A Carnot engine takes `3 xx 10^(6)` cal of heat from a reservoir at `627^(@)C` and gives it to a sink `27^(@)C`. The work done by the engine is

Options

A. `4.2 xx 10^(6) J`
B. `8.4 xx 10^(6) J`
C. `16.8 xx 10^(6) J`
D. zero

Answer

B

Explanation

Here, `T_(1) = 627 + 273 = 900 K, Q_(1) = 3 xx 10^(6) cal`
`T_(2) = 27 + 273 = 300 K`
As `(Q_(1))/(Q_(2)) = (T_(1))/(T_(2)) :. Q_(2) = (T_(2))/(T_(1)) xx Q_(1) = (300)/(900) xx 3 xx 10^(6) = 1 xx 10^(6) cal`
`:.` Work done `= Q_(1) - Q_(2) = 3 xx 10^(6) - 1 xx 10^(6) = 2 xx 10^(6) cal`
`= 2 xx 4.2 xx 10^(6) J = 8.4 xx 10^(6) J`