Linear Expansion Homework: Mercury Depth at 100°C

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Homework Help Overview

The problem involves determining the depth of a mercury column in a glass tube as the temperature changes from 12°C to 100°C. The coefficients of linear expansion for both mercury and glass are provided, and the relationship between temperature change and volume expansion is central to the discussion.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the need to consider both the volume expansion of mercury and the expansion of the glass tube. Questions arise about how to incorporate the linear expansion coefficients into the calculations, particularly regarding the lack of a volumetric expansion coefficient for mercury.

Discussion Status

Participants are exploring different aspects of the problem, including the relationship between linear and volumetric expansion. Some have suggested approximating volumetric expansion using linear coefficients, while others are clarifying the need to account for both the expansion of the mercury and the glass tube. There is an ongoing dialogue about the correct approach to find the final height of the mercury.

Contextual Notes

Participants note that the problem does not provide a coefficient of volume expansion for mercury, which leads to discussions about using linear coefficients for approximations. The initial conditions, including the radius of the tube and the depth of mercury, are acknowledged as critical for determining the final volume and height.

mizzy
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Homework Statement


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.


Homework Equations


L - Lo = alpha * Lo (T - To)


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 coefficient of the glass was also given. How do I put that into account??

Can someone guide me please??
 
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You're going to 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?
 
How can we look at the volume of mercury when we don't have the coefficient of volume expansion?
 
Yeah, that's an excellent point... they really should have given you that, but they might just want to use the linear one to approximate
<br /> 3 \alpha_L \approx \alpha_V<br />

... Either that or they gave you the coefficient for glass just to mess with you :P
 
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?
 
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.
 
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??
 
mizzy said:
but Vo and delta V are unknown right??

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.
 
ok. i think i got it now.

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

THANKS!
 

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