A Carnot's 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

Efficiency `eta = (T_(1) - T_(2))/(T_(1)) = 1 -(T_(2))/(T_(1))`
`therefore eta = 1-((273 + 27))/((273 + 127)) = 1 - (300)/(400) = 1 -3/4 = 1/4`
`therefore eta` = (work done by engine)/(Heat supplied by source) = `(W)/(40(kJ))`
`therefore W = 40 eta = 40 xx (1)/(4) = kJ`
`therefore` Workdone = 10kJ.