The efficiency of a Carnot's engine is $25 \%$, when the temperature of sink is 300 K . The increase in the temperature of source required for the efficiency to become $50 \%$ is

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Aicharya · Accepted Answer

for carnot engine, $\begin{aligned} & \eta_1=1-\frac{T_2}{T_1} \Rightarrow \frac{1}{4}=1-\frac{300}{T_1} 	herefore T_1=400 \mathrm{~K} \ & 	ext { Again, } \eta_2=1-\frac{T_2}{T_1+x} \Rightarrow \frac{1}{2}=1-\frac{300}{400+x} \ & 	herefore \quad x=200 \mathrm{k} \end{aligned}$

The efficiency of a Carnot's engine is $25 \%$, when the temperature of sink is 300 K . The increase in the…

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