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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°C, the efficiency of the engine is doubled. The temperatures of the source and sink are

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90°C, 37°C

80°C, 37°C

95°C, 28°C

99°C, 37°C

answer is D.

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Detailed Solution

Initially,

$η = (1 - \frac{T_{2}}{T_{1}}) = \frac{W}{Q} = \frac{1}{6}$ ......(i)
Finally,

$η^{'} = (1 - \frac{T_{2}^{'}}{T_{1}}) = (1 - \frac{(T_{2} - 62)}{T_{1}}) = 1 - \frac{T_{2}}{T_{1}} + \frac{62}{T_{1}}$
$\Rightarrow η^{'} = η + \frac{62}{T_{1}}$ ......(ii)
It is given that $η^{'} = 2 η .$

using equation (i) and (ii),
$\Rightarrow T_{1} = 372 K = 99^{\circ} C and T_{2} = 310 K = 37^{\circ} C$

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