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  and  respectively, the efficiency of Carnot engine is given by, .

Initially

$0.4=1-\frac{T_{2}}{T_{1}}$

i.e $\frac{T_{2}}{T_{1}}=\frac{3}{5}$

Finally

$0.5=1-\frac{T_{2}-100}{T_{1}-100}$

SO we get

$T_{1}=500 K$  and $T_{2}=300 K$.

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