655_Physics ProblemsTechnical Physics

655_Physics ProblemsTechnical Physics - 657 Chapter 22...

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Chapter 22 657 *P22.63 Like a refrigerator, an air conditioner has as its purpose the removal of energy by heat from the cold reservoir. Its ideal COP is COP K 20 K Carnot = == T TT c hc 280 14 0 . (a) Its actual COP is 0 400 14 0 5 60 .. . a f = Q QQ Qt c c ∆∆ 560 Q t Q t Q t c −= 560100 660 . kW af = Q t c and Q t c = 848 . k W (b) QW Q =+ eng : W t Q t Q t eng kW kW kW =−= = 10 0 8 48 1 52 . (c) The air conditioner operates in a cycle, so the entropy of the working fluid does not change. The hot reservoir increases in entropy by Q T h h = × 10 0 10 3 600 300 120 10 3 5 . . Js s K JK ej bg The cold room decreases in entropy by S Q T c c =− × × 848 10 3600 280 109 10 3 5 . . s K The net entropy change is positive, as it must be:
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