a carnot engine whose low temperature reservoir is at 200k has an efficiency of 50% it is desired to increase is 75% by how many degree temperature of higher temperature reservoir remains constant?
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Answer:
A Carnot engine whose low temperature reservoir is at 7∘C has an efficiency of 50%. It is desired to increase the efficiency to 70% By low many degrees should the temperature of the high temperature reservoir be increased. =933-560=373K=380K.
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Answer:
Efficiency of Carnot engine,
η=1−
T
hot
Tcold
×100
When η=50%
then =
100
50
=1−
T
hot
T
cold
⇒0.5=1−
T
hot
350
or T
hot
=700K
When η=60%
then =
100
60
=1−
T
hot
T
cold
⇒0.6=1−
T
hot
350
Hence, the temperature of high reservoir is increased by
T
hot
′
=T
hot
=875K−700K=175K
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