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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 × 10 6 J kg - 1 and latent heat of fusion of water, L f = 3 . 36 × 10 5 J kg - 1 ]

Options

  1. A20 g
  2. B130 g
  3. C35   g
  4. D150 g

Correct answer

A. 20 g

Step-by-step solution

Let us assume the mass of the water evaporated is m , then m L v = 150 - m L f m × 2 . 1 × 10 6 = 150 - m × 3 . 36 × 10 5 On solving we get m ≈ 20   g

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