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A Carnot engine whose heat sinks at 27°C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

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Increases by 18°C

Increase by 200°C

Increase by 120°C

Increase by 73°

answer is B.

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Detailed Solution

$\begin{aligned} 1 - \frac{300}{T} = 0.25 \\ \frac{300}{T} = 0.75 \\ T = 400 K \end{aligned}$

If efficiency increased by 100% then new efficiency $\Rightarrow$ n’ = 50%

$\begin{aligned} 1 - \frac{300}{T^{'}} = 0 .5 \\ T^{'} = 600 K \end{aligned}$

Increase in temp = 600 – 400

= 200 K or 200°C

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