In a Carnot engine,if the absolute temperatur | vedclass

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(C) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T

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$20$

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(C) The efficiency $\eta$ of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.Given that the source temperature $T_1$ is $25 \%$ more than the sink temperature $T_2$,we have: $T_1 = T_2 + 0.25 T_2 = 1.25 T_2$.Substituting $T_1$ into the efficiency formula: $\eta = 1 - \frac{T_2}{1.25 T_2} = 1 - \frac{1}{1.25}$.Since $1.25 = \frac{5}{4}$,we get: $\eta = 1 - \frac{4}{5} = \frac{1}{5}$.Converting this to percentage: $\eta = \frac{1}{5} 	imes 100 \% = 20 \%$.

In a Carnot engine,if the absolute temperature of the source is $25 \%$ more than the absolute temperature of the sink,then the efficiency of the engine is (in $\%$)

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