Question

A Carnot engine whose heat sinks at 27°C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

• A. Increases by 18°C
• B. Increases by 200°C
• C. Increases by 120°C
• D. Increases by 73°C

Answer 1

Answer: (B)

Increases by 200°C

Answer 2

Initially :
$\frac{1}{4}$=1−$\frac{300}{T_H}$
⇒ T H = 400 K
Finally, : Efficiency becomes $\frac{1}{2}$
∴$\frac{1}{2}$=1−$\frac{300}{T_H}$
∴TH=600 K
⇒ The temperature of the source increases by 200°C.
The correct option is (B) : Increases by 200°C.