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])(adsbygoogle = window.adsbygoogle || []).push({});

[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?

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Strange behavior of thermal expansion and resistivity equations

**Physics Forums | Science Articles, Homework Help, Discussion**