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Thermal Physics Expansion Coefficient Problem

  1. Jan 24, 2014 #1
    1. The problem statement, all variables and given/known data
    When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 550,000. The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, [itex]\beta[/itex]:

    [itex]\displaystyle \beta \equiv \frac{\Delta V/V}{\Delta T}[/itex]

    (where V is volume, T is temperature, and A signifies a change, which in this case should really be infinitesimal if 0 is to be well defined). So for mercury, (3 = 1/550,000 K"1 = 1.81 x 10~4 K_1. (The exact value varies with temperature, but between 0°C and 200° C the variation is less than 1%.)

    (a) 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
    the thermometer to work as required. Assume that the thermal expansion
    of the glass is negligible.


    2. Relevant equations



    3. The attempt at a solution

    Suppose at the temperature [itex]T_0[/itex] all of the mercury is in the bulb of the thermometer; and suppose the radius of the bulb is [itex]r_0[/itex], from which we can get the volume of the bulb, assuming it to be spherical, and consequently the volume of the mercury (the space it occupies). Suppose, by some external means, the temperature of the mercury increases from [itex]T_0[/itex] to [itex]T[/itex], [itex]\Delta T = T - T_0[/itex]. Due to the fact that things expand when their temperature changes, the mercury will begin to diffuse upwards into the cylinder portion of the thermometer, but some will remain in the bulb.

    Now, suppose that when the mercury reaches the temperature [itex]T[/itex], the mercury occupies all of the volume that the thermometer can afford. So, when [itex]T_0[/itex], [itex]V_0 = \frac{4}{3} \pi r^3_0[/itex]; and when the temperature changes to [itex]T[/itex], [itex]V = \underbrace{V_0}_{Initial~volume} + \underbrace{l_0 \pi r_1^2}_{Volume~ due~ to~ expansion}[/itex], where [itex]l_0[/itex] is the length of the thermometer, and [itex]r_1[/itex] is the radius of the cylinder, the variable of which we are solving for.

    The change in volume: [itex]\Delta V = (V_0 + l_0 \pi r^2_1) - V_0 = l_0 \pi r^2_1[/itex]

    Rearranging the thermal expansion coefficient, and substituting,

    [itex]\Delta V = \beta V_0 (T-T_0)[/itex]

    [itex]l_0 \pi r^2_1 = \beta V_0 (T-T_0) [/itex]

    [itex]\displaystyle r_1 = \sqrt{\frac{\beta V_0 (T-T_0)}{l_0 \pi}}[/itex]

    At this point I would make assumptions regarding the values of the radius of the bulb and the length of the thermometer, as per the directions of the problem statement

    If I understand this question properly, then if one was to build a thermometer, such that at temperature [itex]T_0[/itex], the mercury occupied the volume of the bulb. And because of these conditions, these would determine what radius of the of the cylinder portion of the thermometer has to be. These seems to make some sense, but I would appreciate if someone could logically explain these, to clarify this issue. Another thing I am dealing with is, notice the formula that I derived: the radius of the cylinder does not depend on the [itex]T[/itex] or [itex]T_0[/itex] individually, but on the change in temperature. Why is that so, why does it not depend on the beginning and ending temperature, aside from it being mathematically true? I am having a little difficulty grasping that.
     
    Last edited: Jan 24, 2014
  2. jcsd
  3. Jan 24, 2014 #2
    In reality, it does depend on the beginning and end temperatures, but, as an approximation, you are linearizing with respect to some average temperature, and even although the volume depends on the temperature, the change in volume is a linear function of the temperature change (to an excellent approximation) over the range of temperatures.

    I don't understand what else you are asking. Basically, you are trying to get an understanding of the steps involved in designing a thermometer. You start out by choosing a convenient length for the tube over the temperature range of interest. You don't want the thermometer to be 10 ft long, and you don't want it to be 1" long. So you choose something like a foot, maybe. You also choose a convenient size for the reservoir. You've seen thermometers, so you have some idea what this might be. You probably want to minimize the amount of mercury you use for environmental reasons. You then choose a radius so that the mercury in the tube goes from the initial temperature at the bottom to the final temperature at the top. After you're done with this, you then play around with the dimensions a little more, until they satisfy your comfort.

    Chet
     
  4. Jan 24, 2014 #3

    BvU

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    Your last result for r1 is excellent. Your intended next step too. As you see, for a given r1 and V0 the value (T-T0)/L0 is determined. So raise of mercury level per degree is fixed. Part of the exercise was to 'get a mercury thermometer'. The idea is to give you a feeling of how very small the inside diameter of the tube has to be. Especially for e.g. an old-fashioned fever thermometer (full scale around 0.1 m for 5 degrees C).

    Since the OP is in decent units, stick to meters would be my advice...
     
  5. Jan 29, 2014 #4
    I have another question. Suppose we were to implement the conditions I set-forth in my first post, and find the radius that corresponded to having the mercury only reside in the bulb at T = 0 deg C, and then occupying the entire volume of the thermometer at T = 100 deg C. Suppose that value were r0. Now, if I were to build a thermometer with the given specifications, except make the radius of the cylinder such at it's actually greater than r0, when the mercury reaches 100 deg C, it would not occupy the entire thermometer, right? The radius r0 is fine-tuned?
     
  6. Jan 29, 2014 #5

    BvU

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    This sounds like a practical question. Don't know much about making thermometers. I suppose tolerances in r0 are so tight that calibrating the scale only once is good enough for a batch of glass tubes. There will be no fine tuning: narrowing is difficult and widening even worse. The instrument would be too expensive.
    But you are right about not occupying the whole thing.
     
  7. Jan 29, 2014 #6
    Why don't you just try to do an actual design and see what you get? It's occupying more of your speculating about such questions than it would to actually try some design iterations and see what happens.
     
  8. Jan 29, 2014 #7
    And how would I design one, with a computer?
     
  9. Jan 29, 2014 #8

    BvU

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    No, by applying your last expression for r1. Design an old fashioned fever thermometer (10 cm delta height for a range of 36-41 degrees C) and then a school thermometer (30 cm scale for -30 to +120 C).
     
  10. Jan 29, 2014 #9
    I would use a slide rule, but you can use a calculator.
     
  11. Jan 29, 2014 #10
    I am not familiar with the slide rule. What is it?
     
  12. Jan 29, 2014 #11
    It's a wooden mechanical device that we used to use to do arithmetic before they had calculators.
     
  13. Jan 30, 2014 #12

    SteamKing

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    Behold! The slide rule (or slip stick as it is sometimes known):

    http://en.wikipedia.org/wiki/Slide_rule

    The circle of my life is now complete. I've gone from slide rules, desktop calculators, and mechanical adding machines to electronic calculators and microcomputers to someone asking, "What's a slide rule?"
     
  14. Jan 30, 2014 #13

    BvU

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    Come on folks, this is serious business. Bashy, get to work and show us how you calculate r1 for the two thermometers I proposed. I'll postpone my next post until you've posted that. ;-)
     
  15. May 1, 2016 #14
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