Question

A Carnot engine with efficiency $50\%$ takes heat from a source at $600K$. In order to increase the efficiency to $70\%$, keeping the temperature of sink same, the new temperature of the source will be :

• A. $900K$
• B. $300K$
• C. $1000K$
• D. $360K$

Answer 1

Answer: (C)

But $\eta =1-\frac{T_2}{T_1}$
$\therefore \frac{1}{2}=1-\frac{T_2}{600}$
$\Rightarrow \frac{T_2}{600}=\frac{1}{2}\Rightarrow T_2=300K$
Now efficiency is increased to $70\%$ and $T_2=300$
$K$, Let temp of source $T_1=T$
$\Rightarrow \frac{7}{10}=1-\frac{300}{T}$
$\Rightarrow \frac{300}{T}=1-\frac{7}{10}$
$\Rightarrow \frac{300}{T}=\frac{3}{10}$
$\therefore T=1000K$