A Carnot engine takes `3xx10^6cal`. 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. `16 B xx10^6 J`
B. `2 xx10^6` cal
C. `4.2 xx10^6 J`
D. zero

Answer

B

Explanation

Here, `theta = 3 xx10^6` cal where `theta` is the amount of heat required.
`T_1 = 627^@C`, where `T_1` is the temperature of reservoir.
`T_2 = 27^@ C`, where `T_2` is the temperature at sink.
We have to find the work done:
First of all, we find `theta_2` : We know that `theta_2/theta_1 = T_2/T_1`
where `T_1 = 627 + 273 = 900 K. T_2 = 27 + 273 = 300 K, theta_2/theta_1 = 300/900 rArr theta_2 = theta_1 xx3/9 = 3 xx 10^6 xx 3/9 = 10^6`
Work done ` theta_1 - theta_2 = (3 xx10^6) - 10^6 = 2 xx10^J`.