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The efficiency of a carnot heat engine has increased from 40% to 50% which the temperature of source and sink are reduced by 100 degree centigrade find the

source and sink temperature


When the temperatures of source and sink are T_{1} and T_{2} respectively, the efficiency of Carnot engine is given by, \eta=1-\frac{T_{2}}{T_{1}}.
From the information provided, we get two equations,
\frac{T_{2}}{T_{1}}=\frac{3}{5} → 1
\frac{T_{2}-100}{T_{1}-100}=\frac{1}{2} → 2
Solving them, you get T_{1}=500 K and T_{2}=300 K.
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