1.

A Carnot engine operates between temperatures 600 K and 390 K. Ir absorbs 120 cal of heat from the source. Calculate eta and the heat rejected to the sink.

Answer»

SOLUTION :We have `eta=(T_(2)-T_(1))/(T_(2))=(600-390)/(600)=0.25` or `25%`
Now, again we have,
`eta=(W)/(q_(2))=(q_(2)-q_(1))/(q_(2))`
where `q_(2)` is the HEAT absorbed by the system from the source and `q_(1)` is the heat rejected to the sink.
`:.(120-q_(1))/(120)=0.25`
`:.q_(1)=90` cal


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