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Bizaare question

  1. Oct 28, 2003 #1
    The length of the mercury coloumn in a mercury thermometer is 5cm when the bulb is immersed in water at the triple point (273.16k). What is the temperature if it reads 6.0cm? What will the length of coloum be if immersed in boiling water (at steam point)?

    I'm new to this forum, but would really appreciate any help you guys can give (need help asap!).

    I've tried using pv = nKT etc, but can't apply it. All your hints would be welcome!
  2. jcsd
  3. Oct 28, 2003 #2

    Chi Meson

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    You want to examine the "linear expansion" formula. (since no dimension was given for "the bulb" I'm assuming that it's just one end of the colum and volumetric expansion is not necessary) You'll have to look up the coefficient of linear expansion for mercury.

    edit: did the problem really say "triple point"? Nevermind, only the temperatures are important.
  4. Oct 28, 2003 #3

    Yes, in the question it didn't actually mention the temperature, just that it was the triple point.
    Its supposed to be a pretty basic thermal physics question, any ideas what formula I would need to use?

    Thanks, rob.
  5. Oct 28, 2003 #4

    Chi Meson

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    OK, the triple point is at a certain temperature, but it's not 273 K (that's the melting/freezing point). FInd that temperature and determine the "delta" T between each situation. Have you found the linear expansion formula yet?
  6. Oct 28, 2003 #5
    The triple point of water has to be very near freezing. To get ice, liquid water and steam to exist in equilibrium (constant amounts of each) you must have ice (so very near freezing) and you must have low pressure (to get steam at cold temperatures). Daniel Schroeder's "Thermal Physics" gives the triple point of water as 273.16 K and 0.006 bar pressure ( 1 bar = 1/ 1.013 atmosphere).

    Even my thermal physics professor forgot to mention that the triple point requires low pressure.
  7. Oct 29, 2003 #6
    Ok, thanks guys.
    I'm still unsure as to where to go from here.

    Any chance of a worked example anyone?


    Thanks, rob.
  8. Oct 29, 2003 #7


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    Wouldn't it be more accurate to speak of water vapour rather then steam?
  9. Oct 29, 2003 #8
    I see why you say that but Schroeder's text calls it steam. Our concept of steam is at 1 atmosphere and the triple point is at very low pressure so it would be cold steam in any case. Water vapor exists at most temperatures and pressures but the phase diagram for H2O has a definite region labeled steam.
  10. Oct 29, 2003 #9


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    Water vapour is liquid water in small droplets. Steam is the correct term for gaseous H2O.

    By the way, I still don't see how you can calculate the change in height of the column if you don't now the radius of the cylinder.
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