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A heat engine works between the temperature of a hot reservoir, T_h, and a low t

ID: 2059249 • Letter: A

Question

A heat engine works between the temperature of a hot reservoir, T_h, and a low temp reservoir that is at the environmental temperature T_e . An inventor suggests that the efficiency of this heat engine can be increased by using a refrigerator to lower the temperature of the cold reservoir to T_l (the hot reservoir for the fridge is effectively the environment). This refrigerator will use some but obviously not all of the work produced by the engine. Assume the theoretical maximum efficiency for the both the engine and the fridge, and assume the refrigerator consumes a fraction a of the work done by the engine where 0

Explanation / Answer

initial work output from heat engine, Wi=Qh-Qe
when the refrigerator is introduced,
so now the work output of heat engine increases to Wf=Qh-Ql

work input to the refrigrator, Wr=aWf

so net available work, W=Wf-Wr=(1-a)Wf=(1-a)(Qh-Ql)
Thus ratio of net available work and the heat extracted from hot reservoir is
W/Qh=(1-a)(T_h-T_l)/T_h if all cycles operate at maximum efficiency (Carnot cycles)
=(1-a)(1-T_l/T_h)

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