An engine runs between a reservoir at temperature 200 K and a hot body which is initially at temperature of 600 K. If the hot body cools down to a temperature of 400 K in the process, then the maximum amount of work that the engine can do (while working in a cycle) is (the heat capacity of the hot body is 1 J/K)

Options

A. 200(1 – ln 2) J
B. 200(1 – ln 3/2) J
C. 200(1 + ln 3/2) J
D. 200 J

Answer

B

Explanation

`eta =(W )/( Q_("in"))`
` implies W = eta W_("in")`
` Q = eta Q_("in")` ` Q = int C dt `
For maximum amount of work, efficiency should be maximum, means we have to assume carnot engine
` therefore eta =1 - (T_2)/( T_1) =1 -(200)/(T)`
` therefore W = int n Q_("in") = - int_(600 ) ^(400) ( 1- (200)/(T ))CdT`
`=-C [T - 200 l n T ]_(600 )^(400 )`
`W =-C [-200 + l n (3/2) 200 ]`
C=1 ( Given )
` therefore W= 200 - 200 ln (3/2)`
`W = 200 ( 1- ln (3)/(2))`