A Carnot engine, whose efficiency is 40% takes heat from a source maintained at temperature 600K. To have an efficiency 60%, the intake temperature for the same exhaust temperature should be

Step-by-step solution

The efficiency of a Carnot engine is given by = 1 - T₂ T₁ Given ₁ = 40 % = 0.4 and T₁ = 600 K 0.4 = 1 - T₂ 600 T₂ 600 = 0.6 T₂ = 360 K For the new efficiency ₂ = 60 % = 0.6 with the same exhaust temperature T₂ = 360 K, let the new intake temperature be T₁' 0.6 = 1 - 360 T₁' 360 T₁' = 0.4 T₁' = 360 0.4 = 900 K Answer: 900 K