1.

A Carnot engine works between `120^(@)C` and `30^(@)C`. Calculate the efficiency. If the power produced by the engine is `400W`, calculate the heat abosorbed from the source.

Answer» Efficiency, `eta =(T_(2)-T_(1))/(T_(2))`
Here `T_(2) = 273 +120 = 393K`
`T_(1) = 273 +30 = 303 K`
`:. eta = (393-303)/(393) = 0.229 = 22.9%`
Again, `eta = (Q_(2)-Q_(1))/(Q_(2)) = (W)/(Q_(2))`
Thus, heat absorbed form the source.
`Q_(2) = (W)/(eta) = (400)/(0.229) = 1747W`


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