# Linear expansion

1. Aug 8, 2010

### mizzy

1. The problem statement, all variables and given/known data
A glass tube of radius 0.80cm contains liquid mercury to a depth of 64.0cm at 12degrees. Find the depth of the mercury column at 100degrees. Assume that the linear expansion coefficient of the glass is 10x10^-6K-1 and the linear expansion coefficient of mercury is 0.61x10^-4K-1.

2. Relevant equations
L - Lo = alpha * Lo (T - To)

3. The attempt at a solution

With the knowns, I was able to figure out the depth of the mercury at 100degrees using the above equation, however, the radius and the linear expansion coefficent of the glass was also given. How do I put that into account??

2. Aug 8, 2010

### zhermes

You're gonna have to look at the volume of the mercury, and the volume of the glass.

What will the volume of mercury be at the new temperature?
What will be the new dimensions of the glass tube?
What height will the mercury reach?

3. Aug 8, 2010

### mizzy

How can we look at the volume of mercury when we don't have the coefficient of volume expansion?

4. Aug 8, 2010

### zhermes

Yeah, thats an excellent point... they really should have given you that, but they might just want to use the linear one to approximate
$$3 \alpha_L \approx \alpha_V$$

... Either that or they gave you the coefficient for glass just to mess with you :P

5. Aug 8, 2010

### mizzy

I've read over notes again. They gave us the linear expansion coefficient of mercury and glass, but we are to use beta which is equal to 3*alpha.

When i read over the question again, this is what i'm getting. Please someone tell me if i'm going in the right direction:

We have a tube with mercury at a certain depth with an initial temperature. As temperature increases, the volume of mercury will increase and therefore rise up. In addition, the glass tube will also change. What we are looking for is that increase in volume of mercury. Is that right?

6. Aug 8, 2010

### zhermes

Right, the coefficient of volumetric expansion is approximately 3 times that of linear.
And you've almost got all of it for what to do.

The mercury expands to some final volume, so its height would increase in the tube. But the tube also expands, allowing it to contain more volume for a given height--so you have to combine those effects to find the final height.

7. Aug 9, 2010

### mizzy

To find the final volume of mercury:
delta V = beta * V0 * delta T

but Vo and delta V are unknown right??

i think it's this part that i'm not clear. Ok, yes mercury expands and increases in height...yes, tube expands and increases in volume. Do I find the difference of the two??

8. Aug 9, 2010

### zhermes

Vo is given in the initial conditions (the initial radius of the tube, and the initial depth of mercury). So find the final volume.

Once you have the final volume, figure out how much the radius of the tube changes.

Once you know the new final volume of the mercury, and the new radius of the tube, you can figure out the final height of the mercury--which is what you're looking for.

9. Aug 9, 2010

### mizzy

ok. i think i got it now.

I have to use the coefficient of area expansion for the glass which is 2 * alpha.

THANKS!!