I noticed a colleague had a tall narrow cup for his coffee, and it got me thinking about whether it would retain heat for longer or not.

Assume two cups, both are cylindrical, and both hold the same volume. One is taller and more narrow, and the other has even width and height dimensions.

My first thought was that assuming the cup casing encompassed the liquid entirely, a perfectly spherical cup would retain heat best, as it has the smallest surface area.

However, this doesn't account for heat rising to the top of the cup, does that make a difference?

Also, the top surface and the sides and bottoms are different. The top of a cup has no casing, while the sides and the bottom are ceramic or porcelain.

How would we determine which cup keeps the coffee warm longest?


In general, yes, for a given volume a sphere has the least surface area and therefore the least heat loss to the environment.

But we have the complication that the top of the vessel must be open and uninsulated.

For a fixed temperature of the liquid and the ambient air, a simple model would say the three types of surfaces on a cylinder each have a fixed heat flux (power per unit area). Say these are $q_\mathrm{top}$, $q_\mathrm{bottom}$, and $q_\mathrm{side}$. The total energy per unit time lost to the environment is $$ Q = \pi r^2 (q_\mathrm{top} + q_\mathrm{bottom}) + 2\pi rh q_\mathrm{side} $$ for a cylinder of radius $r$ and height $h$. If we fix the volume $V$ as a constraint, then $\pi r^2h = V$, so we have $$ Q = \pi r^2 (q_\mathrm{top} + q_\mathrm{bottom}) + \frac{2V}{r} q_\mathrm{side}. $$

The optimal container has as its radius the $r$ that minimizes $Q$. This occurs for $$ 0 = \frac{\mathrm{d}Q}{\mathrm{d}r} = 2\pi r (q_\mathrm{top} + q_\mathrm{bottom}) - \frac{2V}{r^2} q_\mathrm{side}, $$ or $$ r = \left(\frac{Vq_\mathrm{side}}{\pi(q_\mathrm{top}+q_\mathrm{bottom})}\right)^{1/3}. $$

Qualitatively, we can note a few things. For instance, the ideal radius grows proportional to the cube root of the desired volume, which is not unexpected. In fact, for fixed $q$'s, the ideal container simply scales with $V$ - it's shape doesn't change.

Moreover the ideal container in this model is neither too wide nor too narrow. Too wide, and the $\pi r^2 (q_\mathrm{top} + q_\mathrm{bottom})$ term means all the heat is lost through the top (and bottom), while too narrow and the $(2V/r) q_\mathrm{side}$ term becomes larger than it needs to be.

I don't have an actual number, though, because I don't know the values for the $q$'s. One could presumably look up the properties of the ceramic involved for $q_\mathrm{bottom}$ and $q_\mathrm{side}$. Even then, though, I treat these terms separately because they could depend on the shape of the vessel: the sides will be more efficiently cooled by convection than the bottom, unless the cylinder is very narrow. On top of this, the heat transfer through the top needs to be modeled, and that itself is a difficult fluid dynamics problem.

  • $\begingroup$ Bravo! Upvote from me. (Should I mention radiative cooling?) $\endgroup$ – C. Towne Springer Jul 8 '14 at 19:16
  • $\begingroup$ I suspect there are far more egregious approximations I've made (like doing everything in my power to avoid convection) than ignoring radiative cooling. $\endgroup$ – user10851 Jul 8 '14 at 19:20

Coffee would cool via heat being lost to the surroundings. This heat loss would be larger when the exposed surface area is larger. Now, whatever material makes up the cup, it would insulate heat somewhat, as against being held completely open, like the top surface. (That's one reason why ceramic cups are so common - if the cup doesn't insulate, its like holding the hot coffee right in your hand!)

So, the taller and narrower cup (an example here) exposes lesser area to air and heat loss through a larger volume of coffee, in the interior, would be reduced.

In comparison, an ''even dimensioned'' one (for example, this) has a relatively larger surface area exposed to air, which means more cooling.

  • $\begingroup$ The taller one will have stronger convection currents which will stir the coffee and more vigorously expose the hottest liquid to air where it cool faster by evaporation. But the surface area is less. The air on the outside will also convect and at a higher velocity for the tall cup. I suspect there is an optimal size for a cylinder and fixed volume and the calculation is far from simple. $\endgroup$ – C. Towne Springer Jul 8 '14 at 6:37
  • $\begingroup$ I don't disagree with the ''optimal'' issue, but the question is how big a difference would these convection effects make to this problem. I would be very surprised if inclusion of convection effects reverses this conclusion, since we know about this experimentally - by drinking coffee everyday! Anyways, please go ahead with the full calculation :) $\endgroup$ – 299792458 Jul 8 '14 at 6:41
  • $\begingroup$ If I recall right, convection has a term proportional to the two thirds power of g so the answer can change from planet to planet. Anyway, take an extreme - a tall ceramic straw. Certainly something shorter will cool slower and something taller than a saucer will cool slower. There is a sweet spot in between. No? $\endgroup$ – C. Towne Springer Jul 8 '14 at 6:53
  • $\begingroup$ Right, my answer applies to the coffee I'm sipping here on Mars! I never disagreed about the issue of optimality and completely buy the intermediate sweet spot. But the point I'm trying to make is that convection effects won't make a huge difference on this scale (size of a typical cup, that we have here on Mars :P), that's why we can neglect convection. They do in the case of water boiling in a pan, because there is significantly higher temperature difference involved. In a nutshell, all I'm advocating is - let's not use General Relativity to describe the motion of a car moving on a road. :) $\endgroup$ – 299792458 Jul 8 '14 at 7:02
  • $\begingroup$ Except for the fact that we use saucers instead of cars, here on Mars :P $\endgroup$ – 299792458 Jul 8 '14 at 7:03

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