1
$\begingroup$

If a much hotter object is attached to the tube from the outside, and the heat in question is transferred to the part of the tube that is below the source of heat, so there is no convection. The conductivity of glass is a bit higher than that of water (0.8 to 0.6), but the area of water perpendicular to the flow of heat is way larger than that of the glass, so I think that almost all of the heat will be transferred through the water column. Is my reasoning correct? Is Fourier's law applicable in this case?

$\endgroup$
1
  • 1
    $\begingroup$ Remember an experiment as a kid at school, a test tube had water with an ice cube weighted with metal mesh so it was at the bottom of the test tube, then the top was heated in the bunsen - water boiled 4” from water frozen... $\endgroup$
    – Solar Mike
    Dec 31 '20 at 13:26
0
$\begingroup$

This is a comment. I do not yet have a "reputation of 50" and this is currently the only way I may leave a thought.

I would say that it depends on the thickness of the glass. A micro-thin tube would likely heat the water fast without much conduction to the lower regions, while I could imagine a very thick tube would allow heat to conduct down its exterior more rapidly, and include a gradiant loss to the water through the thickness.

$\endgroup$
0
$\begingroup$
  1. I think that the assumption that "there is no convection" is invalid. Since the heat is coming from the sides, which will very efficiently set up a nice donut vortex (using liberal terminology!) Therefore, Fourier cannot apply.

  2. With the symmetry around the Z axis, this is a 2-D problem. Applying the Fourier formula in 1-D would be invalid again.

Therefore, I would guess that Fourier's Law is not applicable.

That's only a hypothesis. But it's easily tested. Set up your test tube and leave it to settle. Then gently add a drop of colour. Again leave to settle but wait until the colour spreads to the level where the heat is coming in, so you'll be able to obseverve the effects. Heat it up and behold your swirling tube of chaos.

Of course I could be completely wrong! :-)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.