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Time Needed for a Lake to Freeze Over

  1. Jul 21, 2003 #1
    I have been rather unfortunate to attend a school with a physics department that uses Sears and Zemansky's University Physics textbook. I have been working unsuccessfully on the following problem (Chapter 15, #103 for those who have the book) for 2 days...

    Show that the thickness of the ice sheet formed on the surface of a lake is proportional to the square root of the time if the heat of fusion of the water freezing on the underside of the ice sheet is conducted through the ice sheet

    Here is what I know... or think I know...

    heat current = dQ/dt = (kA(Th - Tc))/L
    That is the only equation I have that involves time; however I was always under the impression that I cannot seperate dQ/dt since dt is part of an expression representing the instantaneous heat flow, and not a single variable.
     
  2. jcsd
  3. Jul 22, 2003 #2

    Tom Mattson

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    Sorry, but I still don't understand why you think it can't be separated. In the equation dv/dt=-g, dt "is part of an expression representing the instantaneous" velocity, and it sure can be separated.
     
  4. Jul 22, 2003 #3
    That makes sense. I was wondering whether or not that was possible. I just remember my Calculus book saying to treat it as a single expression... but if it is bounded by a constant like -g, or in this case, heat flow (which should be constant for a given thickness), then I should be able to treat the dt as a single variable. Is it possible to do that in other situations? For instance, something without constant acceleration, heat flow, etc?
     
  5. Jul 22, 2003 #4

    Tom Mattson

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    Yes, it is possible. Start with constant acceleration:

    dv/dt=-g
    dv=-g dt
    ∫dv=-g∫dt
    v-v0=-gt
    v=v0-gt

    OK, now recall that v=dx/dt.

    dx/dt=v0-gt

    Note that on the right hand side we have not a constant, but a function of time. We can still separate the variables.

    dx=(v0-gt)dt

    and you continue with the integration.

    The same rule (separation of variables) is valid for your problem.
     
  6. Jul 22, 2003 #5
    I think I got it...

    dQ/dt = k((Th-Tc)/x) Q = Hf*m dQ/dm = dQ/dx dm = dx
    dQ = [del]Hf*dx dQ/dt = [del]Hf*(dx/dt)
    dQ/dt = k((Th-Tc)/x) = [del]Hf(dx/dt)
    xdx = k(Th-Tc)dt/[del]Hf
    [inte]xdx = [inte]k(Th-Tc)dt/[del]Hf
    (x^2)/2 = k(Th-Tc)t/[del]Hf + C
    x = [squ]2k(Th-Tc)t/[del]Hf
    x is proportional to [squ]t

    Does this look good?
     
  7. Jul 24, 2003 #6

    Tom Mattson

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    Well, you arrived at the correct answer but it is not clear to me that your solution is correct.

    How is it that dQ/dt=dx?
     
  8. Jul 27, 2003 #7
    dQ/dx = dQ/dm
    dQ/dt [x=] dx

    Is it ok to assume that dx = dm ?
     
  9. Jul 27, 2003 #8

    Tom Mattson

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    What is dm?
     
  10. Jul 27, 2003 #9
    dm is the change in mass, and dx is the change in length. Since a change in length results in a change in mass I set dQ/dm = dQ/dx
     
  11. Jul 28, 2003 #10

    Tom Mattson

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    In that case, you can't do it like that because dm and dx have different units. However, dm/dx is the linear density (call it λ), so you could write dm=λdx.
     
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