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:

A Carnot engine has 50% efficiency at a source temperature of 600 K. Find the new source temperature needed to achieve 70% efficiency while keeping the sink temperature unchanged.

Find the sink temperature.

Carnot efficiency: $$\eta = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}$$

$$0.50 = 1 - \frac{T_{\text{cold}}}{600}$$

$$T_{\text{cold}} = 600 \times 0.50 = 300 \text{ K}$$

Find the new source temperature.

$$0.70 = 1 - \frac{300}{T_{\text{new}}}$$

$$\frac{300}{T_{\text{new}}} = 0.30$$

$$T_{\text{new}} = \frac{300}{0.30} = 1000 \text{ K}$$

The correct answer is 1000 K.