# A thermally isolated vessel contains 100 gram of water at zero degree celsius when air above the water is pumped out some of the water freezes and some water evaporates at zero degree celsius itself then the mass of the ice found if no water is left in the vessel latent heat of vaporization of water is zero degree celsius is equals to 2.10 into 10 to the power 6 joule per kg and latent heat of fusion of ice is equals to 3.36 into 10 to the power 5 joule per kg

Rajdwip biswas
19 Points
5 years ago
Latent heat of vapourisation of water at 0 degree celcius,L1= 2.10×10^6 j/kg =2.10×10^3 j/gLatent heat of fusion of ice,L2=3.36×10^5 j/kg=3.36×10^2 j/gLet mass of ice formed = m gramThen mass of water evaporated = (100-m) gramAs no water left in the vessel,Heat gained by water in evaporation = heat lost by water in freezing or (100-m)L1 = m L2 Or (100-m) ×2.10×10^3 = m × 3.36×10^2Or m = 86.2 g
3 years ago
Dear student,

Latent heat of vapourisation of water at 0oC, L1 = 2.10×106 J/kg
= 2.10×103 J/g
Latent heat of fusion of ice, L2 = 3.36×105 J/kg
= 3.36×102 J/g
Let mass of ice formed = m gram
Then mass of water evaporated = (100 – m) gram
As no water is left in the vessel, Heat gained by water in evaporation = heat lost by water in freezing
or, (100 – m)L1 = mL2
or, (100 – m) × 2.10×103 = m × 3.36×102
Hence, m = 86.2 g

Thanks and regards,
Kushagra