Question:
A Carnot engine having an efficiency of 20 % is used as a refrigerator. If the amount of heat absorbed from the reservoir at low temperature is 200 J, then the work done on the refrigerator is:
A Carnot engine having an efficiency of 20 % is used as a refrigerator. If the amount of heat absorbed from the reservoir at low temperature is 200 J, then the work done on the refrigerator is:
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For a refrigerator, Work = Heat Absorbed / COP.
Updated On: Jun 6, 2026
- 150 J
- 50 J
- 100 J
- 75 J
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Coefficient of Performance ($COP$) of a refrigerator.
Step 2: Meaning
$COP = \frac{T_2}{T_1 - T_2} = \frac{Q_2}{W}$. Efficiency $\eta = 1 - \frac{T_2}{T_1} = 0.2$.
Step 3: Analysis
$\frac{T_2}{T_1} = 0.8 \rightarrow T_1 = \frac{T_2}{0.8} = 1.25 T_2$. $COP = \frac{T_2}{1.25 T_2 - T_2} = \frac{1}{0.25} = 4$. $W = \frac{Q_2}{COP} = \frac{200}{4} = 50$ J.
Step 4: Conclusion
The work done is 50 J.
Final Answer: (B)
Coefficient of Performance ($COP$) of a refrigerator.
Step 2: Meaning
$COP = \frac{T_2}{T_1 - T_2} = \frac{Q_2}{W}$. Efficiency $\eta = 1 - \frac{T_2}{T_1} = 0.2$.
Step 3: Analysis
$\frac{T_2}{T_1} = 0.8 \rightarrow T_1 = \frac{T_2}{0.8} = 1.25 T_2$. $COP = \frac{T_2}{1.25 T_2 - T_2} = \frac{1}{0.25} = 4$. $W = \frac{Q_2}{COP} = \frac{200}{4} = 50$ J.
Step 4: Conclusion
The work done is 50 J.
Final Answer: (B)
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