Let's assume I have a bottle of cold (> 0°C, but close) water and the environment has about 25°C.

I am eager to drink this water and I am moving the bottle to reach at least 15°C. I am thinking about the following related to temperature increase:

  • water inside gets some kinetic energy from movement and its slowing down should generate some heat (I cannot figure out what else happens to the energy lost due to slowing down)
  • water hits the somewhat warmer bottle walls and should get some heat

I am interested if this phenomena contribute to the heating of the water and if this can significantly help the water heat up (e.g. it may help reach the desired temperature two times faster than just let the water heat up due to environment).

Question: How to evaluate how fast water in a bottle is heating when moved rapidly?

Note: I am not interested in exact equations, but rather in physical phenomenon explanations and some order of magnitude.

  • $\begingroup$ You need to be more specific on move rapidly. Temperature is a measure of random kinetic energy. $\endgroup$ – paparazzo Jan 25 '18 at 18:34
  • $\begingroup$ @Paparazzi - by "rapidly" I mean as fast as possible a human being can perform without using a device. $\endgroup$ – Alexei Jan 25 '18 at 18:46
  • $\begingroup$ Apparently you did not get the random point. Can't help you. $\endgroup$ – paparazzo Jan 25 '18 at 18:48

The first mechanism you described is called irreversible conversion of mechanical energy to internal energy, and is the result of the water being viscous. Viscous flow is what slows down the velocity of the deforming water, and mechanistically is the result of the water velocity being zero at the wall of the bottle. This mechanism does not contribute much to the heating of the water in your case because the viscosity of water is pretty low.

The second mechanism you described is called convective heat transfer. Ordinarily, if you did not stir the water, the heat transfer would occur near the wall of the bottle by conduction. But this process is slowed down by the increasing thickness of the region near the wall where the temperature has changed. The latter causes the temperature gradient near the wall to decrease, which, in turn, lowers the rate of heat conduction into the water.

Agitating the bottle causes water in the interior of the bottle to exchange places with the warmer water near that wall, and thus bring more colder water into contact with the wall to allow it to heat up. The thickness of the region near the wall where the temperature has changed is thereby reduced, and this causes the temperature gradient near the wall to remain high. This then maintains a higher rate of heat conduction into the water. The net result is that the water in the bottle heats up faster.

  • $\begingroup$ What about when water (because of being moved) hits the wall? Doesn't that induce heat in it? To what degree? $\endgroup$ – MaDrung Jan 25 '18 at 13:42
  • 1
    $\begingroup$ The velocity of the water at the wall is equal to the wall velocity. This is called the "no-slip boundary condition." So it is not really bouncing off the wall. It is the deformation of the water that generates the viscous heat. And, as I said, in practice, this is not nearly as important mechanism as convective heat transfer. $\endgroup$ – Chet Miller Jan 25 '18 at 13:50

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