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Carnot’s cycle operates between T and 300 K (T > 300K). it is 25% efficient when temperature of source (T) is

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Detailed Solution

$T_{2} = TK ?; T_{1} = 300 K; η = 25 % = \frac{25}{100} = 0.25$

$η = 1 - \frac{T_{1}}{T_{2}} Or \frac{T_{1}}{T_{2}} = 1 - η = 1 - 0.25 = 0.75$

$T_{2} = \frac{300}{0.75} = 400 K$

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