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Mixing ice and water

  1. Jun 22, 2017 #1
    1. The problem statement, all variables and given/known data

    1Kg ice at -10° C is mixed with 1 Kg water at 100°C .Then find equilibrium temperature.

    2. Relevant equations

    Specific heat of water = 1Cal/g/°C
    Specific heat of ice = 0.5 Cal/g/°C
    Latent heat of fusion of ice = 80 Cal/g
    3. The attempt at a solution

    Heat required by ice to reach at 0°C= 1000x0.5x10
    = 5000Cal

    Heat required by ice at 0°C to convert in water = 1000x80
    = 80000Cal

    Heat released by water when it moves from 100°C to 0°C = 1000x1x100= 100000Cal

    Surplus heat = 15000 Cal

    This surplus heat increases the temp. of 2Kg water at 0°C

    15000Cal = 2000x1x∆t

    Increase in temperature = 15000/2000

    = 7.5°C

    Is my answer correct ?
     
  2. jcsd
  3. Jun 22, 2017 #2

    haruspex

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    Looks good.
     
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