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Mixing of substances of different temperature with change of state

  1. Jun 12, 2012 #1
    Two questions:
    1.1. The problem statement, all variables and given/known data
    Steam at 100 °C is pumped into an insulating pot containing a mixture of ice and water at 0°C. There are 0.4kg of ice and 1.6kg of water in the pot. Assume there is no heat lost to the surroundings, how much steam (in kg) is needed so that all ice is just melt?

    specific heat capacity of water = 4200 Jkg-1°C-1
    specific latent heat of fusion/vaporization = 334000/2260000 Jkg-1

    1.2. Relevant equations
    E = mcΔT
    E = ml

    1.3. The attempt at a solution
    Energy released by steam going to 0°C (per kg of steam)
    =(2260000 + 4200*100) J kg-1
    =2680000 J kg-1
    Energy needed to turn 0.4kg of ice into water
    =0.4*334000 J
    =133600 J
    Amount of steam needed
    =133600/2680000 kg
    =0.0499kg
    However, the answer is 0.591kg. What have I done wrong?


    2.1. The problem statement, all variables and given/known data
    4kg of ice at -10°C is mixed with 0.1kg of water at 10°C. What is the final state of the mixture? What is the final temperature of the mixture?
    specific heat capacity of ice = 2100 Jkg-1°C-1


    2.2. Relevant equations
    Same as above


    2.3. The attempt at a solution
    Energy required to freeze the water
    =0.1*10*4200+334000*0.1 J
    =37600 J
    [tex]4(T+10)\times 2100 = 0.1(-T)\times 2100 + 37600[/tex]
    [tex]T=-5.39[/tex]
    So final state is solid and final temperature is -5.39°C. Answer says that final temperature is 0°C. What have I done wrong?
     
    Last edited: Jun 12, 2012
  2. jcsd
  3. Jun 12, 2012 #2
    Hi dalcde! :smile:

    For question (1) you have unnecessarily considered the heat being used to raise the temperature of water. The heat from the steam will only be used to change the state of ice.


    For question (2), you need to use the fact that the amount of heat lost by the ice will be equal to the gain in heat by water to change its state, before there is any change in its temperature.
     
  4. Jun 12, 2012 #3
    Where did I consider the heat used to raise the temperature? I've only considered the energy released by cooling the steam to 0°C.

    What do you mean? The specific heat capacity of ice and water are different, so they have to be separated in one way. In the equation, the left hand side is energy gained by the ice, and the right hand side is the energy lost by the thing that was originally water.
     
  5. Jun 13, 2012 #4
    I misunderstood your solutions. After solving it, I get the same answers.

    The given answer for the first question is obviously wrong, though. 0.591 kg steam is way too much to melt 0.4 kg ice.
     
  6. Jun 13, 2012 #5
    Sorry, it was 0.0591kg (still wrong anyways). I'll edit it.

    EDIT: Oh! Somehow I can't.
     
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