Thermal expansion depth

[gomdul2]

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

 

[Redbelly98]

Welcome to Physics Forums. It looks like you are on the right track.

A couple of questions to help you think about what to do next:

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?

 

[gomdul2]

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]

The radius of the glass tube is not fixed. It changes due to thermal expansion.