Question:
A Carnot's engine works as a refrigerator between \(250\ \mathrm{K}\) and \(300\ \mathrm{K}\). 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 :
A Carnot's engine works as a refrigerator between \(250\ \mathrm{K}\) and \(300\ \mathrm{K}\). 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 :
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Coefficient of performance of a Carnot refrigerator is \(\frac{T_L}{T_H-T_L}=\frac{Q_L}{W}\). Convert calories with 1 cal = 4.2 J.
Updated On: Nov 11, 2026
- 420 J
- 772 J
- 2100 J
- 2520 J
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The Correct Option is A
Solution and Explanation
COP \(=\frac{Q_L}{W}=\frac{T_L}{T_H-T_L}=\frac{250}{300-250}=5\).
\(W=\frac{Q_L}{5}=\frac{500}{5}=100\ \mathrm{cal}\).
With \(1\ \mathrm{cal}=4.2\ \mathrm{J}\): \(W=100\times4.2=420\ \mathrm{J}\).
Answer: (1) 420 J
\(W=\frac{Q_L}{5}=\frac{500}{5}=100\ \mathrm{cal}\).
With \(1\ \mathrm{cal}=4.2\ \mathrm{J}\): \(W=100\times4.2=420\ \mathrm{J}\).
Answer: (1) 420 J
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