1cc of water at ${100}^{\circ }\mathrm{C}$ is given 540 cal of heat and the steam formed occupies 1671cc. at 1 atm pressure. Then workdone against atmospheric pressure in this process is nearly(in Cal)

Given: 1cc of water at ${100}^{\circ }\mathrm{C}$ is given 540 cal of heat and the steam formed occupies 1671cc. at 1atm pressure.

Formula Used:

$W=P∆V \left(W=workdone\right)$

Explanation: 

    $$\begin{gathered} W=P∆V \\ {V}_1=1c{m}^2={10}^{-6}{m}^3, V_2=1671c{m}^2=1671\times {10}^{-6}{m}^3 \\ W=P\left(V_2-V_1\right) \\ W=167J \\ W=\frac{167}{4.2}\approx 40cal \end{gathered}$$

Hence, the correct answer is Option C.