The efficiency of an ideal Carnot engine working between temperatures $T_1$ and $T_2$ is $1 / 3$. If the…
The efficiency of an ideal Carnot engine working between temperatures $T_1$ and $T_2$ is $1 / 3$. If the temperature of the sink is reduced by $40 \%$, then its efficiency will be
$50 \%$
$25 \%$
$60 \%$
$75 \%$
Solution
Efficiency of Carnot engine, $\eta=\frac{1}{3}$
But $\eta=1-\frac{T_2}{T_1}$
where, $T_1=$ and temperature of fource
$T_2=$ temperature of sink.
$
\therefore \quad \frac{1}{3}=1-\frac{T_2}{T_1} \Rightarrow \frac{T_2}{T_1}=1-\frac{1}{3} \Rightarrow \frac{T_2}{T_1}=\frac{2}{3}
$
When temperature of sink is reduced by $40 \%$, then its new temperature, $T_2^{\prime}=T_2-40 \%$ of $T_2$
$
=T_2-0.4 T_2
$
$\Rightarrow$
$
T_2^{\prime}=0.6 T_2
$
Now, efficiency of Carnot's engine
$
\begin{aligned}
\eta^{\prime} & =\left(1-\frac{T_2^{\prime}}{T_1}\right) \times 100=\left(1-0.6 \frac{T_2}{T_1}\right) \times 100 \\
& =\left(1-0.6 \times \frac{2}{3}\right) \times 100=0.6 \times 100=60 \%
\end{aligned}
$