The efficiency of a Carnot engine working between $127 C$ and $77 C$ is:

10.5%

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11.5%

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12.5%

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13.5%

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Solution

The correct option is C 12.5%
$T_{L} = 77 ° C = (77 + 273) K = 350 K T_{H} = 127 ° C = (127 + 273) K = 400 K η = (1 - \frac{T_{L}}{T_{H}}) \times 100 = (1 - \frac{350}{400}) \times 100 = \frac{50}{400} \times 100 = 12.5 %$

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