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

1. 90 J

2. 99 J

3. 100 J

4. 1 J

Doubtly Asked 30 Oct 2026 NEET UG 2017 neet Save WhatsApp Report

1 answer

Best answer

Answer: 1. 90 J

Key idea: as a refrigerator, coefficient of performance \(=\dfrac{Q_2}{W}=\dfrac{1-\eta}{\eta}\).

\(\eta=\tfrac1{10}\Rightarrow\dfrac{Q_2}{W}=\dfrac{9/10}{1/10}=9\).

\(Q_2=9\times10=90\) J absorbed from the cold reservoir.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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