Heat and Thermodynamics 5 Question 1
1. A Carnot engine has an efficiency of $1 / 6$. When the temperature of the sink is reduced by $62^{\circ} C$, its efficiency is doubled. The temperatures of the source and the sink are respectively,
(2019 Main, 12 April II)
(a) $62^{\circ} C, 124^{\circ} C$
(b) $99^{\circ} C, 37^{\circ} C$
(c) $124^{\circ} C, 62^{\circ} C$
(d) $37^{\circ} C, 99^{\circ} C$
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Solution:
- Efficiency of a Carnot engine working between source of temperature $T _1$ and sink of temperature $T _2$ is given by
$$ \eta=1-\frac{T _2}{T _1} $$
Here, $T _2$ and $T _1$ are absolute temperatures.
Initially,
$$ \begin{aligned} & \eta=\frac{1}{6} \\ \therefore \quad & \frac{1}{6}=1-\frac{T _2}{T _1} \Rightarrow \frac{T _2}{T _1}=\frac{5}{6} \end{aligned} $$
Finally, efficiency is doubled on reducing sink temperature by $62^{\circ} C$.
$\therefore \quad \eta=\frac{2}{6}, T _{\text {sink }}=T _2^{\prime}=T _2-62$
So, $\quad \eta=1-\frac{T^{\prime}{ } _2}{T^{\prime}{ } _1}$
$$ \begin{aligned} & \Rightarrow \quad \frac{2}{6}=1-\frac{T _2-62}{T _1} \Rightarrow \frac{T _2-62}{T _1}=\frac{4}{6} \\ & \Rightarrow \quad \frac{T _2}{T _1}-\frac{62}{T _1}=\frac{4}{6} \\ & \Rightarrow \quad \frac{5}{6}-\frac{62}{T _1}=\frac{4}{6} \\ & \Rightarrow T _1=6 \times 62=372 K=372-273=99^{\circ} C \\ & \Rightarrow \quad T _2=\frac{5}{6} \times T _1 \simeq 310 K \\ & =310-273=37^{\circ} C \\ & {\left[\because \frac{T _2}{T _1}=\frac{5}{6}\right]} \end{aligned} $$
