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Q.

Efficiency of a heat engine whose sink is at temperature of 300 K is 40%. To increase the efficiency to 60%, keeping the sink temperature constant, the source temperature must be increased by

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a

150 K

b

200 K

c

250 K

d

300 K

answer is C.

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Detailed Solution

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$\begin{array}{l} \frac{T_{2}}{T_{1}} = 1 - η = 1 - \frac{40}{100} = \frac{3}{5} \\ T_{1} = \frac{5 T_{2}}{3} = \frac{5 \times 300}{3} = 500 K \\ η = 60 % \frac{T_{2}}{T_{1}} = 1 - \frac{60}{100} = \frac{2}{5} \\ T_{1}^{1} = \frac{5}{2} \times 300 = 750 K \\ ΔT = T_{1}^{1} - T_{1} = 750 - 500 = 250 K \end{array}$

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