A Carnot engine with efficiency 50 % takes heat from a source at 600 K . In order to increase the efficiency to 70 % , keeping the temperature of sink same, the new temperature of the source will be:

Step-by-step solution

Efficiency of Carnot engine is given by η = 1 - T sink   T source = 1 - T 2 T 1 Given: Initial efficiency η = 1 2 ∴   1 2 = 1 - T 2 600 ⇒ T 2 600 = 1 2 ⇒ T 2 = 300   K When efficiency is increased to 70 % and T 2 = 300   K , Let T ' be new temperature of source ⇒ 7 10 = 1 - 300 T ' ⇒ 300 T ' = 1 - 7 10 ∴   T ' = 1000   K .