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A Carnot engine absorbs $1000 \mathrm{~J}$ of heat energy from a reservoir at $127^{\circ} \mathrm{C}$ and…
A Carnot engine absorbs $1000 \mathrm{~J}$ of heat energy from a reservoir at $127^{\circ} \mathrm{C}$ and rejects $600 \mathrm{~J}$ of heat energy during each cycle. The efficiency of engine and temperature of sink will be:
$20 \%$ and $-43^{\circ} \mathrm{C}$
$40 \%$ and $-33^{\circ} \mathrm{C}$
$50 \%$ and $-20^{\circ} \mathrm{C}$
$70 \%$ and $-10^{\circ} \mathrm{C}$ Solution
Given: $\mathrm{Q}_1=1000 \mathrm{~J}$
$
\begin{aligned}
&\mathrm{Q}_2=600 \mathrm{~J} \\
&\mathrm{~T}_1=127^{\circ} \mathrm{C}=400 \mathrm{~K} \\
&\mathrm{~T}_2=? \\
&\eta=?
\end{aligned}
$
Efficiency of carnot engine,
$
\begin{aligned}
&\eta=\frac{W}{Q_1} \times 100 \% \\
&\text { or, } \eta=\frac{Q_2-Q_1}{Q_1} \times 100 \%
\end{aligned}
$
or, $\eta=\frac{\mathrm{Q}_2-\mathrm{Q}_1}{\mathrm{Q}_1} \times 100 \%$
or, $\eta=\frac{1000-600}{1000} \times 100 \%$
$
\eta=40 \%
$
Now, for carnot cycle $\frac{\mathrm{Q}_2}{\mathrm{Q}_1}=\frac{\mathrm{T}_2}{\mathrm{~T}_1}$
$
\begin{aligned}
&\frac{600}{1000}=\frac{\mathrm{T}_2}{400} \\
&\mathrm{~T}_2=\frac{600 \times 400}{1000}=240 \mathrm{~K}=240-273 \\
&\therefore \mathrm{T}_2=-33^{\circ} \mathrm{C}
\end{aligned}
$
Asked in: JEE Main 2014 (12 Apr Online)
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