Carnot engine operating between temperatures T₁ and T₂ has efficiency 0.2. When T₂ is lowered by 45 K, its efficiency becomes 0.5. Temperatures T₁ and T₂ are respectively (a) 150 K, 120 K (b) 120 K, 150 K (c) 60 K, 80 K (d) 80 K, 60 K

Answer: (a). 0.2 = 1 - (T₂)/(T₁) ⇒ T₂ = 0.8T₁ 0.5 = 1 - (T₂ - 45)/(T₁) ⇒ T₂ - 45 = 0.5T₁ Subtract: 0.3T₁ = 45 ⇒ T₁ = 150 K, T₂ = 120 K So the answer is (a).

Carnot engine operating between temperatures T₁ and T₂ has efficiency… | MHT CET 2026 (11 Apr, Shift 2)

Q7MHT CET 2026 · 11 APR · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $0.2$ . When $T_2$ is lowered by $45$ K, its efficiency becomes $0.5$ . Temperatures $T_1$ and $T_2$ are respectively

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Answer: (a)

$0.2 = 1 - \frac{T_2}{T_1} \Rightarrow T_2 = 0.8T_1$
$0.5 = 1 - \frac{T_2 - 45}{T_1} \Rightarrow T_2 - 45 = 0.5T_1$
Subtract: $0.3T_1 = 45 \Rightarrow T_1 = 150$ K, $T_2 = 120$ K
So the answer is (a).

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