# Thermal expansion depth

## Homework Statement

A glass tube of radius 0.80cm contains liquid mercury to a depth of 64.0cm at 12deg. Find the depth of the mercury column at 100 deg.
Assume that the linear expansion coefficient of the glass is 10 X 10^-6 K-1 and the linear expansion coefficient of mercury is 0.61 X 10^-4 K-1.

## Homework Equations

volume expansion coefficient = 3 X linear expansion coefficient.

delta V = V- Vo = (vol.exp.coefficient)(Vo)(delta T)

## The Attempt at a Solution

Ok, So I multiplied the linear expansion coefficient of mercury and glass by three to calculate volume expansion coefficient.
Since the glass tube has radius of 0.80cm = 0.008m
Therefore, volume of mercury = (pi)(r^2)(h)
= (pi)(0.008^2)(0.64) = 0.000128679 = 1.287X10^-4m^3.

I don't know if I am on the right track,
what about the volume of the glass tube?
Thanks

## The Attempt at a Solution

Redbelly98
Staff Emeritus
Homework Helper
Welcome to Physics Forums. It looks like you are on the right track.

1. What is the change in the volume of the mercury, after it is heated to 100 deg?

2. What happens to the cross-sectional area?

Thanks Redbelly 98 :)

change in the volume of mercury =
delta V = V-Vo = (vol.coeficient)(Vo)(delta T)
= V - (1.287 X 10^-4) = (1.83 X 10^-4)(1.287X10^-4)(100-12deg)
2.072 X 10^-6 = V - (1.287 X 10^-4)
Therefore, V = 1.30772 X 10^-4 m^3.
Since V = (pi)(r^2)(h)
and r is fixed (r is still going to be 0.008m.)
But what about the change in volume of the glass tube?

I am confused @_@

Redbelly98
Staff Emeritus