If a Carnot engine works between $127^{0} C$ and $527^{0} C,$ then the efficiency is:

25%

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37.5%

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50%

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75%

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Solution

The correct option is C 50%
$T_{1} = 127 ° C = 400 K$
$T_{2} = 527 ° C = 800 K$
Efficiency= $η = 1 - \frac{T_{1}}{T_{2}}$
= $1 - \frac{400}{800}$ = $\frac{1}{2}$
Percentage efficiency= $\frac{1}{2} \times 100 = 50$ %

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