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Question: Answered & Verified by Expert
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 °C itself. The mass of evaporated water will be closest to [Take the latent heat of vaporization of water, Lv=2.10×106 J kg-1 and latent heat of fusion of water, Lf=3.36×105 J kg-1]
PhysicsThermal Properties of MatterJEE Main
Options:
  • A 20 g
  • B 130 g
  • C 35 g
  • D 150 g
Solution:
1791 Upvotes Verified Answer
The correct answer is: 20 g

Let us assume the mass of the water evaporated is m, then

mLv=150-mLf

m×2.1×106=150-m×3.36×105

On solving we get

m20 g

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