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Heat and change of phase problem

  1. Sep 12, 2008 #1
    1. The problem statement, all variables and given/known data
    In an insulated vessel, 250g of ice at 0 C is added to 600g of water at 18.0 C.

    A) What is the final temperature of the system?
    B) How much ice remains when the system equal equilibrium?

    2. Relevant equations
    [tex]Q = mc \Delta T + mL [/tex]

    3. The attempt at a solution

    I can't figure out what is wrong when I do it but I am sure its my concept. The answer in the back for temperature final is 0 C. I have tried setting up this equation multiple ways but I still can't seem to get the answer. I think the part I don't understand is what do I do for the ice side of the equation. If the final answer is 0 then the ice never finishes its phase change so what do I do for that side of the equation? Here are a couple of ones I have tried:

    [tex] m_{ice}L{f} + m_{water}c_{water} \Delta T - m_{water}L_{f} = 0[/tex]

    Basically I have just tried various versions of that, with the latent heat part all over the place, and the the ice switching to water as well, but none seem to work. Can anyone help me get this started?
  2. jcsd
  3. Sep 12, 2008 #2


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    The issue is how much heat can be transferred to the ice from the 600 g of water? Then what latent energy is available to be absorbed by the 250 g? Whatever portion of that then is left should let you know how much ice is left shouldn't it?
  4. Sep 12, 2008 #3
    So would I set the latent heat of ice = to mcdeltaT of water?

    [tex]m_{ice}L_{f} + m_{water}c_{water} \Delta T = 0[/tex]
  5. Sep 12, 2008 #4
    I still don't understand
  6. Sep 12, 2008 #5


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    What number are you using for the specific heat for ice to go to 0 degree water? That times its mass is how much heat it will absorb before it ALL melts. If you can supply only half the heat, then you may suppose that half is melted - all other things being equal.

    So what is your heat budget? How much heat is in the water? There is 18 degrees of warmth in the water and the water can give up 18 degrees worth to the ice. When they are in equilibrium then the water is all at 0 degrees as of course is the ice. But not all of the initial ice is still ice. Just what ice has its specific heat requirement still unfulfilled.
    Last edited: Sep 12, 2008
  7. Sep 13, 2008 #6
    I haven't gotten past the first part yet. I am trying to find what the final temperature of the whole system would be once in equilibrium. So I don't know that the final temperature is 0 yet, and thats what I can't figure out how to solve.
  8. Sep 13, 2008 #7
    I don't see how you can solve it since you don't know whether the ice made it through the latent fusion part, or whether it didn't.
  9. Sep 13, 2008 #8


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    How many calories are in 600 g of water at 18 degrees - relative to 0 degrees C?

    That identifies your heat budget. I send you to the store and I only give you so much money to buy ice. If I give you more than what the store has, then you come home with all the ice and some money left over. If you can't buy all the ice, then what's left in the store is what the question is asking. In the case where you come home with money, then that's how many calories you have left still in the original 600 g of water and the now additional 250 g of melted ice, so you calculate it back then according to the 850 g with whatever calories are in your pocket. In the case where you can't buy it all, then the system must be at 0 degrees.
  10. Sep 13, 2008 #9
    Guess I was making it way more complicated then it needed to be. Thanks for the help.
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