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

Options

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

Answer

C

Explanation

Efficiency of the Carnot cycle , `eta=1-(T_2)/(T_1)`
`:.eta=1-((273+27))/(273+127)=1-300/400=1-3/4=1/4`
Also `eta = ("work done by engine")/("Heat supplied by source") = W/(DeltaQ) = W/(40kJ)`
`W = 40eta = 40xx1/4 = 10kJ`.