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Strange behavior of thermal expansion and resistivity equations

  1. Aug 15, 2010 #1
    The equation of linear thermal expansion is [tex]l=l_0\left[1 + \alpha(t - t_0)\right][/tex] (it is similar also to the equation of resistivity, [tex]\rho=\rho_0\left[1 + \alpha(t - t_0)\right][/tex])

    [tex]\alpha[/tex] is a constant dependent only on the material we are speaking about.

    Now, let's say I have a bar 100 m long at 0°C. The material I'm using has a coefficient of thermal expansion of 0.01 1/°C. I'm heating it to 10°C
    So [tex]l_0 = 100 m[/tex], [tex]t_0 = 0°C[/tex], [tex]\alpha = 0.01 °C^{-1} [/tex], t = 10 °C

    [tex]l=(100 m)\left[1 + 0.01 °C^{-1} (10 °C)\right] = 1.1(100 m) = 110 m[/tex]

    Now, I have a bar 110 m long at 10°C and I want to return it to 0°C.
    So [tex]l_0 = 110 m[/tex], [tex]t_0 = 10°C[/tex], [tex]\alpha = 0.01 °C^{-1} [/tex], [tex]t = 0 °C [/tex]

    [tex]l=(110 m)\left[1 + 0.01 °C^{-1} (-10 °C)\right] = 0.9(110 m) = 99 m[/tex]

    With resistance there are almost the same problems. Where am I wrong?
     
    Last edited: Aug 15, 2010
  2. jcsd
  3. Aug 16, 2010 #2
    The formula you are using are only approximations, roughly correct around room temperature. You can't expect precise results.
     
  4. Aug 16, 2010 #3
    Rewrite it in canonical linear function
    [tex]l = \left( {{l_0} - \alpha {l_0}{t_0}} \right) + \alpha {l_0}t[/tex]
    This should hold for all [tex]{l_0}[/tex] in an ideal situation,so this suggest that [tex]\alpha [/tex] is dependent on [tex]{l_0}[/tex],or more specifically,inverse proportional.
     
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