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A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by $62^{\circ} C$ , the efficiency of the engine is doubled. Then the temperature of the source and sink are

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$90^{\circ} C, 37^{\circ} C$

$99^{\circ} C, 37^{\circ} C$

$80^{\circ} C, 37^{\circ} C$

$95^{\circ} C, 28^{\circ} C$

answer is D.

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Detailed Solution

$η = \frac{W}{Q} = 1 - \frac{T_{2}}{T_{1}} \Rightarrow η = \frac{(\frac{1}{6} Q)}{Q} = 1 - \frac{T_{2}}{T_{1}} \Rightarrow η = \frac{1}{6} = 1 - \frac{T_{2}}{T_{1}}$

$\Rightarrow η = \frac{1}{6} and \frac{T_{2}}{T_{1}} = \frac{5}{6} - - - (1)$

$η^{'} = 2 η = 1 - \frac{(T_{2} - 62)}{T_{1}}$

$\Rightarrow 2 (\frac{1}{6}) = 1 - \frac{T_{2}}{T_{1}} + \frac{62}{T_{1}}$

$\Rightarrow \frac{1}{3} = \frac{1}{6} + \frac{62}{T_{1}} \Rightarrow \frac{62}{T_{1}} = \frac{1}{3} - \frac{1}{6} = \frac{2 - 1}{6} = \frac{1}{6}$

$T_{1} = 372 K - - - (2)$

$\frac{1}{6} = 1 - \frac{T_{2}}{372} \Rightarrow \frac{T_{2}}{372} = 1 - \frac{1}{6} = \frac{5}{6}$

$T_{2} = \frac{5}{6} \times 372 = 310 - - - (3)$

$T_{1} = 372 - 273 = 99^{\circ} C; T_{2} = 310 - 273 = 37^{\circ} C$

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