Question:

A Carnot engine has an efficiency of \(\frac{1}{6}\). When the temperature of the sink is reduced by \(62^\circ\mathrm{C}\), its efficiency is doubled. The temperatures of the source and the sink are, respectively

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Use \(\eta = 1 - T_2/T_1\) in kelvin for both situations and solve the two equations.
Updated On: Nov 11, 2026
  • \(99^\circ\mathrm{C},\ 37^\circ\mathrm{C}\)
  • \(124^\circ\mathrm{C},\ 62^\circ\mathrm{C}\)
  • \(37^\circ\mathrm{C},\ 99^\circ\mathrm{C}\)
  • \(62^\circ\mathrm{C},\ 124^\circ\mathrm{C}\)
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The Correct Option is A

Solution and Explanation

\(\eta = 1-\frac{T_2}{T_1} = \frac{1}{6}\Rightarrow T_2 = \frac{5}{6}T_1\).
After lowering the sink by 62 K: \(1-\frac{T_2-62}{T_1} = \frac{1}{3}\Rightarrow T_2 - 62 = \frac{2}{3}T_1\).
Subtract: \(\frac{5}{6}T_1 - \frac{2}{3}T_1 = 62\Rightarrow \frac{T_1}{6} = 62\Rightarrow T_1 = 372\ \mathrm{K} = 99^\circ\mathrm{C}\).
\(T_2 = \frac{5}{6}\times 372 = 310\ \mathrm{K} = 37^\circ\mathrm{C}\).
Answer: (1) \(99^\circ\mathrm{C},\ 37^\circ\mathrm{C}\)
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