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Q.

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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a

$80^{0} C, 37^{0} C$

b

$95^{0} C, 28^{0} C$

c

$90^{0} C, 37^{0} C$

d

$99^{0} C, 37^{0} C$

answer is D.

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Detailed Solution

$η = 1 - \frac{T_{2}}{T_{1}} (or) 1 - \frac{T_{L}}{T_{H}}$

Given initially $η = \frac{1}{6}$

$1 - \frac{T_{L}}{T_{H}} = \frac{1}{6} \Rightarrow 5 T_{H} - 6 T_{L} = 0$ ...(i)

$η_{2} = 1 - \frac{T_{L} - 62}{T_{H}} = 2 η_{1} = \frac{1}{3}$

$2 T_{H} - 3 T_{L} = - 186$ ...(ii)

On solving (i) and (ii).

We get, $T_{H} = 372 K = 99 ° C$

$T_{L} = \frac{5}{6} T_{H} = \frac{5}{6} \times 372 K = 310 K = 37 ° C$

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