant discharges 9 1.913 10 J × of energy into the ocean. Since a typical American house needs 4 2.0 10 J × of energy per second for heating, the number of houses that could be heated with the waste heat is ( ) ( ) 94 96,000 ×× = .
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19.8. Solve: The amount of heat removed from the water in cooling it down in 1 hour is C water water . Qmc T The mass of the water is () ( )( ) 33 3 3 water water water 1000 kg/m 1 L 100 kg/m 10 m 1.0 kg mV ρ == = = ( )( ) 4 C 1.0 kg 4190 J/kg K 20 C 5 C 6.285 10 J Q ⇒= ° °= × The rate of heat removal from the refrigerator is 4 C 17.46 J/s 3600 s Q × The refrigerator does work 8.0. W 8.0 J/s W to remove this heat. Thus the performance coefficient of the refrigerator is 17.46 J/s 2.2 8.0 J/s K
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19.9. Model: Process A is isochoric, process B is isothermal, process C is adiabatic, and process D is isobaric. Solve: Process A is isochoric, so the increase in pressure increases the temperature and hence the thermal energy. Because th s s and 0 J EQW W Δ=− = , Q increases for process A. Process B is isothermal, so T is constant and hence th 0 E Δ= J. The work done W s is positive because the gas expands. Because st h QW E =+ Δ , Q is positive for process B. Process C is adiabatic, so Q = 0 J. W s is positive because of the increase in volume. Since h 0 J QW E ==+ Δ , th E Δ is negative for process C. Process D is isobaric, so the decrease in volume leads to a decrease in temperature and hence a decrease in the thermal energy. Due to the decrease in volume, W s is negative. Because h Δ , Q also decreases for process D.
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Chapter 19 - 19.1. Solve: (a) The engine has a thermal...

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