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I have a selection of mugs that are different shapes. Some are taller elongated cylinders, others are similar in height to their diameter, and some are flatter, wider vessels.

I was wondering, which shape of mug should I use if I want my tea or coffee to stay hottest for the longest time?

All are ceramic and have no lid (obviously a travel-style mug would beat other styles, but let's not worry about that; I'm just interested in "standard" mugs).

Do I want something that is closest in shape to a sphere to minimise the overall surface area of the drink? Or do I want something with the smallest area exposed to the air to minimise convection? Or is there some other consideration that wins out?

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    $\begingroup$ Hint: en.wikipedia.org/wiki/Nusselt_number $\endgroup$ – John Alexiou Oct 23 '18 at 20:22
  • $\begingroup$ Quick guess (assuming that all the mugs can hold the same volume of liquid): The mug with the thickest walls and with the smallest area of liquid-air interface. $\endgroup$ – user190081 Oct 23 '18 at 20:29
  • $\begingroup$ All other things being equal, I would think you would want a shape that minimizes the ratio of exposed surface area to volume. $\endgroup$ – Bob D Oct 23 '18 at 20:35
  • $\begingroup$ @BobD I think that that would be the case if the heat flux was the same between the liquid-air and liquid-ceramic interfaces, which I dont think is the case. Anyway, it would be interesting to build optimal-space-partition mugs (iopscience.iop.org/article/10.1088/1367-2630/18/10/103008/pdf). $\endgroup$ – user190081 Oct 23 '18 at 20:44
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    $\begingroup$ A spherical one with no opening, but it has practical issues. $\endgroup$ – my2cts Oct 23 '18 at 22:04
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Consider a cylindrical mug with radius $r$ and height $h$. It will hold volume $V=\pi r^2h$ and have area $A=2\pi (r h + r^2)$. If we assume it loses heat energy proportional to temperature and area we get the equation $$CV T' = -kAT$$ where $C$ is the heat capacity, $k$ is a factor containing heat conductivity plus convection and we assume $T$ is the temperature above the ambient environment. The solution is $$T(t)=T_0 e^{-(k/C)(A/V)t}.$$ So to make this as slow as possible we should let $A/V \rightarrow 0$, or simply make the volume go to infinity.

That is a bit boring. For a given volume (one cannot drink an arbitrarily large amount of drink), what minimizes $A/V$? If we set $V=1$ (since we only care about shape) $h=1/\pi r^2$ and $A/V=2 (1/ r + \pi r^2)$. Taking the $r$ derivative and setting it to zero we get $-1/r^2 + 2\pi r=0$, which is solved by $r_*=(1/2\pi)^{1/3}\approx 0.5419$. The corresponding height will be $h_*=1/\pi r_*^2=2^{2/3}\pi^{-1/3}\approx 1.0839 = 2r_*$. That is unsurprising, since we should expect the perfect cup to be as close to spherical as possible, and here the height equals the diameter. However, the optimum is pretty shallow: most mugs work fairly well - volume and material will matter more than shape.

Plot of area to volume ratio for different cup radii

What about a top surface conducting or convecting heat at a different rate from the sides? If the heat loss is $T' = -(1/CV)[k_1(2\pi r h + \pi r^2)+k_2(\pi r^2)]T$ we want to minimize $[k_1(2\pi r h + \pi r^2)+k_2(\pi r^2)]$ subject to $V=1$. Let's set $k_1=1$ too, to reduce the number of variables. $f(r)=2/r + \pi (1 + k_2) r^2$, $f'(r)=-2/r^2+2\pi(1+k_2)r$ and we get $r_*=(1/\pi(1+k_2))^{1/3}$. So if $k_2$ is getting larger than 1 we want a tall, narrow cup. But even for $k_2=10$ this just means $r_*\approx 0.3070$ - the change is pretty modest.

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  • $\begingroup$ BTW Anders has just derived the optimum height-to-diameter ratio for minimizing the cost of making steel soup cans. that aspect ratio minimizes the surface area (i.e., the amount of steel needed) for any given volume displacement of cylindrical can. $\endgroup$ – niels nielsen Oct 23 '18 at 22:20
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To me, the answer is obvious by observation. I can't do the math, but if someone would care to dope it out, here is my theory.

If we eliminate designs that are specific to heat conservation, such as double walls and lids, and make it strictly a single-walled material with an open top: a mug which angles inward to a narrow neck and top from a wider bases keep coffee warmer longer--a mug with a conical base incorporated into it.

This shape takes advantage of the common property of heat and steam-- they RISE. In a standard cylinder, most of the steam and heat escape upwards into the air. But the conical body allows for more steam and heat to make contact with the upper body of the mug and heat the material, which in turn warms the coffee.

As a standard cylinder mug's coffee is consumed, the same upward trajectory for heat remains constant because of the uniform circumference. More inner wall is exposed, but the wall is also making more contact with cool air and is ineffective at retaining and transferring heat back to the coffee, either thru transference to the deeper liquid layer or the surface layer.

But in a conical mug made of the same material, the steam/heat from the outer edge of the coffee is channeled along the acute angle of the inner wall. More heat and steam is retained longer, utilized within a more contained chamber before eventually reaching the narrow neck and opening. Some steam might also begin to condense into drops which return to the coffee.

Other advantages: Less likely to tip because whether full or empty, it is always bottom heavy. Disadvantages: Harder to get clean, and most likely won't be cleaned at all in a dishwasher.

Conical Based Mug

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