A Carnot engine has efficiency 1/6. It becomes 1/3, when the temperature of sink is lowered by 57 K. The temperature of the source is (a) 171 K (b) 399 K (c) 342 K (d) 285 K

Answer: (c). η = 1 - (T₂)/(T₁), so lowering T₂ by Δ T raises η by (Δ T)/(T₁) 1/3 - 1/6 = 57/T₁ ⇒ 1/6 = 57/T₁ T₁ = 342 K. So the answer is (c).

A Carnot engine has efficiency 1/6. It becomes 1/3, when the… | MHT CET 2025 (21 Apr, Shift 2)

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Heat Engines, Refrigerators & Second Law

A Carnot engine has efficiency $\frac{1}{6}$ . It becomes $\frac{1}{3}$ , when the temperature of $\text{sink}$ is lowered by 57 K. The temperature of the source is

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Answer: (c)

$\eta = 1 - \frac{T_2}{T_1}$ , so lowering $T_2$ by $\Delta T$ raises $\eta$ by $\frac{\Delta T}{T_1}$
$\frac{1}{3} - \frac{1}{6} = \frac{57}{T_1} \Rightarrow \frac{1}{6} = \frac{57}{T_1}$
$T_1 = 342$ K. So the answer is (c).

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