150 g of water at 0°C is contained in a thermally insulated container. Now the air from the vessel is pumped out adiabatically. A fraction of water gets converted into ice and the remaining evaporates at 0° C itself. The mass of evaporated water will be closest to [Take the latent heat of vaporization of water, L v = 2. 10 x 10 6 J kg -1 and latent heat of fusion of water, L f = 3. 36 x 10 5 J kg -1 ]
Text Solution
Verified by ExpertsA
20 g
Let us assume the mass of the water evaporated is m, then
mL v = (150 - m)L f
m x 2. 1 x 10 6 = (150 - m) x 3. 36 x 10 5
On solving we get
m 
20 g
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