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Latent heat of Fusion and transfer problem

  1. Dec 16, 2013 #1
    1. The problem statement, all variables and given/known data
    A 100g (.1kg) cube o ice at 0dC (d=degrees) is placed in 1kg of water that was originally at 80dC. What is the nal temperature of the water after the ice has melted? Answer: 65.5dC


    2. Relevant equations
    Lf = Q/m
    Q=(DelT)mC


    3. The attempt at a solution
    At first I looked up the latent heat of fusion for ice which turned out to be 3.35*10^5J/kg. I just randomly decided to say Lf ice=3.35*10^5Jkg,
    Q/(m ice) = 3.35*10^5J/kg
    Q = 3.35*10^5Jkg * .1kg (ice) and so determined the amount of energy needed to phase change each kilogram of ice was 3.35*10^6J
    Then I just tried a bunch of random manipulations of other equations that didn't work and now my paper has so many eraser marks that it's hard to distinguish what I write from the background erase marks. I hypothesized that at the start of the scenario, the 1kg of water had a specific amount of internal thermal energy, but the only way I know to measure internal energy is using the second equation I listed, but I thought that if I could detirmine the amount of thermal energy the 80 degree water had to start with, I could say that the energy in ice was subtracted from that total amount of thermal energy and I could just simply solve or the change in temperature, but as I said, I don't know how to just automatically know at an instantnanous moment what the magnitude of the thermal energy of a substance is, just the amount of heat it gained from changing to a different temperature. Surely all 1kg 80 degee amounts of water have the same magnitude of kinetic energy, it doesn't matter what the starting temperature was.

    Right off the bat this problem seem contradictory. It makes physically sense that putting an ice cube in would lower the temperature of the water, but somehow I remember and also looked up to confirm that a phase change doesn't change the temperature of the system, just the internal energy. So to begin with I don't know how the temperature could be lowered if that statement is true, assuming the melting point of water is infinitesimally close to a number greater than 0 degrees Celcius, any added energy to the ice will cause it to melt which shouldn't change it's temperature, yet kinetic energy get's taken from the water in order t break apart the crystalline bonds of ice.
     
    Last edited: Dec 16, 2013
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  3. Dec 16, 2013 #2

    SteamKing

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    That's one of the problems keeping you from analyzing this problem correctly: you are just writing down random bits without understanding the physical processes at work.

    When you drop a bit o' ice into liquid water, two things can happen:

    1. the ice melts completely and turns into liquid water.
    2. the water originally in the container and the water formed from the melting ice come to thermal equilibrium.

    In both cases above, heat is transferred from the water in the container to the ice cube dropped into it. That is why people cool their drinks with ice cubes if a refrigerator isn't handy.

    Now, for you to analyze this problem, you should write some equations describing the transfer of heat from the water in the container to the ice cube dropped into it.

    First, melting must occur, so it is logical to start here. It looks like you have calculated the amount of heat it takes to melt the ice cube. You are correct that the melting occurs at constant temperature, so the melt water from the cube will initially be at 0 degrees C. How much does the temperature of the water originally in the container change? Can you write an equation describing this change?
     
  4. Dec 16, 2013 #3
    Well I can hypothesize that the new total mass of the water is now 1.1kg with the added mass of the melted ice, and I would set (1.1kg)(unknown change in temperature)(4.18kj/kgK, the specific heat of water) equal to the initial thermal energy of the water minus the amount of energy it took to melt the ice, but I have no equation to find the initial amount of total energy, just the change in energy.
     
    Last edited: Dec 16, 2013
  5. Dec 16, 2013 #4

    SteamKing

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    You are given the amount of water in the container and its initial temperature. That's a good start.
     
  6. Dec 16, 2013 #5
    But you are not addressing the fact that even if the equation I described exists, I was not taught it. I know the change in thermal energy, but I don't know what it's changing from
     
  7. Dec 16, 2013 #6

    SteamKing

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    Well, you've got a problem. Someone assigned you homework without teaching you the material. I can't help you with that. Perhaps a discussion with your instructor is in order.
     
  8. Dec 16, 2013 #7

    haruspex

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    For completeness:
    2. the water originally in the container and the water formed from the melting ice come to thermal equilibrium... at 0C.
    3. The water completely freezes.

    PhysicsStuff, In general, you don't know immediately which of the three cases above applies. Yo have to assume one and see if the answer is sensible. E.g. if you assume (1) then you can take it that the ice melts completely and the whole ends up at some temperature T. If you then find that T < 0C then you know you can rule out that scenario and try the next one.

    So let's try that here. The ice (again, in general) needs to go through the following stages:
    a) warm up to 0C (but in this case it's already there)
    b) melt at 0C (the phase change bit)
    c) warm up, as water now, from 0C to final temperature T.
    Meanwhile, the warm water merely has to cool to temperature T.
    Given the latent heat of fusion of ice and the specific heats of water and ice you can calculate, in terms of T, how much heat the ice will absorb and how much the water will release. Equate those two and deduce T.
     
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