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JEE Main 2022 (27 Jun, Shift 1) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2022 (27 Jun, Shift 1)

Q38JEE Main 2022 · 27 JUN · SHIFT 1Medium

Heat Engines, Refrigerators & Second Law

In a carnot engine, the temperature of reservoir is 527 $^\circ$ C and that of sink is 200 K. If the work done by the engine when it transfers heat from reservoir to sink is 12000 kJ, the quantity of heat absorbed by the engine from reservoir is ______________ $\times$ 106 J.

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

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Answer: 16

$T_1 = 527 + 273 = 800\text{ K}$ , $T_2 = 200\text{ K}$ : $\eta = 1 - \frac{200}{800} = 0.75$ .
$Q_1 = \frac{W}{\eta} = \frac{12000\text{ kJ}}{0.75} = 16000\text{ kJ}$
$Q_1 = 16 \times 10^6\text{ J}$ , so the answer is 16.

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