A Carnot engine with efficiency $50 \%$ takes heat from a source at $600 \mathrm{~K}$. To increase the…
A Carnot engine with efficiency $50 \%$ takes heat from a source at $600 \mathrm{~K}$. To increase the efficiency to $70 \%$, keeping the temperature of the sink same, the new temperature of the source will be
$360 \mathrm{~K}$
$1000 \mathrm{~K}$
$900 \mathrm{~K}$
$300 \mathrm{~K}$
Solution
$\begin{aligned}
& \mathrm{T}_{\mathrm{H}}=600 \mathrm{k} \\
& \eta=1-\frac{\mathrm{T}_{\mathrm{C}}}{\mathrm{T}_{\mathrm{H}}} \\
& \text { but, } \eta=\frac{1}{2} .... ( Given: \eta=50 \%)
\\
& \Rightarrow \frac{1}{2}=1-\frac{\mathrm{T}_{\mathrm{C}}}{600} \\
\therefore \quad & \mathrm{T}_{\mathrm{C}}=300 \mathrm{~K}
\end{aligned}$
(Given: $\eta=50 \%$ )
With $\mathrm{T}_{\mathrm{C}}=300 \mathrm{~K}$, the efficiency is increased to $70 \%$
$\therefore \quad$ New temperature of the source will be $\mathrm{T}_{\mathrm{Hac}_{\mathrm{a}}}$
$\begin{array}{ll}
\therefore \quad \eta=1-\frac{300}{\mathrm{~T}_{\mathrm{H}_{\mathrm{ec}}}} \\
\quad \frac{300}{\mathrm{~T}_{\mathrm{H}_{\text {sew }}}}=1-\frac{7}{10} \quad \ldots .(\because \eta=70 \%) \\
\therefore \quad \mathrm{T}_{\mathrm{H}_{\text {ecw }}}=\frac{3000}{3}=1000 \mathrm{~K}
\end{array}$