A Carnot engine whose sink is at 280 K has an efficiency of 30%. By how much should the temperature of source be increased so as to increase its efficiency by 25% of original efficiency ?
Answer & explanation
Correct answer: option 3
\(\eta = 1 - \frac{T_2}{T_1}\)
\(\eta = 0.3 \text{ ; } T_2 = 280 K \)
⇒ T1 = 400 K
now \(\eta' = 0.3 + 0.3 × \frac{25}{100} = 0.375\)
\(\eta' = 1 - \frac{T_2}{T_1 + \Delta T}\)
⇒ \(\Delta T = 48 K\)