Question

A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased to as to increase its efficiency by 50% of original efficiency?

• A. 150 K
• B. 250 K
• C. 300 K
• D. 450 K

Answer 1

The correct option is B 250 K

We know,

Efficiency$=1-\frac{T_2}{T_1}$

Where, $T_2$=sink temperature,

$T_1$=source temperature.

$\frac{6}{10}=\frac{300}{T_1}$

$T_1=500K$

On increasing efficiency by 50%, it becomes \dfrac{ 150}{100}\times 40=60%$

$\frac{4}{10}=\frac{300}{T_1^{\prime }}$

$T_1^{\ast }=750K$

$T_1^{\ast }-T_1=250K$

Option $\text{B}$ is the correct answer