Suppose that the specific heat of a material is a function of position and temperature, $ c(x,u) $. Show that the heat energy per unit mass necessary to raise the temperature of a thin slice of thickness $ \Delta x $ from $ 0 $ to $ u(x,t) $ is not $ c(x) u(x,t) $ but $ \displaystyle \int_{0}^{u(x,t)} c(x,\overline{u}) ~ \mathrm{d}{\overline{u}} $ instead.
Can somebody help me out with this question? What is $ c(x,\overline{u}) $ for example?
You can think about the heat energy per unit mass necessary to raise the temperature as sum of all heat energies per unit mass for temperatures from 0 to $u$ as temperatures rise in small steps. This is
$$\sum_{i=0}^u c(x,i) \Delta i$$
where $\Delta i = u/n$ is the change in temperature. I've given the summation variable an arbitrary name $i$. But we could think of $i$ as the mean temperature in the small step and write $i=\bar{u}$ instead. So the following is equivalent
$$\sum_{\bar{u}=0}^u c(x,i) \Delta \bar{u}$$
You can see that $\bar{u}$ is just a summation variables. Hopefully you can see how this might turn into an integral? But can you see why an integral is necessary?
The specific heat is a function of position $x$ and temperature $u$. But why do we have to do a summation over the temperature but not over the position? The clue is that we're told that we have a thin slice, but nothing is said about the temperature range.