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 :

27°C

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