If the ratio of the absolute temperature of t | vedclass

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(A) For a Carnot engine,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_2$ is the temperature of the sink and $T_1$ is the tem

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$25$

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(A) For a Carnot engine,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_2$ is the temperature of the sink and $T_1$ is the temperature of the source.Initially,the ratio $\frac{T_2}{T_1} = \frac{2}{3}$. Therefore,$\eta_1 = 1 - \frac{2}{3} = \frac{1}{3}$.Finally,the ratio $\frac{T_2'}{T_1'} = \frac{3}{4}$. Therefore,$\eta_2 = 1 - \frac{3}{4} = \frac{1}{4}$.The percentage change in efficiency is given by $\frac{\eta_1 - \eta_2}{\eta_1} 	imes 100$.Substituting the values: $\frac{\frac{1}{3} - \frac{1}{4}}{\frac{1}{3}} 	imes 100 = \frac{\frac{4-3}{12}}{\frac{1}{3}} 	imes 100 = \frac{1/12}{1/3} 	imes 100 = \frac{3}{12} 	imes 100 = \frac{1}{4} 	imes 100 = 25 \%$.

If the ratio of the absolute temperature of the sink and source of a Carnot engine is changed from $2:3$ to $3:4$,the efficiency of the engine changes by (in $\%$)

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