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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 10 J, the amount of energy absorbed from the reservoir at lower temperature is

1. 100 J
2. 99 J
3. 90 J
4. 1 J

Correct answer: 90 J

Solution

The efficiency of the Carnot engine as a heat engine is given by η = W/Qh, where W is the work done and Qh is the heat absorbed from the hot reservoir. Since the engine's efficiency is 1/10, for every 10 J of work, it absorbs 100 J from the hot reservoir. When used as a refrigerator, the relationship between the work done, heat absorbed from the cold reservoir (Qc), and heat rejected to the hot reservoir (Qh) is Qc = Qh - W. Thus, Qc = 100 J - 10 J = 90 J.

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