A Carnot engine whose sink is at 300 K has an efficiency of 50%. By how much should the temperature of source be increased so as to increase its efficiency by 40% of original efficiency ?
Answer & explanation
Correct answer: option 4
\(\eta = 1 - \frac{T_2}{T_1}\)
\(\eta = 0.5 \text{ ; } T_2 = 300 K \)
⇒ T1 = 600 K
now \(\eta' = 0.5 + 0.5 × \frac{40}{100} = 0.7\)
\(\eta' = 1 - \frac{T_2}{T_1 + \Delta T}\)
⇒ \(\Delta T = 400 K\)