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Heat of Fusion Question

  1. Jul 28, 2013 #1
    1. The problem statement, all variables and given/known data
    I did an experiment where crushed ice was added to a known volume of warm water and stirred until all the ice melted. My initial water temperature was 40.2°C and the final water temperature was 3.1°C. The known volume of water is 150 mL = 150 g and the melted ice was found to be 93 mL = 85.3 g.

    The question is: What would be the final temperature if the ice was initially at -10°C when it was added to the water?

    2. Relevant equations

    Q = mLf where Lf = heat of fusion = 3.33 * 10^5 J/kg

    C (water) = 4186 J/kg C
    C(ice) = 2100 J/kg C

    3. The attempt at a solution

    This is what I have:

    (heat to raise 85.3 g of ice from -10 to 0°C) + (heat to change 85.3 g of ice to water) + (heat to raise 85.3 g of water (melted ice) from 0°C to final temperature) = (heat lost by the 150 g of water from 40.2°C to final temperature)

    [m(ice) * c(ice) * (0 - (-10))] + [m(ice) * Lf] + [m(ice)* c(water) * (T - 0)] = [m(water) * c(water) * (40.2 - T)]

    T = - 5.03 °C

    Is that right?
     
  2. jcsd
  3. Jul 28, 2013 #2
    For the final temperature to be subzero, some water would have to freeze. You did not account for that. So that cannot be right.
     
  4. Jul 28, 2013 #3
    Ok since I had to stir the contents until all the ice melted - the final temperature has to be positive then?

    - I took into account bringing the ice from - 10 to 0 (its melting point), then changing the ice from solid to liquid, and then the water from 0 to the final temperature

    - this is all equal to the heat energy lost from the warm water from its initial temperature to the final unknown temperature

    where did I go wrong in my equation?
     
  5. Jul 28, 2013 #4
    You assume that the water has enough heat to bring all the ice to melting and melt it. But is that really so?
     
  6. Jul 28, 2013 #5
    I'm not sure now?! I mean in the experiment, maybe it does theoretically? What else would I need to consider in my equation then?
     
  7. Jul 28, 2013 #6
    "I'm not sure" is not a good answer. Find out, that is important. If the water does not have enough heat, what is the end result?
     
  8. Jul 28, 2013 #7
    So that means all the added ice didn't melt. So the final product would be a mixture of ice and water. So the temperature that I calculated is just the final temperature of my end product of ice and water.
     
  9. Jul 28, 2013 #8
    Correct.

    How can water exist in the liquid form at - 5.03 °C (and normal pressure)?
     
  10. Jul 28, 2013 #9
    So I need to add something to my equation or is it all wrong? Since my initial temperature was not warm enough to melt all of the ice does that mean there is no final temperature for the question?
     
  11. Jul 28, 2013 #10
    Find out exactly how much ice melts. What is the temperature of the remaining ice? What is the temperature of the remaining water?
     
  12. Jul 28, 2013 #11

    haruspex

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    This is an actual experiment, right? If so, the conclusion is that you made an experimental error. Either some ice remained or one of your measurements is wrong, one of:
    - there was less ice
    - there was more water,
    - the water was hotter
    Btw, you wrote
    How was the 93mL determined? If it was by the total water at the end then that is indeed the volume of melted ice, so the mass of ice would be 93g. (That only makes the error worse, though.)
     
  13. Jul 28, 2013 #12

    Yea this was an actual experiment. The measured volume of water was 150 mL and the final volume after the ice was added was 243 mL which gave me 93 mL of ice added. So I used the density of ice to calculate the mass of ice
     
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