0
$\begingroup$

Imagine a vacuum chamber with:

  • a glass container inside filled with water;
  • a copper tube that penetrates a little bit in and a little bit out of the chamber and enters the glass container;

After all the air has been removed, and the water boiled, the water is now cold.

Now I heat the copper tube outside of the chamber for 1 second and with a very hot flame.

What happens next? Does the water become cold fast again? Will it boil again and become very cold fast again?

Basically, what I wanted is a fast way to transfer heat and make the copper tube cold again. I think that this method might work because of water convection.

$\endgroup$
2
  • $\begingroup$ Yes. Indeed it looses some steam but that too condenses and become part of ice. And the copper also cools. $\endgroup$ Commented Oct 2, 2021 at 13:48
  • $\begingroup$ Do a google search for "heat pipe". $\endgroup$ Commented Oct 2, 2021 at 19:28

1 Answer 1

0
$\begingroup$

If you pump the air out of the chamber until the water boils vigorously and then seal the chamber shut, you will end up with an atmosphere mostly comprised of water vapor at a pressure somewhere in the neighborhood of 0.01 atmospheres.

If you heat your copper tube, then that will cause the temperature of the water (and therefore its vapor pressure) to rise by a small amount. A bit more water will move into the vapor phase, which will raise the pressure in the chamber a tiny bit; as the water and vapor cool back down, the process will be reversed and the extra water will return to the liquid phase.

Altogether, nothing dramatic or particularly interesting would happen. In particular, the partial vacuum would not play a huge role. Repeating the experiment with the chamber filled with air wouldn't change much except the timing and magnitude of the temperature/pressure swing, because the extra gas would transfer heat to and from the environment more efficiently than the tenuous water vapor atmosphere would.

$\endgroup$
5
  • $\begingroup$ In your second paragraph... Will that be a fast process, or a slow one? $\endgroup$ Commented Oct 2, 2021 at 14:05
  • $\begingroup$ @fabiolinhares It would depend a lot on the details, e.g. on how much water there was in the glass, how hot the copper rod got, etc. I would expect it to take somewhere on the order of a minute (possibly several) to return to normal, but you'd need to actually do the experiment to be sure. $\endgroup$
    – J. Murray
    Commented Oct 2, 2021 at 14:13
  • $\begingroup$ Ok. I'm intending to do the experiment, but only if it is worth it... In your 3rd paragraph... Repeating the experiment like this wouldn't provide me freezing water... Because there is no vacuum to cool the water down... Do you think that If I do cycles of heating and freezing by means of the vacuum itself that the chamber will remain the water cold, or will the chamber star to approach ambient temperature? This is my last question, I promise! $\endgroup$ Commented Oct 2, 2021 at 14:29
  • $\begingroup$ I forgot to put your name in the last comment $\endgroup$ Commented Oct 2, 2021 at 15:03
  • $\begingroup$ @fabiolinhares You don't need to - if you comment on my answer, I'll automatically be notified. And it's important to remember that it's not the vacuum which cools the water down; it's that the highest energy water molecules escape into the vapor phase and are subsequently removed by the vacuum pump, which reduces the average energy of the molecules which remain in the liquid. I'm entirely sure what your goal is - if it's refrigeration, there are far more efficient methods than putting your glass of water in a vacuum chamber. $\endgroup$
    – J. Murray
    Commented Oct 2, 2021 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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