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An ideal Carnot engine whose efficiency is $50 \%$ receives heat at $500 \mathrm{~K}$. If the efficiency is to be $60 \%$, then intake temperature for the same exhaust temperature is
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The correct answer is:
$625 \mathrm{~K}$
The efficiency of Carnot engine,
$$
\begin{aligned}
& \eta=1-\frac{T_2}{T_1}, \text { where } T_1=\text { initial intake }=500 \mathrm{~K}, \\
& T_2=\text { exhaust temperature, }
\end{aligned}
$$
Initial efficiency, $\eta_1=50 \%=\frac{50}{100}$
Final efficiency, $\eta_2=60 \%=\frac{60}{100}$ and let $T_1{ }^{\prime}=$ final intake temperatue.
Now, $\eta_1=1-\frac{T_2}{T_1}$
$$
\frac{50}{100}=1-\frac{T_2}{500} \Rightarrow T_2=250 \mathrm{~K}
$$
and
$$
\begin{aligned}
& \eta_2=\frac{60}{100}=\frac{6}{10} \Rightarrow \frac{6}{10}=1-\frac{T_2}{T_1{ }^{\prime}} \\
& \frac{T_2}{T_1{ }^{\prime}}=1-\frac{6}{10}=\frac{4}{10} \Rightarrow T_1{ }^{\prime}=\frac{10}{4} T_2 \\
& T_1{ }^{\prime}=\frac{10}{4} \times 250=625 \mathrm{~K}
\end{aligned}
$$
$$
\begin{aligned}
& \eta=1-\frac{T_2}{T_1}, \text { where } T_1=\text { initial intake }=500 \mathrm{~K}, \\
& T_2=\text { exhaust temperature, }
\end{aligned}
$$
Initial efficiency, $\eta_1=50 \%=\frac{50}{100}$
Final efficiency, $\eta_2=60 \%=\frac{60}{100}$ and let $T_1{ }^{\prime}=$ final intake temperatue.
Now, $\eta_1=1-\frac{T_2}{T_1}$
$$
\frac{50}{100}=1-\frac{T_2}{500} \Rightarrow T_2=250 \mathrm{~K}
$$
and
$$
\begin{aligned}
& \eta_2=\frac{60}{100}=\frac{6}{10} \Rightarrow \frac{6}{10}=1-\frac{T_2}{T_1{ }^{\prime}} \\
& \frac{T_2}{T_1{ }^{\prime}}=1-\frac{6}{10}=\frac{4}{10} \Rightarrow T_1{ }^{\prime}=\frac{10}{4} T_2 \\
& T_1{ }^{\prime}=\frac{10}{4} \times 250=625 \mathrm{~K}
\end{aligned}
$$
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