Question

A vessel containing 100 g water is thermally isolated at 0?C. The air above the water is pumped out, some of the water freezes and some evaporates at 0?C itself. Mass is left inside the vessel is. [latent heat of vaporisation of water at 0?C $=2.1\times {10}^{6}Jk{g}^{-1}$ latent heat of fusion of ice $=3.36\times {10}^{5}Jk{g}^{-1}$ ]

• A. 86g
• B. 103g
• C. 77g
• D. 63g

Answer 1

Answer: (A)

Given, total mass of water =M= 100 g $L_1=2.1\times {10}^5\text{J}\text{k}{\text{g}}^{-1}$ $L_2=3.36\times {10}^5\text{J}\text{k}{\text{g}}^{-1}$ Suppose mass of ice formed = m the mass of water evaporated = M ? m amount of heat taken by water evaporate $=(M-M)L_1$ On freezing heat given by water $=m{L}_2$ As, $M{L}_2=(M-m)L_1$ $\Rightarrow$ $m=\frac{M{L}_1}{L_1+L_2}$ $\Rightarrow$ $m=\frac{100(21\times {10}^5)}{(21+3.6)\times {10}^5}=86g$