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

Solution

Physics sir · 4 steps

Same engine run backwards: η = 1/10 tells you the work is one tenth of the heat at the hot end.

1. Reverse the engine. A Carnot engine run backwards is a refrigerator with the same reservoirs and the same heats, only the arrows flip. So the relation W = η Q_H still holds, where Q_H is the heat exchanged with the hot reservoir.
2. Heat at hot end. With W = 10 J and η = 1/10, the heat rejected to the hot reservoir is Q_H = W/η = 100 J.

   Q_L = Q_H - W = 100 J - 10 J = 90 J

3. Check with COP. The refrigerator's coefficient of performance is β = (1-η)/η = 0.9/0.1 = 9. Then Q_L = β W = 9 × 10 J = 90 J, which agrees.
4. So. The heat absorbed from the cold reservoir is 90 J, option (a).

The idea to keep. For a reversed Carnot engine, β = (1-η)/η. Energy balance Q_H = Q_L + W always holds, so the hot end gets more heat than the cold end gives.

Answer from the NTA final answer key. Solution by PrepWiser.