# A thermally insulated vessel contains $150\; g$ of water at $0^{o}C.$ Then the air from the vessel is pumped out adiabatically. A fraction of water turns into ice and the rest evaporates at $0^{o}C$ itself. The mass of evaporated water will be closest to :(Latent heat of vaporization of water =$2.10\times 10^{6}\; J\; Kg^{-1}$ and Latent heat of Fusion of water = $3.36\times\; 10^{5}J\; kg^{-1}$)  Option 1)$150\; g$Option 2)$130\; g$Option 3)$35\; g$  Option 4)$20\; g$

Let m = mass of water evaporates in (gram)

$\Delta Q_{req}=mL_{v}$

So M = mass which convert into ice (in gram)

$M =(150-m)$

$So\; \Delta Q_{req}=mL_{f}$

$\\ \Delta Q_{req}=\Delta Q_{rel}\\\; \; \; \; \\mL_{v}=(150-m)^{L_{f}}$

$m(L_{v}+L_{f})=150 L_{f}$

$m=\frac{150\; L_{f}}{L_{v}+L_{f}}=\frac{150\times3.36\times10^{5}}{3.36\times10^{5}+2.20\times10^{+6}}$

$m=20\; g$

Option 1)

$150\; g$

Option 2)

$130\; g$

Option 3)

$35\; g$

Option 4)

$20\; g$

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