# Volume Thermal Expansion

1. Jan 15, 2010

### plfarrell

A 1.55 m long vertical glass tube is half-filled with a liquid at 23.0°C. How much will the height of the liquid column change when the tube is heated to 40.0°C? Take αglass = 1.0 × 10-5 °C-1 and βliquid = 4.0 × 10-5 °C-1.

First I found how much the height of the glass changed by using
$$\Delta$$V=$$\alpha$$glass x 1.55m x 17$$^{o}$$C.
I added that to the original height of the cylinder, and took the ratio of that value to the original, 1.5502635m/1.55m = 1.00017. (Sorry for not using sig. figs. I want to be as exact as possible.)
The textbook we have says that the V$$_{o}$$ = L$$^{3}_{o}$$. Using that relationship, I found what the change in volume was, given the coefficient of the volume expansion, then converted that back into height by taking the cube root. I used the ratio found earlier to find the actual height of the liquid, then took the difference between the two giving me .7753074067m - .775 m = 3.07E-4m.
This answer is apparently incorrect, so can anyone shed some light on this situation? Thanks!

2. Jan 15, 2010

### nasu

The change in the cross section of the tube has an influence on the final height. You cannot assume that the cross section stays the same and just subtract the lengths.

$$\Delta h =\Delta V_{liquid}/A_{tube}$$
where
$$A_{tube}=A_0(1+2\alpha \Delta T)$$
$$\Delta V_{liquid}=V_0 \beta \Delta T$$
and
$$V_0 =A_0 L_0, L_0=1.55m/2$$
A_0 will simplify in the end.

3. Jan 15, 2010

### plfarrell

How do i find the Surface Area of the tube without knowing its radius?

Last edited: Jan 15, 2010
4. Jan 15, 2010

### nasu

It's not the surface area but the cross section area.
And you don't need it. It simplifies in the end, as I already mentioned.