Heat engine operating between temperature T₁ and T₂ has efficiency 1/6. When T₂ is lowered by 62 K, its efficiency increases to 1/3. Then T₁ and T₂ respectively are (a) 372 K, 310 K (b) 372 K, 330 K (c) 330 K, 268 K (d) 310 K, 248 K

Answer: (a). η = 1 - (T₂)/(T₁). First case: T₂ = 5/6T₁. Second case: T₂ - 62 = 2/3T₁. Subtract: T₁/6 = 62 ⇒ T₁ = 372 K, T₂ = 310 K. So the answer is (a) 372 K, 310 K.

Heat engine operating between temperature T₁ and T₂ has efficiency… | MHT CET 2024 (15 May, Shift 2)

Q15MHT CET 2024 · 15 MAY · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

Heat engine operating between temperature $T_1$ and $T_2$ has efficiency $\frac{1}{6}$ . When $T_2$ is lowered by 62 K, its efficiency increases to $\frac{1}{3}$ . Then $T_1$ and $T_2$ respectively are

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Answer: (a)

$\eta = 1 - \frac{T_2}{T_1}$ . First case: $T_2 = \frac{5}{6}T_1$ .
Second case: $T_2 - 62 = \frac{2}{3}T_1$ .
Subtract: $\frac{T_1}{6} = 62 \Rightarrow T_1 = 372\text{ K}$ , $T_2 = 310\text{ K}$ .
So the answer is (a) $372\text{ K},\ 310\text{ K}$ .

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