Expansion of mercury in a thermometer

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Homework Help Overview

The problem involves the expansion of mercury in a thermometer, specifically calculating how far the mercury column rises as the temperature changes from 10°C to 90°C. The context includes the volume of mercury and the dimensions of the cylindrical bore in the thermometer.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the change in volume of mercury using the coefficient of volume expansion and then relates this to the length of the mercury column in the cylindrical bore. Some participants question the reasonableness of the resulting length of the mercury column.

Discussion Status

There is a recognition that the calculations provided by the original poster yield a lengthy mercury column, which raises questions about the assumptions made regarding the volume of mercury and the bore diameter. Some participants confirm the calculations but express concern about the values used.

Contextual Notes

Participants note the seemingly large volume of mercury and the small diameter of the bore, which may affect the interpretation of the results.

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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 0.400cm^{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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I thought you had made a mistake in your calculation but I have used your numbers and got the same as you
 
Last edited:
In addition!...0.4 seems like a large volume and 0.1 mm seems like a very fine bore
 
Thanks for your help.
 

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