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JEE Main 2020 (7 Jan, Shift 1) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2020 (7 Jan, Shift 1)

Q51JEE Main 2020 · 7 JAN · SHIFT 1Easy

Heat Engines, Refrigerators & Second Law

A Carnot engine operates between two reservoirs of temperatures 900 K and 300 K. The engine performs 1200 J of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle, is _______.

Numerical answer type. Work it out, then open the answer.

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Answer: 600

$\eta = 1 - \frac{300}{900} = \frac{2}{3}$
$Q_1 = \frac{W}{\eta} = \frac{1200}{2/3} = 1800$ J
$Q_2 = Q_1 - W = 1800 - 1200 = 600$ J

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