# Expansion of mercury in a thermometer

I think that I’m doing this problem correctly, but the answer seems a bit unreasonable. Can someone else check my work?

A thermometer has a quartz body within which is sealed a total volume of 0.400$cm^{3}$ of mercury. The stem contains a cylindrical hole with a bore diameter of 0.10mm. How far does the mercury column extend in the process of rising from 10deg C to 90deg C? Neglect any change in the volume of the quartz.

I first found the change in volume of the mercury.
$\Delta V=\beta V_{0}\Delta T$
$\beta = 182x10^{-6}K^{-1}$ from a table in the book.
$V_{0}=.400cm^{3}$
$\Delta T=90-10=80 deg$
$\Delta V=.005824cm^{3}=5.824mm^{3}$

This change in volume will fill a portion of the slender cylindrical hole in the quartz.
The volume of a cylinder is $V=L\pi \frac{D}{4}^{2}$ Where L is the length of the cylinder and D is its diameter.
$L =\frac{4V}{\pi D^{2}}$
V is the $\Delta V$ calculated above and D is given as 0.10mm.
L=741mm

This seems like a very long thermometer.

Thanks in advance for any responses.

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