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Question:

A Carnot engine, having an efficiency of η=1/10 as heat engine, is used as a refrigerator. If the work done on the system is 10J, the amount of energy absorbed from the reservoir at lower temperature is:

90J

100J

1J

99J

Solution:

efficiency of carnot cycle is given as η=W/QH where W is work done and QH is amount of heat added to system.
QH=W/η=100J
QC=QH−W=100−10=90J