Question:
A carnot engine having an efficiency of \(\frac{1}{10}\) 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 :
A carnot engine having an efficiency of \(\frac{1}{10}\) 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 :
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For a Carnot cycle \(\eta=1-\frac{T_2}{T_1}\) and the refrigerator COP is \(\frac{T_2}{T_1-T_2}=\frac{1-\eta}{\eta}\).
Updated On: Nov 5, 2026
- 90 J
- 99 J
- 1 J
- 100 J
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The Correct Option is A
Solution and Explanation
\(\eta=1-\frac{T_2}{T_1}=\frac{1}{10}\Rightarrow\frac{T_2}{T_1}=\frac{9}{10}\).
Coefficient of performance as a refrigerator: \(\beta=\frac{T_2}{T_1-T_2}=\frac{9/10}{1/10}=9\).
\(\beta=\frac{Q_2}{W}\Rightarrow Q_2=9\times10=90\) J.
Answer: (1) 90 J
Coefficient of performance as a refrigerator: \(\beta=\frac{T_2}{T_1-T_2}=\frac{9/10}{1/10}=9\).
\(\beta=\frac{Q_2}{W}\Rightarrow Q_2=9\times10=90\) J.
Answer: (1) 90 J
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