An engine has an efficiency of 1/5. When the temperature of sink is reduced by 120°C, its efficiency is tripled. Temperature of the source is :
Answer & explanation
Correct answer: option 1
Since efficiency of engine is : \(\eta = 1 - \frac{T_2}{T_1}\)
When the temperature of the sink is reduced by 27°C, its efficiency is doubled : \(3\eta = 1 - \frac{T_2-120}{T_1}\)
According to ques : \(\eta = \frac{1}{5}\)
⇒ T1 = 300 K ; T1 = 27°C = Temperature of source