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Consider a reversible engine of efficiency $\frac{1}{6}$. When the temperature of the sink is reduced by $62^{\circ} \mathrm{C}$, its efficiency gets doubled. The temperature of the source and sink respectively are
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Verified Answer
The correct answer is:
$372 \mathrm{~K}$ and $310 \mathrm{~K}$
Given
$$
\eta_1=\frac{1}{6}
$$
According to the question,
$$
\begin{gathered}
\eta_1=\frac{T_1-T_2}{T_1} \\
\frac{T_1-T_2}{T_1}=\frac{1}{6} \\
\eta_2=\frac{T_1-\left(T_2-62\right)}{T_1} \Rightarrow 2 \times \eta_1=\frac{T_1-T_2+62}{T_1} \\
2 \times \frac{1}{6}=\frac{T_1-T_2+62}{T_1} \Rightarrow \frac{1}{3}=\frac{T_1-T_2+62}{T_1}
\end{gathered}
$$
(ii)
From Eq. (i), we get
$$
T_1-T_2=\frac{T_1}{6}
$$
Substituting this value in Eq. (ii), we get
$$
\begin{gathered}
\frac{1}{3}=\frac{\frac{T_1}{6}+62}{T_1} \Rightarrow \frac{1}{3}=\frac{T_1+372}{6 T_1} \\
6 T_1=3 T_1+1116 \Rightarrow 3 T_1=1116 \Rightarrow T_1=372 \mathrm{~K}
\end{gathered}
$$
From Eq. (i), we get
$$
\begin{aligned}
\frac{372-T_2}{372} & =\frac{1}{6} \\
6\left(372-T_2\right) & =372
\end{aligned}
$$
$$
\begin{aligned}
2232-6 T_2 & =372 \\
6 T_2 & =2232-372 \\
6 T_2 & =1860 \\
T_2 & =\frac{1860}{6} \\
T_2 & =310 \mathrm{~K}
\end{aligned}
$$
$$
\eta_1=\frac{1}{6}
$$
According to the question,
$$
\begin{gathered}
\eta_1=\frac{T_1-T_2}{T_1} \\
\frac{T_1-T_2}{T_1}=\frac{1}{6} \\
\eta_2=\frac{T_1-\left(T_2-62\right)}{T_1} \Rightarrow 2 \times \eta_1=\frac{T_1-T_2+62}{T_1} \\
2 \times \frac{1}{6}=\frac{T_1-T_2+62}{T_1} \Rightarrow \frac{1}{3}=\frac{T_1-T_2+62}{T_1}
\end{gathered}
$$
(ii)
From Eq. (i), we get
$$
T_1-T_2=\frac{T_1}{6}
$$
Substituting this value in Eq. (ii), we get
$$
\begin{gathered}
\frac{1}{3}=\frac{\frac{T_1}{6}+62}{T_1} \Rightarrow \frac{1}{3}=\frac{T_1+372}{6 T_1} \\
6 T_1=3 T_1+1116 \Rightarrow 3 T_1=1116 \Rightarrow T_1=372 \mathrm{~K}
\end{gathered}
$$
From Eq. (i), we get
$$
\begin{aligned}
\frac{372-T_2}{372} & =\frac{1}{6} \\
6\left(372-T_2\right) & =372
\end{aligned}
$$
$$
\begin{aligned}
2232-6 T_2 & =372 \\
6 T_2 & =2232-372 \\
6 T_2 & =1860 \\
T_2 & =\frac{1860}{6} \\
T_2 & =310 \mathrm{~K}
\end{aligned}
$$
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