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How to determine final temperature of mixture if initial substances are at MP/FP?

  1. Jan 9, 2013 #1
    1. The problem statement, all variables and given/known data
    A 100g piece of ice at 0.0°C is dropped into 500g of water at 100°C. What is the final temperature of mixture.



    2. Relevant equations
    Q=mL (latent heat)
    Q=mcΔt
    Qw=-Qx (energy absorbed by water is equal but opposite to that of substance inserted into water)
    specific heat ice: 2.09 * 10^3 J/kg * °C
    specific heat water: 4.186 * 10^3 J/kg * °C
    specific heat steam: 2.01 * 10^3 J/kg * °C
    latent heat of fusion: 3.33 * 10^5 J/kg
    latent heat of vaporization: 2.26 * 10^6 J/kg

    3. The attempt at a solution
    Do you find the total heat including phase change heat for both substances, using final temperature as an unknown variable? If so:
    Q = mcΔt + mL
    Qw = (0.5 *4.186 * 10^3 *(Tf-100)) + 0.5 * 2.26 * 10^6
    Qx= -((0.5 *4.186 * 10^3 *(Tf-0)) + (0.5 * 3.33 * 10^5))
    I set both equal after simplifying, and obtained a final temperature of 264.14°C, which is too bizarre of an answer.
     
  2. jcsd
  3. Jan 9, 2013 #2

    SteamKing

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    The water at 100C is still liquid, so the heat of vaporization is not needed.
    However, the ice at 0C is solid, and the heat of fusion must be added to the ice to melt it.
     
  4. Jan 9, 2013 #3

    TSny

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    Only the piece of ice undergoes a phase change. The hot water remains water as it cools to the final temperature. So, think about how you wrote Qw. Also, I think you used the wrong mass for the ice in your expression for Qx.
     
  5. Jan 9, 2013 #4
    Thanks for the help. Not including the heat of vaporization, I got 28.08 C. I also did have a mistake in calculating the energy of the piece of ice where I put it the mass of the water instead. Is 28.08 a reasonable answer, or should it be closer to 100 C because the more massive object would have more of an influence on the temperature?
     
  6. Jan 9, 2013 #5

    TSny

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    I get the final temperature to be quite a bit higher, in agreement with your intuition.
     
  7. Jan 10, 2013 #6
    To get 28.08 I did
    Qw=-Qx
    (0.5 *4.186 * 10^3 *(Tf-100)) = -((0.1 *4.186 * 10^3 *(Tf-0)) + (0.1 * 3.33 * 10^5))
    2093Tf - 209300 = -418.6Tf - 33300
    2511.6Tf = 176000
    Tf = 70.07 C

    I think I have it, I might have made some original mistake in calculation.
     
  8. Jan 10, 2013 #7

    mfb

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    That looks good.

    As a rough approximation, ice at 0°C corresponds to (hypothetical) water at -80°C. So you heat 100g by 150° and cool five times this amount by 1/5 of the temperature difference. The .07 are the deviation from this approximation ;).
     
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