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When the absolute temperature of the source of a Carnot heat engine is increased by $25 \%$, its efficiency…
When the absolute temperature of the source of a Carnot heat engine is increased by $25 \%$, its efficiency increases by $80 \%$. The new efficiency of the engine is
$12 \%$ $24 \%$ $48 \%$ $36 \%$ Solution
According to the question, initially
$\begin{aligned}
& \eta=\frac{100-T_2}{100} \\
& =\eta+\eta \times \frac{80}{100}=1.8 \eta=\frac{125-T_2}{125} \\
& \frac{\eta}{\eta^{\prime}}=\frac{100-T_2}{100} \times \frac{125}{125-T_2} \\
& \Rightarrow \quad \frac{1}{18}=\frac{100-T_2}{100} \times \frac{125}{125-T_2} \\
& \text { or, } \frac{5}{9}=\frac{100-T_2}{100} \times \frac{5}{\left(125-T_2\right)} \\
& \text { or, } 9\left(100-\mathrm{T}_2\right)=500-4 \mathrm{~T}_2 \\
& \text { or, } 900-9 \mathrm{~T}_2=500-4 \mathrm{~T}_2 \\
& \text { or, } 400=5 \mathrm{~T}_2 \\
& \text { or, } \mathrm{T}_2=\frac{400}{5}=80
\end{aligned}$
The new efficiency,
$\begin{aligned}
& \eta^{\prime \prime}=1-\frac{T_2}{T_1}=1-\frac{80}{125} \\
& \frac{125-80}{125}=\frac{45}{125}=\frac{45}{125} \times 100=36 \%
\end{aligned}$
Asked in: AP EAMCET 2016
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