A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40, its efficiency increases by 30%. The temperature of the source will be: (a) 166.7 K (b) 255.1 K (c) 266.7 K (d) 367.7 K

Answer: (c). η = 0.5 ⇒ T₂ = 0.5T₁. Efficiency increases by 30% of itself: η' = 0.5 × 1.3 = 0.65, so T₂ - 40 = 0.35T₁. Subtract: 0.15T₁ = 40 ⇒ T₁ = 266.7 K. So the answer is (c) 266.7 K.

A Carnot engine has efficiency of 50%. If the temperature of sink is… | JEE Main 2022 (28 Jul, Shift 1)

Q32JEE Main 2022 · 28 JUL · SHIFT 1Medium

Heat Engines, Refrigerators & Second Law

A Carnot engine has efficiency of $50\%$ . If the temperature of sink is reduced by $40^\circ\text{C}$ , its efficiency increases by $30\%$ . The temperature of the source will be:

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Answer: (c)

$\eta = 0.5 \Rightarrow T_2 = 0.5T_1$ .
Efficiency increases by 30% of itself: $\eta' = 0.5 \times 1.3 = 0.65$ , so $T_2 - 40 = 0.35T_1$ .
Subtract: $0.15T_1 = 40 \Rightarrow T_1 = 266.7\text{ K}$ .
So the answer is (c) 266.7 K.

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