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Question:

A carnot engine having an efficiency of 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 heat engine η= (QH−QL)/QH = 0.1 (Given)
Coefficient of performance of refrigerator β = QL/(QH−QL) (2)
Or β = QL/(QH × η). (3)
Given : W = QH−QL = 10 J
Putting in (2) we get β = QL/10.. (4)
From (4) and (5), we get
∴ QL/10 = QL/QH × 0.1
⇒ QH = 100 J
Thus heat absorbed at lower temperature QL = QH−W
∴QL = 100 − 10 = 90 J