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An engine is working. It takes 100 calories of heat from source and leaves 80 calories of heat to sink. If the temperature of source is $127^{\circ} \mathrm{C}$, then temperature of sink is
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Verified Answer
The correct answer is:
$47^{\circ} \mathrm{C}$
Heat taken from source Heat left to sink
$$
\begin{aligned}
& Q_1=100 \mathrm{cal} \\
& Q_2=80 \mathrm{cal}
\end{aligned}
$$
$\therefore$ efficiency of the engine
$$
\begin{aligned}
\eta=1-\frac{Q_2}{Q_1} & =1-\frac{80}{100} \\
& =20 \%
\end{aligned}
$$
Temperature of the source $T_1=127^{\circ} \mathrm{C}=400 \mathrm{~K}$ Temperature of the sink $T_2=$ ?
We know that
$$
\begin{aligned}
\eta & =1-\frac{T_2}{T_1} \\
\Rightarrow \quad 0.2 & =1-\frac{T_2}{400} \\
\Rightarrow \quad \frac{T_2}{400} & =1-0.2=0.8 \\
\Rightarrow \quad T_2 & =320 \mathrm{~K}=47^{\circ} \mathrm{C}
\end{aligned}
$$
$$
\begin{aligned}
& Q_1=100 \mathrm{cal} \\
& Q_2=80 \mathrm{cal}
\end{aligned}
$$
$\therefore$ efficiency of the engine
$$
\begin{aligned}
\eta=1-\frac{Q_2}{Q_1} & =1-\frac{80}{100} \\
& =20 \%
\end{aligned}
$$
Temperature of the source $T_1=127^{\circ} \mathrm{C}=400 \mathrm{~K}$ Temperature of the sink $T_2=$ ?
We know that
$$
\begin{aligned}
\eta & =1-\frac{T_2}{T_1} \\
\Rightarrow \quad 0.2 & =1-\frac{T_2}{400} \\
\Rightarrow \quad \frac{T_2}{400} & =1-0.2=0.8 \\
\Rightarrow \quad T_2 & =320 \mathrm{~K}=47^{\circ} \mathrm{C}
\end{aligned}
$$
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