I'm a bit confused about thermal expansion in the case in which both a liquid and the container do expand. I will describe an example situation to expose the problem.
Consider a cylindrical glass tube (linear thermal expansion coefficient $\alpha$) that contains liquid (volume thermal expansion coefficient $\beta$). The height of the tube is $h_{t,0}$ and the height of the liquid inside of it is $h_{l,0}$. If the temperature changes of an amount $\Delta T$ what is the new height of the liquid? If the cylindrical tube is provided of a measuring scale, what is the new height of liquid measured from the scale?
The relation I would use is $$\frac{\Delta V}{V_0}\approx\frac{\Delta h}{h_0} +\frac{\Delta A}{A_0}$$
Which comes from $$(V_0+\Delta V)=(h_0+\Delta h) \cdot(A_0+\Delta A)$$
Neglecting higher order terms.
To find the new "absolute" height of the liquid I would simply consider the change in volume $\Delta V_{l}=V_{l,0} \beta \Delta T$, and then the change in the area of the cylinder $\Delta A_{t}=A_{t,0} 2 \alpha \Delta T$. Then I would write
$$\frac{\Delta h_{l}}{h_{l,0}} =\frac{\Delta V_{l}}{V_{l,0}}- \frac{\Delta A_{t}}{A_{t,0}}=(\beta-2\alpha) \Delta T$$
So actually in this case I would not consider the change in height of the tube, since I'm looking for the absolute change in height of the liquid.
To get the new height of liquid "relative to the tube" I would consider the "relative change in volume" $$\Delta V_{l,relative}=\Delta V_{l}-\Delta V_{t}=(V_{l,0} \beta- V_{t,0} 3\alpha)\Delta T$$
Here is my main doubt: does this "relative" change already takes into account the fact that both the area and the height of the tube change? If so, considering this "relative change" I can write
$$\frac{\Delta h_{l,relative}}{h_{l,0}}= \frac{\Delta V_{l,relative}}{V_{l,0}}$$
Because "relative to the tube" the only thing that can change is the height of the liquid and the base area is "constant" (infact the change in area of the liquid is the same of the one of the tube).
I'm not very convinced about this last consideration made.
Are these two processes correct or are there any mistakes (conceptual or of other kind) ? Any suggestion is highly appreciated
Best Answer
You already have the answer when you write $$\frac{\Delta h}{h} = (\beta -2\alpha)\Delta T$$ What you do after that is unnecessary and does not make sense. You have already said that the height of the tube is irrelevant, so the height of the liquid "relative to the tube" is meaningless.
If initially the liquid fills the tube completely and you want to know how much liquid spills out, use $$\frac{\Delta V}{V} = (\beta - 3\alpha)\Delta T$$
In response to your comment :
I think what you are trying to do is calculate the new volume reading of the liquid on the scale on the tube. For this you should use the same formula (for volume), which is marked in units of $cc$ or $cm^3$. So if the reading on the scale was initially $V_0$ cc then after expansion of the liquid and the glass tube the reading will be $V_1$ cc where $$V_1 - V_0 = V_0 (\beta - 3\alpha)\Delta T.$$