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Heat Transfer. Ice and steam in a container (easy)

  1. Jan 19, 2017 #1
    1. The problem statement, all variables and given/known data
    "In a container of negligible mass, equal amounts (in weight) of ice at 0◦C and steam at 100◦C are mixed at the atmospheric pressure. Assuming no heat exchange with the surroundings, what is the temperature when the system reaches equilibrium? What are the fractions of weights of ice, water and steam?"

    I know this question has been asked about 100 times but Ive been working on it for over an hour and I just cant get it. I was even the best in my class at Physics 1 and 2. This should be especially easy because the initial conditions are at saturation conditions.

    2. Relevant equations

    Q=mL (Where L is latent heat of phase change)
    Q=mCΔT
    Lf = 3.34*10^5 J/kg (fusion)
    Lv = 2.256*10^6 J/kg (vaporization)
    Cwater = 4186 J/kg/K

    3. The attempt at a solution

    I assume 1 kg to make the math fun. mice = msteam = m

    Heat absorbed by the ice melting is:
    Qmelt = mLf =3.34*10^5 J

    Heat released by vapor condensing:
    Qcond = mLv = 2.256*10^6 J

    If the steam melts all of the ice, the excess energy would be the difference between the two:
    2.256*10^6 - 3.34*10^5 = 1.931*10^6 J

    And the heat required to get all of the 0°C water up to 100°C is:
    Qheat up = mCΔT = 4186 * 100 = 4.186*10^5 J
    which is less than the excess steam heat, so the steam should in fact melt all of the ice and turn it to 100°C

    Now to find the remaining energy in the steam:
    1.931*10^6 - 4.186*10^5 = 1.5124*10^6 J


    Oh wow, while typing this i thought of what to do lol. I will continue in hopes that my line-by-line explanations will help others. Edit: But I do have one question at the end!


    And the heat required to transform the 100°C hot water to steam would be:
    Q=mLv = 2.256*10^6

    And since the remaining energy in the steam is not sufficient enough to transform all the 100°C hot water to steam, we will have the original steam, hot water, and some newly created steam.
    So my guess is that the ratio of remaining energy to the energy needed to complete a full phase change will be equal to the ratio of the masses that have changed.
    So for this example, the percentage of hot water that turns into steam is
    1.5124 / 2.256 = 67%

    So we started with:
    1) 50% ice........... 50% steam
    2) 50% water....... 50% steam
    3) 16.5% water.... 83.5% steam

    This would be mass percentage, right?
    Thank you in advance :)


     
  2. jcsd
  3. Jan 19, 2017 #2
    A tidier presentation might be this .... Energy to melt and raise 1 kg ice to 100C = 3.34 + 4.186 = 0.7526 MJ

    How much steam required to condense to produce this energy = 0.7526 / 2.256 = 0.333 Kg

    So we have.... 0 ice ....1,333 water .....0.666 steam

    The question specifically asked " What are the fractions of weights of ice, water and steam?"

    So you must present your answer this way 0 ice ....0.666 water .....0.333 steam ...
    or 0% ...66.6% .....33.3%
    or in this case 'common' fractions would be better 0 ....2/3 ....1/3

    Very strange that the answer is almost exactly 0 , 1 and 2 thirds , yes this is mass (or weight if experiment performed under the effects of gravity)...
     
    Last edited: Jan 19, 2017
  4. Jan 19, 2017 #3

    BvU

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    No.
    No water gets evaporated. You want to write a proper energy balance, starting with indeed a mass fraction of the steam condensing. Gets you water at 100 C. That condensing gives you X Joule of energy and you use that to melt all the ice and heat up all the water to 100 C.
    [edit] Oz was faster.
     
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