Question

what is the maximum efficiency of an engine for which the ratio of the energy exhausted to the work done is 0.25?
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Answer 1

Answer

We have η=1−T1T2

0.2=1−T1T2

T1T2=1−0.2=0.8

0.25=1−T1+25T2

T1+25T2=1−0.25=0.75

T1+250.8T1=0.75

0.8T1=0.75T1+0.75×25

0.05T1=0.75×25

T1=0.050.75×25=15×

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Answer 2

see above attachment for your answer

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