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Ice melting problem

  1. Nov 14, 2006 #1
    I have done some calculations so far but i am kind of stuck, here is the problem:

    A jar of tea is placed in sunlight until it reaches an equilibrium temp of 32.4 deg C. In an attempt to cool the liquid, which has a mass of 185g, 113g of ice at 0degC is added. Assume specific heat capacity of the tea to be that of pure liquid water. At the time at which the temp of the tea is 30.9degC, find the mass of the remaining ice in the jar (in grams)
    This is the work i have so far:
    Q = heat gained by ice
    = mcdeltaT
    = 0.113kg x 2090J/kgdegC x (30.9C-0C)
    = 7 297.653 J

    Q = heat lost by water
    = mcdeltaT
    = 0.185kg x 4186J/kgdegC x (30.9C-32.4C)
    = -1 161.615J
    So,
    ice from 0degC -> 30.9degC = 7297.653J
    water from 32.4degC -> 30.9degC = -1161.615J

    In my book it suggests to do Q that is left = Qwater-Qice, this gives me 8459.268J

    I really am not sure what do do from here on, any suggestions? I have to find the grams remaining in the jar:confused:

    Thanks

    Sergio
     
  2. jcsd
  3. Nov 14, 2006 #2
    Ok did a little more work and i think i might have the solution, can somebody tell me if im doing this correctly...

    using the 8459.268J i calculated for the Q that is left, i did:
    mass= Q/Lf(of ice)
    = 8495.268J / 33.5e+4J/kg
    =0.02525kg (this is amount of ice melted?)

    using that then i just subtract my initial mass of ice 0.113kg by 0.02525kg and convert into grams giving my final answer to 87.75grams of ice not melted

    this seems reasonable, can somebody tell me if my thought process is correct?
     
  4. Nov 14, 2006 #3

    OlderDan

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    Ice does not warm up from 0°C to 30.9°C. Ice melts at 0°C and the water that results warms up.
     
  5. Nov 14, 2006 #4
    so this being said it is -7297.653?
     
  6. Nov 14, 2006 #5

    OlderDan

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    I have no idea what this refers to.
     
  7. Nov 14, 2006 #6
    Q = heat gained by ice
    = mcdeltaT
    = 0.113kg x 2090J/kgdegC x (30.9C-0C)
    = 7 297.653 J

    If its not heat gained by the ice...it is lost? therefore the number should be negative?
    I'm kind of loosing grip on the problem...perhaps im thinking to much into it

    I'm assuming you are suggesting to rethink this equation (and that the rest of my calculations are correct?)

    stick in the bicycle wheels :S
    Sergio
     
  8. Nov 14, 2006 #7

    OlderDan

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    The ice temperature cannot go above 0°C.

    What have you learned about the latent heat associated with changes of phase? It takes a lot of heat to melt ice at 0°C to change it to liquid water at 0°C. Then it takes additional heat to change the temperature of that melted water from 0°C to 30.9°C.
     
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