A Carnot engine with efficiency 50% takes heat from a source at 600 K. To increase the efficiency by 20%, keeping temperature of the sink same, the new temperature of the source will be

Step-by-step solution

The efficiency of a Carnot engine is given by = 1 - T₂ T₁ , where T₁ is the source temperature and T₂ is the sink temperature. Given initial efficiency ₁ = 50 % = 0.5 and source temperature T₁ = 600 K . 0.5 = 1 - T₂ 600 T₂ 600 = 0.5 T₂ = 300 K The efficiency is increased by 20 % , so the new efficiency is ₂ = 50 % + 20 % = 70 % = 0.7 . Let the new source temperature be T₁' . The sink temperature T₂ remains the same. 0.7 = 1 - 300 T₁' 300 T₁' = 0.3 T₁' = 300 0.3 = 1000 K Answer: 1000 K