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

Answers (1)

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}}.

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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