Work done by a Carnot engine operating between temperatures 127 and 27 is 2 kJ. The amount of heat transferred to the engine by the reservoir is: (a) 8 kJ (b) 2 kJ (c) 4 kJ (d) 2.67 kJ

Answer: (a). T₁ = 400 K, T₂ = 300 K, so η = 1 - (300)/(400) = 0.25 Q₁ = W/η = 2/0.25 = 8 kJ So the answer is (a) 8 kJ.

Work done by a Carnot engine operating between temperatures 127 and… | JEE Main 2023 (8 Apr, Shift 2)

Q21JEE Main 2023 · 8 APR · SHIFT 2Easy

Heat Engines, Refrigerators & Second Law

Work done by a Carnot engine operating between temperatures $127^\circ\text{C}$ and $27^\circ\text{C}$ is $2\text{ kJ}$ . The amount of heat transferred to the engine by the reservoir is:

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Answer: (a)

$T_1 = 400$ K, $T_2 = 300$ K, so $\eta = 1 - \frac{300}{400} = 0.25$
$Q_1 = \frac{W}{\eta} = \frac{2}{0.25} = 8$ kJ
So the answer is (a) 8 kJ.

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