A thermally insulated vessel contains $150\ \text{g}$ of water at $0^\circ\text{C}$. Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at $0^\circ\text{C}$ itself. The mass of evaporated water will be closest to (Latent heat of vaporization of water $= 2.10\times10^6\ \text{J kg}^{-1}$ and Latent heat of fusion of water $= 3.36\times10^5\ \text{J kg}^{-1}$)
Answer: (B) $20\ \text{g}$
The heat needed to evaporate mass $m$ comes from freezing the rest, $150 - m$:
$$m\times2.10\times10^6 = (150 - m)\times3.36\times10^5$$
$$m = \frac{150\times3.36\times10^5}{2.10\times10^6 + 3.36\times10^5} = \frac{150\times0.336}{2.436} \approx 20.7\ \text{g}$$
So about $20\ \text{g}$ evaporates.
Solution by Sreeraj P, M.Sc Physics