# Strange behavior of thermal expansion and resistivity equations

1. Aug 15, 2010

### 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?

Last edited: Aug 15, 2010
2. Aug 16, 2010

### AJ Bentley

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

3. Aug 16, 2010

### 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.