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JEE Main 2023 (1 Feb, Shift 2) · Heat Engines, Refrigerators & Second Law

Heat Engines, Refrigerators & Second Law question from JEE Main 2023 (1 Feb, Shift 2)

Q26JEE Main 2023 · 1 FEB · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

A Carnot engine operating between two reservoirs has efficiency $\frac{1}{3}$ . When the temperature of cold reservoir is raised by x, its efficiency decreases to $\frac{1}{6}$ . The value of x, if the temperature of hot reservoir is $99^\circ$ C, will be:

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Answer: (b)

$T_1 = 99 + 273 = 372\text{ K}$
$\eta = \frac{1}{3}$ : $T_2 = \frac{2}{3}(372) = 248\text{ K}$
$\eta = \frac{1}{6}$ : $T_2 + x = \frac{5}{6}(372) = 310\text{ K}$
$x = 310 - 248 = 62\text{ K}$ . So the answer is (b) 62 K.

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