A Carnot's engine works as refrigerator between `250 K` and 300K. It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is :

Options

A. 420 J
B. 2100 J
C. 772 J
D. 2520 J

Answer

A

Explanation

Given: Temperature of cold body, `T_(2)= 250 K` temperature of hot body, `T_(1) = 300 K`
Heat received, `Q_(2) = 500` cal work done, W= ?
Efficiency `=1-T_(2)/T_(1) =W/(Q_(2) + W) rArr 1 - 250/300 = W/(Q_(2) + W)`
`W = Q_(2)/5 = (500 xx 4.2)/5 J = 420 J`