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Homework Help: Heat Exchange: Ice added to water

  1. Jan 29, 2012 #1
    1. The problem statement, all variables and given/known data

    A glass containing water is initially of temperature 20oC. The mass of the glass is 100 g and of specific heat 0.16 cal/g-co and water is of mass 200g. If 5 cubes of ice each of mass 15g will be added to the glass of water, solve for
    (a) The temperature of the mixture.
    (b) The mass of ice remaining if any.

    2. Relevant equations


    3. The attempt at a solution

    Ok so used the equation for heat exchange and I arrived at the equation,

    mCΔt of water + mCΔt of glass + mCΔt of ice = 0

    My concern is the Specific heat that I will use for C of ice, because my teacher said to only use 0.5 when below 0 degrees but it doesn't really say its temperature in the problem. Will I use 1 or 0.5?

    And for part b, I need some tips on how to start solving for the possible remaining ice
  2. jcsd
  3. Jan 29, 2012 #2
    well if you want to find the temperature of the mixture, you'll have to do the following:

    [tex]0=\mu_{cup}(\theta_{f}-\theta_{i})+m_{1}C_{w}(\theta_{f}-\theta i)+m_{2}C_{ice}(\theta_{f}-\theta_{i})[/tex]

    after you write that down you'll have to foil everything out then factor out what needs to be factored out with θf and just do simple algebra to find the final temperature.
  4. Jan 29, 2012 #3


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    The ice is probably starting at 0o, so you don't have to worry about it warming up. You do have to account for the ice melting though - you need to use the latent heat of fusion.

    Part (b) suggest not all the ice will be melted, so I would be working out how much heat energy has to be absorbed to cool the glass and water to 0o, and see how much ice that heat would melt.
    Suppose that will melt 55 grams of ice, then the final mix will be at 0o, with 20g [of the original 75g,] still remaining.
  5. Jan 29, 2012 #4


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    SO what about melting the ice!!
  6. Jan 29, 2012 #5

    So you mean that instead of (mCΔt) of ice for the Qloss, i'll just have to use (mLatent) of ice instead and that only?
  7. Jan 29, 2012 #6


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    In most of these problems, yes.

    Sometimes we are told that the ice cubes begin at a negative temperture - like -10o - and you then also need a bit of (mCΔt) of ice to allow for the ice warming up to 0o. If you are not given the beginning temperature of the ice, it is reasonable to assume it begins at 0o.
  8. Jan 30, 2012 #7
    I tried that, making the equation

    mcΔt of water + mcΔt of glass +mLf

    I used this for part a. but i get a negative value,-7.78 degrees is that possible, since i assumed that the ice started from 0 degrees it can't go lower than that can they?
  9. Jan 30, 2012 #8

    some part of the ice melts, and the other doesnt. madali lng yn hw :)
  10. Jan 30, 2012 #9
    Ano sagot? haha!
  11. Jan 30, 2012 #10


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    That negative temperature is an indicator that not all the ice melted. If you considered the ice had a smaller mass, the answer would have been zero degrees.
    That leads to part (b)

    re-read post #3; my first response.
  12. Jan 30, 2012 #11
    Thanks. I get it now. :D Last concern, is the negative temperature acceptable as the final temperature of the mixture? Just so I can clear things out. Haha

    Thanks again! :D
  13. Jan 30, 2012 #12


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    The final temperature will never be higher than the beginning highest temperature, nor lower than the beginning lowest temperature.

    If we assume the ice begins at 0o that means the final temperature will be at, or above 0o.

    The only way to get a negative final temperature would be to begin with ice at a very negative temperature - and then the final mix would be solid [ice].
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