How Does Mercury Expansion in a Thermometer Relate to Temperature Changes?

[Je m'appelle]

Homework Statement

The length of a mercury (Hg) column in a glass thermometer is 15,00 cm when the thermometer is in contact with water at it's triple point (vapor-liquid-solid equilibrium) or 0,01 Celsius. Consider the length of the column as a thermometric property "X" and the empiric temperature measured by this thermometer as \theta.

(a) Find the empiric temperature when the length of the Hg column is 19,00 cm

(b) If this thermometer has an accuracy of 0,01 cm, can it distinguish the normal freezing temperature of the water and the triple point?

Homework Equations

None given by the problem.

The Attempt at a Solution

(a)I have absolutely no idea where to start, all I know is that we could think of this problem as a dilation problem, so that we could use

\alpha = \frac{1}{L}(\frac{\partial L}{\partial T})_F

Or,

L = L_0(1 + \alpha (\theta_f - \theta_i))

But I don't see how I could use them, so I suppose they are not needed? And in this case what should I do?(b) I suppose so, as the triple point is about 0,01 Celsius and the accuracy is up to 0,01 Celsius, it can distinguish, so yes. Correct?

 

[AEM]

Here's a couple of things you might want to consider. (1) What is the temperature of the mixture at the triple point? (2) What is the pressure at the triple point? (3) What is the pressure at the normal freezing point of water? What does that imply about the difference between the triple point temperature and the normal freezing point of water? Is it greater, or less than 0,01 Celsius?

Also, your equation above is the right way to approach the problem. You have the L's and the initial temperature. Just look up alpha.

 

[Je m'appelle]

Well, can anyone check this for me please?

I suppose that as we are talking about mercury dilation, we should use the thermal expansion coefficient \beta instead of the linear expansion alpha,

\beta = \frac{1}{V} (\frac{\partial V}{\partial T})_p

But as I wasn't given any function, I suppose I should just plug-in the \beta value for Mercury which is 181\times10^{-6}

So we can solve it this way,

L - L_0 = L_0 \beta (\theta_f - \theta_i)

\frac{(0,19 - 0,15)}{(0,15)(181\times10^{-6})} = \theta_f - 0,01

\theta_f = 220,99 + 0,01 = 221 C

Is this it? I find it very unusual that the value for \beta wasn't provided on the problem, I had to look out for it on google.

 

[AEM]

I would make the approximation that the expansion of the glass can be ignored. Then you don't have to worry about the expansion of the cross section and the linear formula will give you an acceptable answer.