Q139 / 160 · AP EAMCET 2024 (22 May Shift 2) · Physics
The efficiency of a Carnot heat engine is $25 \%$ and the temperature of its source is $127^{\circ} \mathrm{C}$. Without changing the temperature of the source, if absolute temperature of the sink is decreased by $10 \%$, the efficiency of the engine is
- A$27.5 \%$
- B$17.5 \%$
- Correct answerC$32.5 \%$
- D$22.5 \%$
Solution
For car not engine, $\eta=25 \%=0.25$
$\begin{aligned}
& \mathrm{T}_1=127^{\circ} \mathrm{C}=(127+273) \mathrm{K}=400 \mathrm{~K} \\
& \therefore \quad \eta=1-\frac{\mathrm{T}_2}{\mathrm{~T}_1} \\
& \Rightarrow 0.25=1-\frac{\mathrm{T}_2}{400} \\
& \therefore \quad \mathrm{~T}_2=300 \mathrm{~K}
\end{aligned}$
In second case, $\mathrm{T}_2^{\prime}=0.9 \mathrm{~T}_2=0.9 \times 300$ $\therefore \quad \eta=1-\frac{T_2^{\prime}}{T_1}=1-\frac{0.9 \times 300}{400}=32.5 \%$
In second case, $\mathrm{T}_2^{\prime}=0.9 \mathrm{~T}_2=0.9 \times 300$ $\therefore \quad \eta=1-\frac{T_2^{\prime}}{T_1}=1-\frac{0.9 \times 300}{400}=32.5 \%$
Hence, the correct answer is option: 'C'.