A reversible engine has an efficiency of 1/4. If the temperature of the sink is reduced by 58 ° C, its efficiency becomes double. Calculate the temperature of the sink: (a) 174 ° C (b) 280 ° C (c) 180.4 ° C (d) 382 ° C

Answer: (a). η = 1/4 ⇒ T₂ = 3/4T₁; doubled η = 1/2 ⇒ T₂ - 58 = 1/2T₁. Subtract: T₁/4 = 58 ⇒ T₁ = 232 K. T₂ = 3/4(232) = 174 K So the answer is (a) 174 (the value is 174 K).

A reversible engine has an efficiency of 1/4. If the temperature of… | JEE Main 2021 (31 Aug, Shift 1)

Q44JEE Main 2021 · 31 AUG · SHIFT 1Medium

Heat Engines, Refrigerators & Second Law

A reversible engine has an efficiency of $\frac{1}{4}$ . If the temperature of the sink is reduced by 58 $^\circ$ C, its efficiency becomes double. Calculate the temperature of the sink:

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Answer: (a)

$\eta = \frac{1}{4} \Rightarrow T_2 = \frac{3}{4}T_1$ ; doubled $\eta = \frac{1}{2} \Rightarrow T_2 - 58 = \frac{1}{2}T_1$ .
Subtract: $\frac{T_1}{4} = 58 \Rightarrow T_1 = 232\text{ K}$ .
$T_2 = \frac{3}{4}(232) = 174\text{ K}$
So the answer is (a) 174 (the value is 174 K).

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