An engine has an efficiency of 1/3. When the temperature of sink is reduced by 56°C, its efficiency is halved. Temperature of the source is :
Answer & explanation
Correct answer: option 3
Since, efficiency of engine is : \(\eta = 1 - \frac{T_2}{T_1}\)
When the temperature of the sink is reduced by 56°C, its efficiency is doubled : \(\eta/2 = 1 - \frac{T_2 - 56}{T_1}\)
According to ques : \(\eta = \frac{1}{3}\)
⇒ T2 = 336 K ; T1 = 63°C = Temperature of source