A Carnot engine is operating between the two reservoirs of temperature 227 and 127 respectively. If the engine performs 1.04 × 10⁵ J work per cycle. Then, the heat absorbed from hot reservoir per cycle will be: (a) 2.08 × 10⁴ J (b) 1.04 × 10⁵ J (c) 5.20 × 10⁵ J (d) 2.36 × 10⁵ J

Answer: (c). T₁ = 500 K, T₂ = 400 K, so η = 1 - (400)/(500) = 0.2 Q₁ = W/η = (1.04 × 10⁵)/(0.2) = 5.20 × 10⁵ J So the answer is (c).

A Carnot engine is operating between the two reservoirs of… | JEE Main 2022 (30 Jul, Shift 2)

Q30JEE Main 2022 · 30 JUL · SHIFT 2Easy

Heat Engines, Refrigerators & Second Law

A Carnot engine is operating between the two reservoirs of temperature $227^\circ\text{C}$ and $127^\circ\text{C}$ respectively. If the engine performs $1.04 \times 10^{5}\text{ J}$ work per cycle. Then, the heat absorbed from hot reservoir per cycle will be:

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Answer: (c)

$T_1 = 500$ K, $T_2 = 400$ K, so $\eta = 1 - \frac{400}{500} = 0.2$
$Q_1 = \frac{W}{\eta} = \frac{1.04 \times 10^5}{0.2} = 5.20 \times 10^5$ J
So the answer is (c).

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