I have one problem at hand which asks the following:

Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for thermometer to work as required. Assume that the thermal expansion of the glass is negligible.

It is also given that the thermal expansion coefficient for mercury is $\beta = 1/5500 K^{-1}$, but between $0^\circ C$ and $200^\circ C$ the variation is less than $1$%.

So, if I assume that the radius of the bulb is, say, 0.5 cm, and that the distance between the degrees on the thermometer scale is 0.2 cm then is it correct that to estimate the diameter of the tube, I need to setup the following equation:

$$\frac{4}{3} \pi (0.5)^3 \cdot0.01 = 0.2 \cdot \pi (\frac{x}{2})^2$$

where $x$ is the diameter of the tube?

I'm just trying to see if my understanding of the problem is correct. Thank you for your input.

  • 1
    $\begingroup$ It looks close and I think that your understanding is correct. You're taking the volume of the bulb and calculating how much the volume of the mercury inside it will expand in going from 0 to 200C. That gives you the left side of the equation. Then you want to say that that volume should be equal to the volume of the thin, cylindrical cavity of the thermometer tube. But shouldn't the height of the cylindrical cavity be 0.2 cm times the number of degrees measured (which is 200?) and not just 0.2 cm? $\endgroup$
    – user93237
    May 5, 2017 at 3:18
  • $\begingroup$ @SamuelWeir: Are we given that the expansion is less than 1% per one degree Celsius, or per 200C? I interpreted the statement that the expansion is less than 1% per degree Celsius for the range from 0 to 200C. $\endgroup$
    – sequence
    May 5, 2017 at 3:47
  • $\begingroup$ @sequence, the 1% is the variation in the expansion, not the expansion itself $\endgroup$ May 5, 2017 at 3:53
  • 3
    $\begingroup$ @sequence. The volume expansion coefficent of Mercury changes with temperature, the question is saying that the value of 1/5500 only varies by 1% over the range 0-200C so you can use 1/5500 as a constant approximation $\endgroup$ May 5, 2017 at 3:57
  • 1
    $\begingroup$ @sequence:" I interpreted the statement that the expansion is less than 1% per degree Celsius for the range from 0 to 200C." - A rate of volume expansion of 1% per degree would be enormous and unrealistic. Thermal expansion coefficients for most materials are in the parts per million per degree range. That's why I interpreted your statements as meaning 1% change over the entire 200 ˚C range. That still corresponds to a very high thermal expansion coefficient, but at least it's a bit more believable. $\endgroup$
    – user93237
    May 5, 2017 at 16:31

1 Answer 1


If you are only considering the mercury advancing 0.2cm = 1 deg C, wouldn't it be:

$$\frac{4}{3} \pi (0.5)^3 / 5500 = 0.2 \cdot \pi (\frac{x}{2})^2$$

Since the volume in the bulb expands by 1 part in 5500 for each degC.

edit: To clear up the confusion. Mercury expands by about $1/5500 k^{-1}$ but this factor varies with temperature, making the calculation for a real thermometer complicated. But you are told that the change in this value over the range you are interested in is small. It's a confusing and unnecessary bit of information, a sign of a badly written question.

  • $\begingroup$ Could be this as well, but it is said in the problem statement that there are different expansion coefficients for different ranges, so I just used the "less than 1%" for a thermometer range. Because the thing is that mercury expands by less than 1% in either case, but in the case of other ranges (for the general $\beta$) the expansion appears to be way less than 1%. $\endgroup$
    – sequence
    May 5, 2017 at 3:52

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