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Incremental energy problem

  1. Feb 23, 2007 #1
    1. The problem statement, all variables and given/known data

    A pond of water at 0°C is covered with a layer of ice 4.50 cm thick. If the air temperature stays constant at -11.0°C, how much time does it take for the thickness of the ice to increase to 9.00 cm?


    Hint: To solve this problem, use the heat conduction equation,

    dQ/dt = kA (change in)T/x

    and note that the incremental energy dQ extracted from the water through the thickness x is the amount required to freeze a thickness dx of ice. That is, dQ = LpA dx, where p is the density of the ice, A is the area, and L is the latent heat of fusion. (The specific gravity and thermal conductivity for ice are, respectively, 0.917 is 2.0 W/m/°C.)

    2. Relevant equations

    Heat conduction equation

    3. The attempt at a solution

    I am not sure how to do this, I have done extensive research but am still confused... please help. Also I don't study maths, so please help me with a word equation if possible.

    Many Thanks,
    Last edited: Feb 24, 2007
  2. jcsd
  3. Feb 23, 2007 #2
    This is a fun problem.

    Put your expressions together (I'm not doing anything but math), and you get the following.
    LA \,dx = \frac{kA \Delta T}{x} dt

    Separate variables, and integrate. You've already done all the physics.
    Last edited: Feb 23, 2007
  4. Feb 24, 2007 #3
    I am not sure what this all means- I am doing AS level physics and I don't do maths, would it be possible to give me a word equation?
  5. Feb 24, 2007 #4
    Can someone please help, or is this too challenging?
  6. Feb 24, 2007 #5


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    Is A assumed to be constant? If so, you should be able to plug dQ=LA dx into dQ/dt = kA T/x and integrate.
  7. Feb 24, 2007 #6
    I don't do maths... i.e. Integration means nothing to me... sorry
  8. Feb 24, 2007 #7


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    What do the "d"s mean in your equation then, if you are not using calculus? Are they meant to be [itex]\Delta[/itex]; denoting "change in"?

    edit: you didn't answer my question re A being constant.
    Last edited: Feb 24, 2007
  9. Feb 24, 2007 #8
    Yes, I think so, Webassign just gave me the question to do, and the formula to use.
  10. Feb 24, 2007 #9
    [tex]LA \,dx = \frac{kA \Delta T}{x} dt[/tex]
    Someone gave me this^^, but still confused
  11. Feb 24, 2007 #10


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    Ok, I've seen your other thread-- you're studying AS Physics, so I presume you did moderately well in GCSE maths. Can you see how the equation you give above comes from your two equations in your original post? (simply substitute dQ from the second equation into the first).

    Now, dx means "change in x", so rewrite it as dx=x-x0 where x0 is the value of x when t=t0=0 (since we will take the original time as t=0). Thus we write dt=t (since t0 is zero).

    This should make your equation easier to handle. Try to rearrange it to make t the subject of the equation.
  12. Feb 24, 2007 #11
    okay... I have got this far...


    Is this correct?
  13. Feb 24, 2007 #12


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    No, dx is not the same as x. Let's replace the d's in the equations in post #6, as I suggested in #7. This will give [tex]LA(x-x_0) = \frac{kA T}{x} t[/tex]. Can you rearrange this?
  14. Feb 24, 2007 #13
    I have confused myself with density and difference in etc...

    Is it this? Where is density?

    Change in time = Latent heat of fusion*change in x^2 / thermal conductivity of ice * 11
  15. Feb 24, 2007 #14
    Wait hang on, just recieved your msg.. let me do it
  16. Feb 24, 2007 #15


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    There should be a term for density in your equation for dQ = LAdx in your original post. You sort of mentioned it in the sentence, but left it out of your equation.
  17. Feb 24, 2007 #16


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    You said this:
    Now, have you missed out the density term from the equation? My guess is that you copied the question incorrectly, and there should be a ρ somewhere.
  18. Feb 24, 2007 #17
    Of course! I have seen it now! Sorry, when I pressed copy on the pc it did not copy:

    dQ= LpA , where p is density of ice!
  19. Feb 24, 2007 #18
    Also the delta sign did not copy for the heat conduction equation

    dQ/dt = kA (Change in) T/x
  20. Feb 24, 2007 #19
    So does this mean I am lost?

  21. Feb 24, 2007 #20
    I am lost, please help Cristo!
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