Strange behavior of thermal expansion and resistivity equations

  • Thread starter atat1tata
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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:
667
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The formula you are using are only approximations, roughly correct around room temperature. You can't expect precise results.
 
193
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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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