Question

A reversible engine working between temperatures limits 600 degree kelvin and 1200 degree kelvin receives 50 kj of heat work done by engine is

Answer 1

Given

Temperature of source T₁ = 1200 K

Temperature of sink T₂ = 600 K

Heat supplied Q₁ = 50 J

To determine work done = Heat supplied (Q₁) - Heat Rejected (Q₂) ------(A)

Now as mentioned in the question the engine is reversible, thus

Efficiency η = (Output/ Input )= (T₁ - T₂)/(T₁) = (Q₁ - Q₂)/(Q₁)

1 - (600/1200) = 1 - (Q₂/50)

Q₂ = (600 × 50)/1200 = 25 J  

Reducing the above found value in A we get,

Work done = 50 -25 = 25 J