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

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

Correct answer: 90 J

Solution

The efficiency of the Carnot engine is given as 1/10, which means the ratio of work done to heat absorbed is 1/10. When used as a refrigerator, the heat absorbed (Q2) from the lower reservoir is given by Q2 = W / (1 - efficiency). Substituting W = 10 J and efficiency = 1/10, we get Q2 = 10 / (1 - 1/10) = 10 / (9/10) = 90 J.

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