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JEE Main 2007 (AIEEE) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2007 (AIEEE)

Q80JEE Main 2007 · AIEEEMedium

Heat Engines, Refrigerators & Second Law

A Carnot engine, having an efficiency of $\eta = 1/10$ 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

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Answer: (c)

Carnot refrigerator: $\beta = \frac{1-\eta}{\eta} = \frac{9/10}{1/10} = 9$ .
$Q_2 = \beta W = 9 \times 10 = 90$ J
So the answer is (c) $90$ J.

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