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Heat and Phase Changes

  1. Dec 7, 2008 #1
    I'm working on the following homework problem but have just been stumped:

    A 0.185 kg piece of aluminum that has a temperature of -155°C is added to 1.5 kg of water that has a temperature of 2.1°C. At equilibrium the temperature is 0.0°C. Ignoring the container and assuming that the heat exchanged with the surroundings is negligible, determine the mass of water that has been frozen into ice.

    So I started by writing down the data for the aluminum and water, and tried the following:
    Qalum.+Qwater+Qice = 0
    mcΔT + mcΔT + mLvapor. = 0
    (0.185)(9e2)(0-(-155)) + (1.5)(4186)(0-2.1) + (1.5 - x)(22.6e5) = 0
    x = 1.50558 kg of ice was frozen, which, unfortunately for me, is greater than the amount I started with. :(

    If anyone could point out what I'm doing wrong or if I'm missing some major concept, it would be awesome!
  2. jcsd
  3. Dec 7, 2008 #2


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    Hi jason.maran,

    The number 22.6e5 seems too large. I believe that is the heat of vaporization, and here you need the heat of fusion.

    Also, compare compare the signs of the three terms. Water is losing heat, and so its heat change terms should be negative, so I think you have a sign error here.
  4. Dec 7, 2008 #3
    I'm still not getting the answer correct, I've now got:
    (0.185)(9e2)(155) - (1.5)(4186)(-2.1) - (1.5 - x)(33.5e4) = 0
    x = 1.3836 kg
    I'm wondering if my quantity (1.5 - x) is wrong for finding the amount of ice formed?

    Thanks for your help

    EDIT: I'm getting my values for L from: http://www.webassign.net/CJ/12-03tab.gif if it matters.
  5. Dec 7, 2008 #4


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    You don't need to add the negative sign for the second term; it was already negative. (You've now made the second term positive!)

    The mcΔT terms automatically have the right sign; but when you have an mL term you have to decide if the substance is melting or freezing, and then give it the appropriate sign.

    Do you get the right answer?
  6. Dec 7, 2008 #5
    Thanks alphysicist -- that was it, thanks for the help. I really appreciate it! :)
  7. Dec 7, 2008 #6


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    Glad to help!
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