The temperature of the sink of a Carnot's engine is $300 \mathrm{~K}$ and the efficiency of the engine is 0…
The temperature of the sink of a Carnot's engine is $300 \mathrm{~K}$ and the efficiency of the engine is 0.25 . If the temperature of the source of the engine is increased by $100 \mathrm{~K}$, the efficiency of the engine increases by
$0.50$
$0.25$
$0.15$
$0.40$
Solution
Temperature of the sink, $T_2=300 \mathrm{~K}$ Efficiency of the engine, $\eta=0.25$
$\begin{aligned}
& \eta_1=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1} \Rightarrow 0.25=1-\frac{300}{\mathrm{~T}_1} \\
& \frac{300}{\mathrm{~T}_1}=0.75 \Rightarrow \mathrm{T}_1=\frac{300}{0.75}=400 \mathrm{~K} \\
& \eta_2=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1}=1-\frac{300}{400+100} \\
& =1-\frac{300}{500}=\frac{2}{5} \\
& =0.40
\end{aligned}$
The increase in efficiency, $=\eta_2-\eta_1$ $=0.40-0.25=0.15$