0
$\begingroup$

There are various questions around how cups of tea cool down, or why splitting a cup of tea helps it cool, and things like that.

Thinking about the various mechanisms - heat rising within or escaping from the liquid via convection, cooler liquid dropping, conduction with the base of the cup and underlying medium, etc., it should be reasonable to expect some temperature differential between the bottom and top of a cup of tea.

Assume:

  • The initial liquid would be uniformly $\approx 95 \, ^\circ \text{C}$ once poured into the cup, and, eventually, it would all reach room temperature $\approx 20 \, ^\circ \text{C}$
  • The liquid follows the same properties as pure water
  • The cup is fairly tall ($\geq 10 \, \text{cm}$) and narrow ($\leq 5 \, \text{cm}$) without a lid or significant insulation, resting on a typical non-metallic surface that conducts heat moderately.

During the middle of the cooling process ($\approx 60 \, ^\circ \text{C}$ or wherever is convenient to consider it), how much of a delta should we expect in the temperature difference between the top and bottom of the cup? What's the order of that delta - is it closer to $0.1$ of a degree, $1$ degree or $5$ degrees?

Improve this question
$\endgroup$
1
  • $\begingroup$ The condition at the surface is very important with respect to evaporation rate, heat, and mass transfer. Is the air well-mixed, or is it stagnant. What temperature do you think exists at the interface between the air and the tea (there is no discontinuity of temperature at the interface)? $\endgroup$
    – Chet Miller
    Commented Oct 21 at 14:39

1 Answer 1

Reset to default
1
$\begingroup$

I would expect a temperature difference of around 1 °C or more (between the center and the wall of the cup) based on an experiment I did for my labs at home during the pandemic. I measured a cooling curve using a mercury thermometer, which isn't ideal for this purpose due to its size and thermal capacity. This was the gradient I was able to measure early in the cooling process, when the water was still around 80–90 °C.

The actual difference will depend significantly on the parameters. For example, a lower thermal conductivity of the liquid or cup material could result in a larger gradient, but convection will probably take over at some point, mixing the liquid and lowering the gradient (this means you would also have to account for fluid dynamics, viscosity, etc.).

There are some videos with infrared cameras (e.g., https://www.youtube.com/watch?v=74b5nrekTCw), though IR is not a great tool for seeing what's happening inside. I also found a video of someone making a simulation (https://www.youtube.com/watch?v=74b5nrekTCw) which looks similar to what you're describing.

There's probably no better way to find out the actual value than to make an experiment under your specific conditions or to create a simulation using software like the one in the video.

Improve this answer
$\endgroup$

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.