Answer
Work done is 57.9KJ
Step-by-step explanation
First solve the problem according to work done due to variation in temperature
So W= intergral Cu( 1-Tu/T). at Tu and T
So Given that
C = Heat capacity of the Brick
TEPc= Cold Temperature
TEPh = Hot Temperature
W = C ( TEPh-TEP) - TEPhCln ( TEPh/TEPc)
So
W= (1)-(300-150)-300 (1) ln 2
W= -57.9KJ