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A Carnot's engine works as a refrigerator between 250 K and 300 K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:

1. 420 J
2. 2100 J
3. 772 J
4. 2520 J

Correct answer: 420 J

Solution

For a Carnot refrigerator, W = Q_cold*(Th - Tc)/Tc. Q_cold = 500 cal = 500*4.186 = 2093 J. W = 2093*(300-250)/250 = 2093*0.2 = 418.6 J, approximately 420 J.

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