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A reversible Carnot heat engine converts $\frac{1}{4}$ th of its input heat into work. When the temperature of the sink is reduced by $50 \mathrm{~K}$, its efficiency becomes $33 \frac{1}{3} \%$. The initial temperatures of the source and the sink respectively are
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600 K, 450 K
Given work, $W=\frac{Q}{4}$
where, $Q=$ input heat.
$$
\text { Efficiency, } \eta=\frac{W}{Q}=\frac{Q / 4}{Q} \Rightarrow \eta=\frac{1}{4}
$$
Also, $\quad \eta=1-\frac{T_2}{T_1}$
where, $T_2=$ temperature of sink
and $\quad T_1=$ temperature of source.
$$
1-\frac{T_2}{T_1}=\frac{1}{4}
$$
$\Rightarrow$ When temperature of sink is reduced by $50 \mathrm{~K}$, efficiency becomes $\frac{100}{3} \%$
$$
\eta=\frac{100}{3} \times \frac{1}{100}=\frac{1}{3}
$$
So, $\quad 1-\frac{T_2-50}{T_1}=\frac{1}{3}$
Subtracting Eq (ii) from Eq. (i), we get
$$
\begin{aligned}
& \frac{50}{T_1}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12} \\
& T_1=600 \mathrm{~K} \\
& T_2=\frac{3}{4} \times 600=450 \mathrm{~K}
\end{aligned}
$$
where, $Q=$ input heat.
$$
\text { Efficiency, } \eta=\frac{W}{Q}=\frac{Q / 4}{Q} \Rightarrow \eta=\frac{1}{4}
$$
Also, $\quad \eta=1-\frac{T_2}{T_1}$
where, $T_2=$ temperature of sink
and $\quad T_1=$ temperature of source.
$$
1-\frac{T_2}{T_1}=\frac{1}{4}
$$
$\Rightarrow$ When temperature of sink is reduced by $50 \mathrm{~K}$, efficiency becomes $\frac{100}{3} \%$
$$
\eta=\frac{100}{3} \times \frac{1}{100}=\frac{1}{3}
$$
So, $\quad 1-\frac{T_2-50}{T_1}=\frac{1}{3}$
Subtracting Eq (ii) from Eq. (i), we get
$$
\begin{aligned}
& \frac{50}{T_1}=\frac{1}{3}-\frac{1}{4}=\frac{1}{12} \\
& T_1=600 \mathrm{~K} \\
& T_2=\frac{3}{4} \times 600=450 \mathrm{~K}
\end{aligned}
$$
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