A Carnot engine takes `3xx10^6` cal of heat from a reservoir at `627^@C` and gives it to a sink at `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

`T_(1) = 627 +273 = 900 K , Q_(1) = 3 xx 10^(6) cal`
`T_(2) = 27 +273 = 300 K :. (Q_(1))/(T_(1)) = (Q_(2))/(T_(2))`
`Q_(2) = (T_(2))/(T_(1)) xx Q_(1) = (300)/(900) xx3xx 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`