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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 ?


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.
0 0 0
I really doubt my answers.
I don't think temperatures like that are possible.
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