Question

the efficiency of a carnot heat engine has increased from 40% to 50% . when the temperature of source and sink are increased by 100 degree centigrade .find the source and sink temperatures ?

Answer 1

We have, η = $1-\frac{T_{sink}}{T_{source}}$
Firstly, $\frac{T_{sink}} {T_{source}}=0.6$
then, $\frac{T_{sink}+100} {T_{source}+100}=0.5$.
Solving these equations, you have T(source)=-500K & T(sink)=-300K.