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JEE Main 2019 (12 Apr, Shift 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2019 (12 Apr, Shift 2)

Q58JEE Main 2019 · 12 APR · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

A Carnot engine has an efficiency of $\frac{1}{6}$ . When the temperature of the sink is reduced by 62ºC, its efficiency is doubled. The temperatures of the source and the sink are, respectively

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Answer: (a)

$T_2 = \frac{5}{6}T_1$ and $T_2 - 62 = \frac{2}{3}T_1$ .
Subtract: $\frac{T_1}{6} = 62 \Rightarrow T_1 = 372\text{ K} = 99^\circ\text{C}$ .
$T_2 = 310\text{ K} = 37^\circ\text{C}$
So the answer is (a) $99^\circ\text{C},\ 37^\circ\text{C}$ .

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