How Does Temperature Affect the Depth of Mercury in a Glass Tube?

[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.

 

[rwisz]

for any help in advance.I would like to clarify a few things before providing a response to this problem.

Firstly, "thermal expansion depth" is not a commonly used term in the scientific community. It is important to use correct terminology, such as "thermal expansion coefficient" and "depth of the mercury column."

Secondly, it is important to mention the units for all the given values. In this case, the radius of the glass tube is given in centimeters, while the linear expansion coefficients are given in inverse kelvin (K^-1). It would be more accurate to convert the radius to meters (0.008 m) to maintain consistency in units.

Now, moving on to the solution, you are correct in calculating the volume of the mercury column using the given radius and depth. However, you also need to consider the volume of the glass tube. The volume of the glass tube can be calculated using the same formula, but with the radius of the tube and the total height of the tube (64.8 cm = 0.648 m).

Once you have calculated the volumes of both the mercury and the glass tube, you can use the formula delta V = V - Vo = (volume expansion coefficient)(Vo)(delta T) to calculate the change in volume of the mercury and the glass tube separately.

Finally, you can use the relationship between volume and depth (V = (pi)(r^2)(h)) to calculate the change in depth of the mercury column at 100°C.

I hope this helps. As a scientist, it is important to pay attention to detail and use accurate terminology and units in problem-solving.