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JEE Main 2021 (17 Mar, Shift 1) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2021 (17 Mar, Shift 1)

Q41JEE Main 2021 · 17 MAR · SHIFT 1Easy

Heat Engines, Refrigerators & Second Law

A Carnot's engine working between 400 K and 800 K has a work output of 1200 J per cycle. The amount of heat energy supplied to the engine from the source in each cycle is:

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Answer: (a)

$\eta = 1 - \frac{400}{800} = 0.5$
$Q_1 = \frac{W}{\eta} = \frac{1200}{0.5} = 2400$ J
So the answer is (a) 2400 J.

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