Q107 / 160 · JIPMER 2011 · Physics
In Carnot engine efficiency is $40 \%$ at hot reservoir temperature $T$. For efficiency $50 \%$, what will be temperature of hot reservoir?
- A$\frac{T}{5}$
- B$\frac{2 T}{5}$
- C$6 T$
- Correct answerD$\frac{6 T}{5}$
Solution
The efficiency of a heat engine is defined as the ratio of work done to the heat supplied, ie,
$\eta=\frac{\text { Work done }}{\text { Heat input }}=\frac{W}{Q}$
or $\quad \eta=1-\frac{T_2}{T_1}$
where $T_2$ is temperature of sink, and $T_1$ is temperature of hot reservoir
$\therefore \quad \frac{40}{100}=1-\frac{T_2}{T_1}$
$\Rightarrow \quad \frac{T_2}{T_1}=0.6$
$\Rightarrow \quad T_1=0.6 T_1$
Again, $\quad \frac{50}{100}=1-\frac{T_2}{T_1^{\prime \prime}}$
$\Rightarrow \quad \frac{T_2}{T_1{ }^{\prime}}=0.5$
$\begin{array}{ll}\Rightarrow & \frac{0.6 T_1}{T_1{ }^{\prime}}=0.5 \\ \therefore & T_1{ }^{\prime}=\frac{0.6}{0.5} T_1=\frac{6}{5} T\end{array}$
Hence, the correct answer is option: 'D'.