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# Sol-Ch21 - CHAPTER 21 E xercises 1(a Q H = 4 W = 800 J(b Q...

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Exercises 1. (a) Q H = 4W = 800 J; (b) Q C = Q H – W = 600 J 2. W = Q H – Q C = 250 J, P = 250/0.4 = 625 W 3. (a) P = W/T =500 W; (b) COP = Q C /W = 100/25 = 4 4. (a) W = Q C /CPO = 80/3.5 = 22.9 J; (b) Q H = 102.9 J 5. Q H = 4W = 40 kWh 6. dQ H /dt = (dW/dt)/ ϵ = 333.3 MW, and dQ C /dt = 233.3 MW 0.8(233.3) = (dm/dt)cΔT, thus dm/dt = 5560 kg/s 7. (a) dQ H /dt = 30/0.22 = 136 kW; (b) dQ C /dt = dQ H /dt – dw/dt = 106 kW; (c) (1.36 ×   10 8 J/s) (3600 s/h) (1.3 × 10 8 J/gal) = 3.77 gal/h 8. (a) dW/dt = 1 kW; dQ C /dt = 1.758 kW; since COP = Q H /W, we find COP = 2.76; (b) 2.76 kW 9. ϵ C = 1 – 283/363 = 22% 10. ϵ C = 1 – 300/400 = 1/4, so Q H = 4W = Q C + W, thus W = 110 J 11. (a) ϵ C = 1 – 278/295 = 5.76 %; (b) Q H = W/ ϵ C = Q C + W, thus dQ C /dt = 16.4 MW 12. ϵ C = 1 – 350/600 = 0.417 ; dQ H /dt = W/ ϵ C = 500/0.417 = 1200 W, thus dQ C /dt = 700 W 13. (a) \$12.00 ; (b) ϵ C = 1 – 273/293 = 0.0683. dW/dt = ϵ C dQ H /dt = 0.341 kW, thus W = 8.18 kW.h, which costs \$0.82 14. Q H /Q C = 293/273, thus Q H = 1073 J CHAPTER 21

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15. (a) Q H /Q C = 293/273, thus Q H = 1073 J (b) W = Q H – Q C = 73 J 16. (a) Q H = Q C / (1 – ϵ C ) =200/0.65 = 308 J (b) T H = T C /(1 – ϵ C ) = 300/0.65 = 462 K 17. ϵ C =1 – 293/472 = 0.381 ; thus dQ H /dt = (dW/dt)/ ϵ C = 945 W, and dQ C /dt = 945 – 360 = 585 W. For a 0.2 s cycle, Q H =1 89 J, Q C =117 J 18. (a) Carnot COP = QH/W = T H /(T H – T C ) = 298/25 = 11.92 so actual COP = 7.152 = (Q C + W)/W, thus W = 16.3 J (b) Q H = Q C + W = 116.3 J 19. (a) ϵ C = 1 –313/553 = 0.434; thus W = ϵ C Q H = 434 J; (b) COP = Q H /W = 1/ ϵ C = 2.30 20. ϵ 1 = [(T H – T C )+5]/(T H + 5), so Δ ϵ 1 = 5/(T H + 5) ϵ 2 = [(T H – T C )+5]/T H , so Δ ϵ 2 = –5/T H Thus a decrease in T c has a greater effect on 21. (a) ΔS = mL/T = 6.06 kJ/K; (b) ΔS = mL/T = 1.22 kJ/K; (c) ΔS = mc ln(373/273) = 1.31 kJ/K 22.
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