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:
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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:

G1 Physics · Accepted Answer

For a Carnot engine, η = 1 - T₂/T₁. Initially 0.5 = 1 - T₂/600, so T₂ = 300 K. For η = 0.7 with the same sink, 0.7 = 1 - 300/T₁, giving T₁ = 1000 K.