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A thermometer has a quartz body within which is sealed a total volume of 0.400[itex]cm^{3}[/itex] 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.

[itex]\Delta V=\beta V_{0}\Delta T[/itex]

[itex]\beta = 182x10^{-6}K^{-1}[/itex] from a table in the book.

[itex]V_{0}=.400cm^{3}[/itex]

[itex]\Delta T=90-10=80 deg[/itex]

[itex]\Delta V=.005824cm^{3}=5.824mm^{3}[/itex]

This change in volume will fill a portion of the slender cylindrical hole in the quartz.

The volume of a cylinder is [itex]V=L\pi \frac{D}{4}^{2}[/itex] Where L is the length of the cylinder and D is its diameter.

[itex]L =\frac{4V}{\pi D^{2}}[/itex]

V is the [itex]\Delta V[/itex] calculated above and D is given as 0.10mm.

L=741mm

This seems like a very long thermometer.

Thanks in advance for any responses.