# Linear expansion over temperature change

From the equation $$\Delta L=aL\Delta T$$, $$\frac{\Delta L}{\Delta T} = aL$$

For small change $$\frac{dL}{dT}=aL$$

Solving we get $$L(T)=e^{aT}$$ (as differentiating it with respect to temperature gives $$a \times e^{aT}=aL$$)

Putting $$L(t)= l$$, $$l=e^{aT}$$ Giving $$\frac{\ln(l)}{a} = T$$

But as $$a$$ is different for different temperatures. Different materials must have different lengths at same temperature.

I'm just a highschooler, but where did it go all wrong?

• I forgot that differentiating two different functions can yield same results. Thank you very much for the answer:-) – gaurang agarwal May 18 at 2:23

$$L=L_{0}e^{a\left(T-T_{0}\right)}$$
So you need to measure what is the length of your material at temperature $$T_{0}$$, and you can use the solution to predict the length at all other temperatures. Even if you use the same material, you can still cut it and have different length at the same temperature.
To actually quantify differences between materials, you would want to look at $$L/L_{0}=e^{a\left(T-T_{0}\right)}$$ instead. Such quantities that do not depend on the size of your system are known as intensive quantities in thermodynamics. In this case the relative expansion of different materials would be different, independent of their length at temperature $$T_{0}$$.