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Mixing ice, steam, and water and find equilibrium

  1. Feb 20, 2007 #1
    1. The problem statement, all variables and given/known data
    .15 kg of steam at 100 degree Celcius is mixed with .2kg of water at 50 degree and .2 kg of ice at 0 degree. Determine the final temperature and phase(s) of the final state. Assume that no energy is lost to the surroundings.

    2. Relevant equations

    3. The attempt at a solution
    I am thinking that since energy is conserved, the steam will lose energy while the ice and water will gain.


    -Q_steam + Q_water + Q_ice = 0

    I know Q = mc*deltaT

    The answer key I have says that the end result will be 100 degree with .485 kg water, and .065 kg steam. I would try to test if the all of the steam and ice condense and melt if I didn't know this.

    So ice will all melt

    Q_ice = .2*2100*(T-0) + .2*L_f

    Q_water = .2*4190*(T-50)

    Q_steam = .15*c*(T-100) + x*L_v

    There are 3 possibilities, 1) none of the water boils 2) some of the water boils 3) all of the water boils. In this case, it is option 2. T must be 100

    Q_ice = .2*2100*100 + .2*3.34*10^5 = 108800
    Q_water = 41900
    Q_steam = x*2.256*10^6

    x = .0667 kg of steam turns into water

    If this is the case, I have 1.43 kg of steam left and .4667 kg of water after the ice melt. This answer is however incorrect. Where did I go wrong?
  2. jcsd
  3. Feb 20, 2007 #2
    i dont remember the specific values but for the heat capacities - did you use the values which had units J/g C or J/g K??

    If you used the latter then you should have converted all your celcius figures to kelvins.

    p.s. none of the water did boil after all, did it?
  4. Feb 21, 2007 #3


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    No, if the final answer has .485 kg of water and .065 kg of steam, then none of the water boiled. So it's option 1. In fact, some of the steam condenses.

    Qsteam = mLcond
    Qwater = (0.2)cwater(50)
    Qice = (0.2)Lfus + (0.2)cwater(100)

    Wher Lcond is the latent heat of condensation for steam to water, Lfus is the latent heat of fusion (melting) for ice to water, and cwater is the specific heat capacity of water. Note Lcond is a negative quantity. You want to solve for m in the following equation:

    mLcond + (0.2)Lfus + 30cwater = 0

    It's easy, just plug in numbers and isolate, or better yet isolate then plug in numbers. You then argue that since the value for m is between 0 and 0.15 (you should get this if the book is right), your solution is consistent thus the only possibility. If you want a less ad hoc approach, you can compare




    You should find that the first quantity is much bigger, hence in this set up, all of the ice will melt. So either all the steam condenses, none of it condenses, or some of it does. If all of it condenses, then it takes at least |0.15Lcond| away from the total energy, and the ice and water contribute at most 0.2Lfus + 30cwater to it, and you should find that there's no way in this situation for the total energy to be zero (it will have to be negative). If none of it condenses, well then it loses no energy and the ice and water strictly gain energy, so again there's no way for the total energy to be zero (it will have to be negative). Therefore, some of the steam must condense. Now you're justified in setting up the equations the way I did.
  5. Feb 21, 2007 #4
    Thank you very much for the help
  6. Apr 9, 2010 #5
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