Strange behavior of thermal expansion and resistivity equations

[atat1tata]

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

\alpha 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 l_0 = 100 m, t_0 = 0°C, \alpha = 0.01 °C^{-1}, t = 10 °C

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

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

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

With resistance there are almost the same problems. Where am I wrong?

 

[AJ Bentley]

The formula you are using are only approximations, roughly correct around room temperature. You can't expect precise results.

 

[netheril96]

Rewrite it in canonical linear function
l = \left( {{l_0} - \alpha {l_0}{t_0}} \right) + \alpha {l_0}t
This should hold for all {l_0} in an ideal situation,so this suggest that \alpha is dependent on {l_0},or more specifically,inverse proportional.