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Question
A thermally isolated vessel contains 100 g of water at 0∘C. When air above the water is pumped out, some of the water freezes and some evaporates at 0∘C itself. Calculate the mass of the ice formed if no water is left in the vessel. Latent heat of vaporization of water at 0∘C = 2.10 × 105 J kg−1 and latent heat of fusion of ice = 3.36 × 105 J kg−1
A thermally isolated vessel contains 100 g of water at 0∘C. When air above the water is pumped out, some of the water freezes and some evaporates at 0∘C itself. Calculate the mass of the ice formed if no water is left in the vessel. Latent heat of vaporization of water at 0∘C = 2.10 × 105 J kg−1 and latent heat of fusion of ice = 3.36 × 105 J kg−1
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Solution
The correct option is C 86g
Total mass of the water = 100g.
Latent heat of vaporization of water at 0∘C = L1 = 21.0 × 105 J kg−1
Suppose, the mass of the ice formed = m.
Then the mass of water evaporated = M−m.
Heat taken by the water to evaporate = (M−m)L1 and heat given by the water in freezing = mL2.
Thus, mL2 = (M−m)L1
or, m = ML1L1 + L2
= (100 g)(21.0 × 105 J kg−1)(21.0 + 3.36) × 105 J kg−1 = 86 g.
86g
Total mass of the water = 100g.
Latent heat of vaporization of water at 0∘C = L1 = 21.0 × 105 J kg−1
Suppose, the mass of the ice formed = m.
Then the mass of water evaporated = M−m.
Heat taken by the water to evaporate = (M−m)L1 and heat given by the water in freezing = mL2.
Thus, mL2 = (M−m)L1
or, m = ML1L1 + L2
= (100 g)(21.0 × 105 J kg−1)(21.0 + 3.36) × 105 J kg−1 = 86 g.
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