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One Dimensional Heat Equation

  1. Jun 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Show that the heat energy per unit mass necessary to raise the temperature of a thin slice of thickness [itex]\Delta[/itex]x from [itex]0^o{}[/itex] to [itex]u(x,t)[/itex] is not [itex]c(x)u(x,t)[/itex]. but instead [itex]\int_0^uc(x,\overline{u})d\overline{u}[/itex].

    2. Relevant equations

    According to the text, the relationship between thermal energy and temperature is given by

    [itex]e(x,t) = c(x)p(x)u(x,t)[/itex],

    which states that the thermal energy per unit volume equals the thermal energy per unit mass per unit degree times the temperature time the mass density.

    When the specific heat [itex]c(x)[/itex] is independent of temperature, the heat energy per unit mass is just [itex]c(x)u(x,t)[/itex].


    3. The attempt at a solution

    The only hint really is that this is related to the area, from the solution. How can I go about this geometrically and/or algebraically?

    Any help/pointers will be much appreciated. Thank you!
     
  2. jcsd
  3. Jun 15, 2011 #2

    lanedance

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    that integral doesn't make sense based on what you've posted, can you check it... c has changed to a function of x only or x and u (position and temperature)...?
     
    Last edited: Jun 15, 2011
  4. Jun 15, 2011 #3

    lanedance

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    so i think you do need to assume c = c(x,u), then try and find the change in energy de, for a small change in temp du and integrate.
     
  5. Jun 15, 2011 #4
    Argh of course!! Thank you very much! :approve:
     
  6. Jun 16, 2011 #5
    I have used your advice and went about it in the following way:

    For a small slice of thickness [itex]\Delta{x}[/itex] a small change in energy will be given by

    [itex]de = c(x,u)du[/itex]

    Dividing by [itex]du[/itex] I obtain [itex]e_{u} = c(x,u)[/itex].

    From the Fundamental Theorem of Calculus, this really says that

    [itex]e(x,t) = \int_0^uc(x,t)dt[/itex].

    Am I correct in my thinking? It feels a bit messy somehow...
     
  7. Jun 16, 2011 #6

    lanedance

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    you can just start from differentials
    [tex] de = c(x,u)du [/tex]
    [tex] \delta e = \int_{e_0}^{e_f}d\bar{e} = \int_{0}^{u} c(x,\bar{u})d\bar{u} [/tex]
     
  8. Jun 16, 2011 #7
    Your equation did not even occur to me but simplifies it a lot. Thank you very much for your help. I can now get some sleep again :-)
     
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