Question

A Carnot engine whose sink is at a temperature of 300 K has an efficiency of 40%. By how much should the temperature of the source be increased, so as to increase the efficiency by 60% ?

• A. 250 K
• B. 275 K
• C. 300 K
• D. 325 K

Answer 1

Answer: (A)

250 K

Efficiency of a Carnot's engine:
$\eta =1-\frac{T_c}{T_h}$

$0.4=1-\frac{300}{T_1}$

$T_1=\frac{300}{0.6}=500K$

$0.6=1-\frac{300}{T_2}$

$T_2=\frac{300}{0.4}=750K$

$\therefore$change in temperature is $750-500=250K$