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

Options

A. 30 kJ
B. 10 kJ
C. 4 kJ
D. 1 kJ

Answer

B

Explanation

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`