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Heat of fusion for water

  1. Aug 14, 2010 #1
    Hello again,

    I did a lab where ice was added to 150 mL of warm water and the temperature was observed and recorded. (The mass of the ice was not important; but after it all melted the total volume of the water in the cup was 218 mL)

    Here's the question I have, "Assuming that the paper cup is a good insulator and that no heat energy was lost to the outside environment, estimate the heat of fusion of water. Note that part of the heat energy was absorbed by the melted ice to raise its temperature from 0°C
    to the final temperature. "

    How would I go about doing this? Is it just Q = mL = 0.150 kg x 333 KJ/kg ?

    Here's what I tried:

    If I find the heat lost by the water using the "Q=mc/\T" equation I get 20 kJ for "Q".

    Then using Q=mL,
    I get L = 20kJ/0.15kg =133 kJ/kg

    Does that look right, based on what they are asking? Since the actual heat of fusion is 333 kJ/kg, my experimental data must be off?
  2. jcsd
  3. Aug 14, 2010 #2
    Is it really 0.15kg? Or some other value?
  4. Aug 14, 2010 #3
    I'm not sure? I put that based on the idea that I started with 150mL of water and since water's density is 1g/mL, 150 grams or .150 kg

    Is that not right?
  5. Aug 14, 2010 #4
    The "m" you should be using should be that of the ice instead of the water?
    Also, you appear to have neglected the heat gained by the melted ice.
  6. Aug 14, 2010 #5
    Oook, I think I need the mass of the ice to determine this don't I? Since the final volume was 218 mL, I subtracted that from the initial volume of 150mL to give me 68 mL. Then since water's density is 1g/mL, I used .068kg for the above equation and came out with a value of 294 kJ/kg for L.

    Is that correct?
  7. Aug 14, 2010 #6
    Do I need that? And what calculation do I need for that? Is it Q = mc(delta)T?
  8. Aug 14, 2010 #7
    Yes. The melted ice doesn't stay at 0°C, it gets heated up till the equilibrium temperature. (Besides, the question explicitly reminded "Note that part of the heat energy was absorbed by the melted ice to raise its temperature from 0°C to the final temperature" :p)
    Yup, same equation, but just take note of the values.
  9. Aug 14, 2010 #8
    Q = mc/\T = 0.068kg x 0.5 x (0-7.1) = -0.24 kcal x 4.186 kJ/kcal = -1.01kJ

    Does that say that the ice lost 1.01 kJ? Should I subtract that value from the 294 value?
  10. Aug 14, 2010 #9
    The melted ice lost heat?!
    Basically, energy lost by original water = energy used to melt ice + energy used to heat melted ice. Use that equation (essentially conservation of energy) to compute. Why would you mix energy with specific latent heat of fusion?
  11. Aug 14, 2010 #10
    I know, it doesn't make sense! Lol.

    Well I originally calculated the heat lost by the water using (Q = mc/\T)
    Q = 150g x 1.00 cal/g-C x (39.6 - 7.1) = 4875 cals
    Q = 20 kJ

    So, 20kJ = mc/\T + mL ?? And everything right of the equal sign are the values for ice (ie: mass= 68 g, c = 0.5 cal/g-C, T1 = 0 degrees Celsius, T2 = 7.1 degrees Celsius)?

    This question seems like it should be easy; or am I just stupid? lol
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