Question

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

Answer 1

Answer

We have η=1−

T

1

T

2

0.2=1−

T

1

T

2

T

1

T

2

=1−0.2=0.8

0.25=1−

T

1

+25

T

2

T

1

+25

T

2

=1−0.25=0.75

T

1

+25

0.8T

1

=0.75

0.8T

1

=0.75T

1

+0.75×25

0.05T

1

=0.75×25

T

1

=

0.05

0.75×25

=15×25=375K

T

2

=0.8T

1

=0.8×375=300K