A Carnot engine with efficiency 50% takes heat from a source at 600 K. In order to increase the efficiency to 70%, keeping the temperature of sink same, the new temperature of the source will be :
Select the correct option:
A
360 K
B
1000 K
C
900 K
D
300 K
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Sol.
$$
\begin{array}{ll}
\text { But } & \eta=1-\frac{T_2}{T_1} \\
\therefore & \frac{1}{2}=1-\frac{T_2}{600} \\
\Rightarrow & \frac{T_2}{600}=\frac{1}{2} \quad \Rightarrow T_2=300 \mathrm{k}
\end{array}
$$
Now efficiency is increased to $70 \%$ and $\mathrm{T}_2=300$ K , Let temp of source $\mathrm{T}_1=\mathrm{T}$
$$
\begin{array}{ll}
\Rightarrow & \frac{7}{10}=1-\frac{300}{T} \\
\Rightarrow & \frac{300}{T}=1-\frac{7}{10} \\
\Rightarrow & \frac{300}{T}=\frac{3}{10} \\
\therefore & T=1000 \mathrm{~K}
\end{array}
$$
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