A Carnot engine whose sink is at 300 K has an efficiency of 40 % . By how much should the temperature of source be increased so as to increase its efficiency by 50 % of original efficiency?

Step-by-step solution

T 2 T 1 = 1 - η = 1 - 40 100 = 3 5 ∴ T 1 = 5 3 T 2 = 5 3 × 300 = 500 K Increase in efficiency = 50 % of 40 % = 20 % ∴ New efficiency η ′ = 40 + 20 = 60 % ∴ T 2 T ′ 1 = 1 - η ′ = 1 - 60 100 = 2 5 T 1 ′ = 5 2 × 300 = 750 K Increase in temperature of source = T 1 ′ - T 1 = 750 - 500 = 250 K