A Carnot engine takes 5000 kcal of heat from a reservoir at 727 ° C and gives heat to a sink at 127 ° C. The work done by the engine is (a) 3 × 106 J (b) Zero (c) 12.6 × 106 J (d) 8.4 × 106 J

Answer: (c). T₁ = 1000 K, T₂ = 400 K: η = 1 - (400)/(1000) = 0.6. W = η Q = 0.6 × 5000 = 3000 kcal W = 3000 × 4200 J = 12.6 × 10⁶ J So the answer is (c) 12.6 × 10⁶ J.

A Carnot engine takes 5000 kcal of heat from a reservoir at 727 ° C… | JEE Main 2022 (24 Jun, Shift 2)

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Heat Engines, Refrigerators & Second Law

A Carnot engine takes 5000 kcal of heat from a reservoir at 727 $^\circ$ C and gives heat to a sink at 127 $^\circ$ C. The work done by the engine is

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Answer: (c)

$T_1 = 1000\text{ K}$ , $T_2 = 400\text{ K}$ : $\eta = 1 - \frac{400}{1000} = 0.6$ .
$W = \eta Q = 0.6 \times 5000 = 3000\text{ kcal}$
$W = 3000 \times 4200\text{ J} = 12.6 \times 10^6\text{ J}$
So the answer is (c) $12.6 \times 10^6$ J.

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