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Thermodynamics- conservation of heat and energy

  1. Nov 11, 2015 #1
    1. The problem statement, all variables and given/known data
    A 3.60 kg block of copper at a temperature of 92 °C is dropped into a bucket containing a mixture of ice and water whose total mass is 1.50 kg. When thermal equilibrium is reached the temperature of the water is 5 °C. How much ice was in the bucket before the copper block was placed in it?
    (ci = 2000 J/(kg.°C), cw = 4186 J/(kg.°C), Lf=3.35 × 105 J/kg, Lv=2.26 × 106 J/kg, ccopper = 387 J/(kg.°C). Neglect the heat capacity of the bucket.)

    2. Relevant equations
    Q=mcΔT
    Q=mL

    3. The attempt at a solution
    Because of conservation of heat, the energy the copper block loses must be gained by the ice/water system.

    So; QLOST= QGAINED

    Expanded it should become something like;

    (cmΔT)COPPER =
    (heat to raise temperature of ice) + (heat to raise temperature of water) + (latent heat to melt ice) + (latent heat to vaporize water)


    However I think there are too many unknowns at this point.

    Any idea on what a possible next step could be / look like, or if I'm even doing this remotely correctly?
    All help appreciated!!
     
  2. jcsd
  3. Nov 11, 2015 #2

    JBA

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    Based upon the temperatures given, why are you including the latent heat of vaporization of water?
     
  4. Nov 11, 2015 #3

    SteamKing

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    Since thermal equilibrium is reached at 5° C, you don't have to worry about vaporizing any water.

    This is essentially a book keeping problem, where you are trying to balance the heat given up by the copper block to the heat gained by the water/ice mixture.

    I would start by figuring out how much heat the copper block gives up. Then I would see how much ice would be needed to absorb this heat given up by the copper.
     
  5. Nov 11, 2015 #4
    I figured out how much energy the copper gives up, its 121208 J.
    So when I set them equal to each other I get:

    121208 = (1.5-mwater)(2000)(ΔT) + (1.5 -mice)(4186)(ΔT) + (1.5- mwater)(3.35x105)
    energy lost by copper = energy gained by ice +energy gained by water + latent heat to melt ice

    I have 3 unknown variables to solve for
     
  6. Nov 11, 2015 #5

    JBA

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    Remember you are given the total mass of the ice and water combined.with the mass of ice as the unknown.
     
  7. Nov 11, 2015 #6
    Right, so for the last part (latent heat) it would just be (1.5)(3.35x105) But for the first two i don't see how I can separate the mass since i dont know the original of either. I know the combined, but not individual.
     
  8. Nov 11, 2015 #7

    JBA

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    If you know the total heat input and the total heat absorption of the combined ice and water mass then what is the remaining unknown and what part of that unknown do you know about the ice.
     
  9. Nov 11, 2015 #8

    SteamKing

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    No, this is not correct. Some, but not all of the water/ice mixture, is water already, so the latent heat is required only to melt the ice.

    Look, if the mass of water + ice = 1.5 kg, it stands to reason that by letting x be the amount of ice in the mixture, (1.5 - x) will be the amount of water you start with.
     
  10. Nov 11, 2015 #9
    Ok, that makes sense.
    re-written:
    121208= (1.5-mwater)(2000)(ΔT) + (mwater)(4186)(ΔT) + (1.5 - mwater)(3.35x105)
    I could simplify some more but still ΔT and one of the masses is unknown.
    I'm having a real mental block with this one. Normally with two variables I would do some kind of substitution but I can't think of anything to substitute.
     
  11. Nov 11, 2015 #10

    JBA

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    Once the of the total heat capacity of the combined ice and water is determined and subtracted from the total heat capacity of the copper block, what is left? With that information, what do you know about the two heat absorption rates of the ice mass and the temperatures at which those rates occur.
     
  12. Nov 11, 2015 #11
    I think the thing you're missing is the initial temperature of the ice-water mixture. Also, once the ice melts, it turns into water.
     
  13. Nov 11, 2015 #12

    SteamKing

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    You're making some extra work for yourself here.
    If you assume x = mass of the ice, then you can replace (1.5 - mwater) with x and save some unnecessary calculation. mwater = (1.5 - x), of course.
    Perhaps this article can answer the question about the initial temperature of the water/ice mixture:

    https://en.wikipedia.org/wiki/Frigorific_mixture

    Just take things slow and steady. Work through the arithmetic and see if you can't wind up with x = some expression.
     
  14. Nov 11, 2015 #13
    Thank you so much! using x made a lot more sense, and simplified it down. Once again, I appreciate the help
     
  15. Nov 11, 2015 #14
    Don't forget that after the ice melts, it's all water!
     
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