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A thermally insulated vessel containing 100 grams of ice at 0 degrees Celsius . Water at 55 degrees celsius is poured and after some time , the system reaches the equilibrium at a temperature of 10 degrees Celsius. That amount of water in grams was placed . ( heat fusion ice 80cal / g . The specific heat capacity of water is about 1cal / ( gr.K )

asked by David
  1. Here suppose mass of ice formed=m gram
    Therefore, mass of water evaporated= 100-m gram
    As, no water is left in he vessel ,
    Amt. Of heat taken by water in evaporation = amt. Of heat given out by water in freezing
    (100-m)×2.1×10^6/10^3=m×3.36×10000/1000
    m=2100/24.36
    m=86.2g

    posted by Nisha

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