How to calculate overflow of liquid being expanded thermally?

Question

There is a question on OpenStax college physics that is phrased:

If a 500-mL glass beaker is filled to the brim with ethyl alcohol at a temperature of 5.00ºC, how much will overflow when its temperature reaches 22.0ºC?

The answer of 9.35 mL makes sense if the container was closed and all the ethyl alcohol in it increased temperature. Using $dV = \beta VdT$ where for ethyl alcohol $\beta= 1.100 \times 10^{-3}$ 1/°C yields that solution.

However, assuming that as the liquid was expanding, whatever overflowed from the top of the container fell into another container where it stopped heating up and was gradually returned to 5.00ºC? How would I approach computing this or similar problems?

Answer 1

Say we did have a container that could contain the expansion from 5.00 °C to 22 °C without overflow, but we did the temperature raising in four steps.

Step 1: 5 °C -> 10 °C which has $dT\ =\ 5\ °C$
Step 2: 10 °C -> 15 °C which has $dT\ =\ 5\ °C$
Step 3: 15 °C -> 20 °C which has $dT\ =\ 5\ °C$
Step 4: 20 °C -> 22 °C which has $dT\ =\ 2\ °C$

The initial volume $V$ at the beginning of Step 2 would be more than 500 mL because of the increase of volume from the temperature increase of Step 1. The same applies for subsequent steps:

$dV_1 = (1.1 \times 10^{-3})(500)(5) = 2.75\ mL$
$dV_2 = (1.1 \times 10^{-3})(500 + 2.75)(5) = 2.77\ mL$
$dV_3 = (1.1 \times 10^{-3})(500 + 2.75 + 2.77)(5) = 2.78\ mL$
$dV_4 = (1.1 \times 10^{-3})(500 + 2.75 + 2.77 + 2.78)(2) = 1.12\ mL$

$dV_{total} = dV_1 + dV_2 + dV_3 + dV_4= 9.42\ mL$

Which is slightly higher than the 9.35 mL you get if you do the calculation as "one step." What does this actually mean? Let me quote Wikipedia's section on Volume Expansion:

[Using this equation] assumes that the expansion coefficient did not change as the temperature changed and the increase in volume is small compared to the original volume. This is not always true, but for small changes in temperature, it is a good approximation. If the volumetric expansion coefficient does change appreciably with temperature, or the increase in volume is significant, then the above equation will have to be integrated.

So the surprising answer to your question is that the equation is an approximation that works best when the liquid does overflow, maintaining the volume at 500 mL throughout the entire expansion. I would guess the very reason the question is phrased like that is to allow the student to avoid getting involved with integrals.

Answer 2

This sort of problem is rather nasty in my opinion. It is hard to know where to cutoff the analysis. For instance the ethanol might well rise above the lip of the container. It wasn't specified if the contact angle were positive or negative.

There was a previous question How far can water rise above the edge of a glass? which explored this notion. You can also look at this youtube video at 40 seconds in.

So the gist is that you have to be careful and not overthink such problems. It is also a very good idea to explicitly state any assumptions which you made to solve the problem. So for this problem:

(1) The liquid and the container are both in complete equilibrium with the temperature as the temperature changes.

(2) The container is originally filled to the geometric plane of the lip.

(3) Any liquid which goes above the geometric plane of the lip overflows.

(4) The volume of the beaker stays constant with the temperature change (from @DJohnM's comment).

(5) The overflow is at the same temperature as the beaker and ethanol.

Answer 3

There is no difference in the volume leaving the beaker in the two scenarios you describe. Once the overflowing liquid leaves the beaker, it has no effect on the liquid in the beaker.

Consider the second, step-by-step expansion.

Assume the liquid warms up from $T_0$ to $T_1$, where the change in temperature can be as small as you like.

Then the overflow for this step, $\Delta V_1$, is given by:$$\Delta V_1=\beta V(T_1-T_0)$$

This overflow, well, overflows, leaving the exact same original volume, $V$, in the beaker.

So, when the second small temperature change, from $T_1$ to $T_2$, takes place, the next overflow is:$$\Delta V_2=\beta V(T_2-T_1)$$We can continue this process as long as we want, until the small temperature changes bring us to the final temperature. Add up all the small volume overflows, noting that in each expression, the second, negative term in each expression cancels the first, positive expression in the preceding expression. The result of all this cancelling, $$\Delta V_{Total}= \beta(T_{final}-T_{initial)}$$ is exactly the same as the first scenario...