An engine operating between temperatures T₁ and T₂ has efficiency 1/5. When T₂ is lowered by 45 K, its efficiency becomes 1/2. Temperatures T₁ and T₂ are respectively (a) 100 K, 70 K (b) 160 K, 120 K (c) 140 K, 110 K (d) 150 K, 120 K

Answer: (d). η = 1 - (T₂)/(T₁). First case: (T₂)/(T₁) = 4/5 ⇒ T₂ = 0.8T₁. Second case: (T₂ - 45)/(T₁) = 1/2 ⇒ T₂ - 45 = 0.5T₁. Subtract: 0.3T₁ = 45 ⇒ T₁ = 150 K, T₂ = 120 K. So the answer is (d) 150 K, 120 K.

An engine operating between temperatures T₁ and T₂ has efficiency… | MHT CET 2025 (21 Apr, Shift 1)

Q10MHT CET 2025 · 21 APR · SHIFT 1Medium

Heat Engines, Refrigerators & Second Law

An engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac{1}{5}$ . When $T_2$ is lowered by 45 K, its efficiency becomes $\frac{1}{2}$ . Temperatures $T_1$ and $T_2$ are respectively

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Answer: (d)

$\eta = 1 - \frac{T_2}{T_1}$ . First case: $\frac{T_2}{T_1} = \frac{4}{5} \Rightarrow T_2 = 0.8T_1$ .
Second case: $\frac{T_2 - 45}{T_1} = \frac{1}{2} \Rightarrow T_2 - 45 = 0.5T_1$ .
Subtract: $0.3T_1 = 45 \Rightarrow T_1 = 150\text{ K}$ , $T_2 = 120\text{ K}$ .
So the answer is (d) $150\text{ K},\ 120\text{ K}$ .

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