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JEE Main 2020 (8 Jan, Shift 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2020 (8 Jan, Shift 2)

Q53JEE Main 2020 · 8 JAN · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

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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Answer: (a)

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 (a) 90 J.

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