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Explaining an experimental oddity

  1. Mar 22, 2012 #1
    I had to do an experiment to measure the latent heat of vaporization of liquid nitrogen. This was done by placing a canister of liquid nitrogen on a scale, and inserting a piece of aluminum of known mass into the canister. Knowing the initial temperature and heat capacity of aluminum, it's possible to figure out how much heat is added into the system. Then, using the recordings of the scale to figure out the rate of mass loss, it's not hard to calculate the latent heat of vaporization.

    Now, some of the mass loss will be due to heat entering the system from the surroundings. I tried to account for this by measuring the rate of mass loss before and after the aluminum was adding heat to the nitrogen. Here, there's an interesting effect: for about half a minute after the aluminum was at the same temperature as the liquid, the rate of mass loss was quite small but constant. Then, abruptly, the rate of mass loss increases to a value close to what it was before the experiment started. I suspect that's it's related to the fact that, at the end the period over which aluminum is dissipating heat, the rate of vaporization shoots up (because the layer of nitrogen created by the Leidenfrost effect disappears) and a noticeable decrease in mass occurs over a very short time interval.

    This is important because I need to estimate the rate of mass loss due to the surroundings while the piece of aluminum heating the nitrogen, and I'll get a different answer for the latent heat of vaporization if I say that the rate of mass loss is what is was right after the aluminum stops adding heat (when it was very small) or if I say that the rate of mass loss is what it was after it shoots up. So, which is it?
  2. jcsd
  3. Mar 22, 2012 #2
    You may describing what happens incorrectly as you refer to "when the aluminium was at the same temperature as the liquid" - How do you know this? Do you instead mean to say " when the aluminium was inserted into the liaquid"

    And, "at the end the period over which aluminum is dissipating heat, the rate of vaporization shoots up" - Again, how do you know the dissipation of heat has ended? I would think that if the rate of vaporization increases then the aluminium is still not at thermal equilibrium with the liquid, but has just fallen below the Leidenfrost temperature. ( if that is what is happenning with the liquid and aluminium which seems very plausible )
  4. Mar 22, 2012 #3
    I've included a sketch of a typical graph of the mass of nitrogen versus time. It's pretty clear when the aluminum stops dissipating heat: the rate of mass loss reduces substantially.

    Attached Files:

    Last edited: Mar 22, 2012
  5. Mar 22, 2012 #4
    There's no time scale so it's difficult to offer specific possibilities.

    My one thought is obviously the 'steady' heating process is probably some sort of convective state in the liquid. Putting a block in changes this - now the block is the main source of heat. Once it's cooled the convective heat transfer process will take time to restart and heat transfer is probably less efficient until it does. It's a guess though.
  6. Mar 22, 2012 #5


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    Staff: Mentor

    Extend the lower sloped line backwards and see where it hits the steeply sloped line...see if that tells you anything.
  7. Mar 22, 2012 #6
    OK. A picture is worth a thousand words.

    Is the piece of aluminium still completely covered by the liquid throughout the whole experiment. That might account for the lessor rate of mass loss if the vapour is recondensing onto any aluminium exposed above the surface.

    Or as Reasonableman states the the thermal currents set up from heat loss to the surroundings have to re-start after being disturbed by the aluminium.

    In any case the steeper part of the curve is where the aluminium was transferring heat to the liquid, and that mass loss should be be used to calculate your heat of vaporization. Why there is a type of hysterisis, if it can be called that, after the aluminium and liquid are in thermal equilibrium, could be a source of error in your experiment and could be further studied as to why. In this experiment one is assuming that the temperature of the liquid is the same before and after the experiment.

    Is this your expriment
    http://www.physicslabs.umb.edu/Physics/spring06/Exp3_182_Spr06.pdf [Broken]
    Last edited by a moderator: May 5, 2017
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