# Thermal expansion of both liquid and glass tube

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).

Are these two processes correct or are there any mistakes (conceptual or of other kind) ? Any suggestion is highly appreciated

• I believe you wrote some additions instead of multiplications, for instance, the second equation is inconsistent in the units – Wolphram jonny Jul 9 '16 at 0:16

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 $$\frac{\Delta V}{V} = (\beta - 3\alpha)\Delta T$$
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 $$V_1 - V_0 = V_0 (\beta - 3\alpha)\Delta T.$$
• Thanks for the reply! Actually I wanted to neglet the change in height of the tube just in the first point. In the second point I do not neglet the expansion of the tube (in particular is change in height). The formula for second point in my question does not differ a lot from your second formula $$\Delta h_{l,rel}=( \beta \cdot h_{l,0}- 3 \alpha \cdot h_{t,0}) \Delta T$$ The difference is that there are the two (possibly different) heights of tube and liquid. Can this formula be correct if I do not neglet the change in height of the tube? – Sørën Jul 11 '16 at 7:50
• Thanks for the add to your answer, thats what I'm trying to do! I understood the formula but my doubt was in considering the case where $V_{0,l}$ (initial volume of liquid) is not the same as $V_{0,t}$ (initial volume of tube) (which also means that the initial heights are different, or, in other words, the liquid does not fill the tube completely initially). In this case I don't think that is correct to evaluate the (relative) change in volume of the tube as $V_{0,l} 3\alpha \Delta T$, since $V_{0,l}\neq V_{0,t}$ but I would say $V_{0,t} 3 \alpha \Delta T$. Would that make sense? – Sørën Jul 11 '16 at 10:50
• In my answer $V_0$ is the volume of the liquid as indicated by the scale on the tube, so by definition $V_{0,l}=V_{0,t}$. $V_1$ is the new volume reading opposite the liquid level after expansion of the liquid and the tube. This assumes (of course) that the liquid does not spill out of the tube. $V_0$ and $V_1$ are not the volumes of the tube, which could be much bigger - how much bigger is irrelevant. – sammy gerbil Jul 11 '16 at 11:14