Q64 / 90 · JEE Main 2023 (25 Jan Shift 1) · Physics

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:

Solution

Efficiency of Carnot engine is given by $η = 1 - \frac{T_{sink}}{T_{source}} = 1 - \frac{T_{2}}{T_{1}}$

Given: Initial efficiency $η = \frac{1}{2}$

$∴ \frac{1}{2} = 1 - \frac{T_{2}}{600}$

$\Rightarrow \frac{T_{2}}{600} = \frac{1}{2}$

$\Rightarrow T_{2} = 300 K$

When efficiency is increased to $70 %$ and $T_{2} = 300 K$ ,

Let $T^{'}$ be new temperature of source

$\Rightarrow \frac{7}{10} = 1 - \frac{300}{T^{'}}$

$\Rightarrow \frac{300}{T^{'}} = 1 - \frac{7}{10}$

$∴ T^{'} = 1000 K$ .

Hence, the correct answer is option: 'B'.