# Thermal expansion of liquid in a tube

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1. Jul 7, 2016

### Soren4

1. The problem statement, all variables and given/known data
A cylindrical glass tube (linear thermal expansion coefficient $\alpha$) 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? Do not neglet any thermal expansion.

2. Relevant equations
Thermal expansion coefficients

3. The attempt at a solution
I get a bit confuse in considering all the expansions. The relation I would use is $$\frac{\Delta V}{V}\approx\frac{\Delta h}{h} +\frac{\Delta A}{A}$$
• 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 did not consider the change in height of the tube, since, what is asked is simply the (absolute) change in height of the liquid.​
• While, 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 the area of the tube changes and that the height of the tube changes? If so, considering this "relative change" I can write​
$$\frac{\Delta h_{l,relative}}{h_{l,0}}= \frac{\Delta V_{l,relative}}{V_{l,0}}$$
And get the new height relative to the tube, but I'm not convinced about this last equation I wrote.

Are these two processes correct or are there any mistakes (conceptual or of other kind) ?

2. Jul 8, 2016

### 256bits

Hey Soren4,
The absolute looks OK.

Still thinking about the relative equation.
But as a hint, you have the absolute height of the liquid.
You can determine the absolute height of the marking on the tube.
Does your equation confirm the relative change in height tube/liquid?

By the way, do you have an understanding about how
say, the equation of ΔV/V that you have listed comes about

ie from assuming small changes in temperature
linear ( 1 + α )
planar ( 1 + α ) ( 1 + α )
volumetric ( 1 + α )( 1 + α )( 1 + α )

3. Jul 8, 2016

### Irene Kaminkowa

Soren4,
For the volumes of liquid before and after changing T we have $$V_0 = h_0A_0$$ and $$V_1=h_1A_1$$ The second equation can be rewritten as
$$(V_0+\Delta V)=(h_0+\Delta h)(A_0+\Delta A)$$
$$V_0\left ( 1+\frac{\Delta V}{V_0} \right )=h_0A_0\left ( 1+\frac{\Delta h}{h_0} \right )\left ( 1+\frac{\Delta A}{A_0} \right )$$
$$1+\frac{\Delta h}{h_0}=\frac{1+\frac{\Delta V}{V_0}}{1+\frac{\Delta A}{A_0}}=\frac{1+\beta \Delta T}{1+2 \alpha \Delta T}$$
The new height
$$h_1=\frac{1+\beta \Delta T}{1+2 \alpha \Delta T}h_0$$

You can use the same approach in the second part of your problem.