A Carnot engine receiving heat at 400K has an efficiency of 25%. The C.O.P. of a Carnot refrigerator working between the same temperature limits is

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Quizplus · Accepted Answer

The efficiency of a Carnot engine is given by 1 - (Tc/Th). Given the efficiency is 25%, we have 0.25 = 1 - (Tc/400), which gives Tc = 300K. The C.O.P. of a Carnot refrigerator is given by Tc/(Th-Tc) = 300/(400-300) = 3.

A Carnot Engine Receiving Heat at 400K Has an Efficiency