1.

A Carnot engine operates with source at 127^(@)C and sink at 27^(@)C. If the source supplies 40 kJ of heat energy the work done by the engine is

Answer»

30 kJ
10 kJ
4 kJ
1 kJ

Solution :The efficiency of a Carnot engine is `eta = 1 - (T_(2))/(T_(1))`
Here, `T_(1) = 127^(@)C = 127 + 273 = 400 K`
`T_(2) = 27^(@)C = 27 + 273 = 300 K`
`:. eta = 1 - (300)/(400) = 1 - (3)/(4) = (1)/(4)`
Also `eta = ("work done by engine")/("heat supplied by source") = (W)/(40 kJ)`
`:. W = 40 eta = 40 xx (1)/(4) = 10 kJ`


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