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Heat exchange in an isolated system

  1. Jan 8, 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 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?

    2. Relevant equations
    Q = mCΔT
    Q = mLf
    Q = mLv

    3. The attempt at a solution
    1) First I began by calling the mass M.
    2) Then I calculated the heat required to melt the ice since it is already at 0°C:
    Q = MLf = M(3.34×105)​
    3) Then I did the same but for the steam at 100°C:
    Q = MLv = M(2.256×106)​
    4) You can see it requires more heat loss to condense the steam versus melt the ice so now we have a solution consisting of just water and steam.
    5) This is where I run into trouble. I know we have all the ice (mass M) now in the form of water plus some of the steam that was condensed to water. I assume that the equilibrium temperature is under 100°C therefore the solution would be 100% water but I don't know how to show that or find the equilibrium temperature.
  2. jcsd
  3. Jan 8, 2017 #2
    How much heat would it take to melt all the ice and then raise the temperature of the water produced to 100 C?
  4. Jan 9, 2017 #3
    That's what I was thinking but since some of the steam has condensed to water I thought I would need to consider the change in mass for both the water and steam.
  5. Jan 9, 2017 #4
    You do, in a way. Condensing of the steam at 100 C produces more water, and reduces the amount of steam.
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