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I have several thermal mugs, two of them by the same brand, have the same look, shape and size except that one is in steel (inside and outside) and the other one, plastic. Both have an insulating layer of air or vacuum.

I do not need a thermometer to tell me that the steel one is much better at keeping liquids cool/hot than the plastic one. With water near 90°C inside the mugs, I cannot feel any heat by placing my hand on the steel one (same as a thermos), while the plastic one feels almost like burning.

However steels conducts heat much better than plastic (about one order of magnitude greater thermal conductivity). Therefore intuition tells me that a metallic mug should be less efficient than a plastic one to keep things cool/hot, but the reverse seems true!

What am I missing?

I am adding pictures of the two mugs. On the transparent one, we can see the insulating gap. I think it is an enclosed volume so there shouldn't be any air transfer between it and the surroundings. Transparent mug close up.

Here's the other picture:

The two mugs next to each other

OK, so far the two answers given mention two different reasons. One is that there might be air between the 2 plastic surfaces in the transparent mug, and vacuum in the steel one. Air has a thermal conductivity of about $3\times 10^{-2}\,\frac{\mathrm{W}}{\mathrm{Km}}$, so the thermal conductance should be about $3\times 10^{-2}\,\frac{\mathrm{W}}{\mathrm{Km}} \times \text{surface area} / (5\times10^{-3} \mathrm{m})$ for an air gap of $5\,\mathrm{mm}$. So $6A\,\frac{\mathrm{W}}{\mathrm{Km^2}}$.

For radiation, $P=A\sigma\varepsilon(T^4-T_\text{room}^4) \approx A\varepsilon\, 462\,\mathrm{W}/\mathrm{m}^2$. That steel is not polished so its emissivity is probably not as low as 0.1, but let's take it as zero for simplicity. And 1 for plastic, to get orders of magnitude. It looks like radiation might play the bigger role (the answer should run the numbers!) at high temperatures. But at lower temperature differences (when the liquid has partly cooled), conduction may play the bigger role.

Thus, plugging numbers for a temperature difference of 80C, radiation losses are about 460 W/m^2 while conduction through the air about 480 W/m^2. In the comments, I am told that air convection should also transfer more heat, hence I conclude that even at "high" temperatures, the conduction through air might play the bigger role. At temperature differences below 60C, the role of radiation should lower faster than the one of conduction, because of the 4th power dependence on temperature.

I note that I took the air's thermal conductivity at 1 atm. If the air has a lower pressure I guess the thermal conductivity would be lower and in that case radiation may play the bigger role at high temperatures.

4 Answers4

46

Note that double-wall plastic cups aren't strong enough to withstand a vacuum and so they usually just contain a thin layer of air in which convection currents (absent in the case of the vacuum) can easily get started. This will bleed the heat out of the contents of the cup much faster than in the vacuum case.

niels nielsen
  • 99,024
34

There is a reason steel is better for thermos flasks than plastic: lower emissivity. The way a thermos flask works is that it has two walls ideally separated by vacuum. Heat passes between them as blackbody radiation, with each unit of area transmitting $\sigma \epsilon T^4$ watt per square meter. Here $\epsilon$ is the emissivity, which for polished steel can be around 0.07. Plastics have $\epsilon\approx 0.90 - 0.97$. Hence a plastic surface will radiate more heat than a metal one, and more heat can leak between the inner and outer surface.

A small calculation: the heat flow per unit area will be $\sigma(\epsilon_i T_i^4 - \epsilon_o T_o^4)$ where $i$ and $o$ denote the inner and outer surfaces. For 80C coffee and 20C room temperature I get 46.3193 W/m$^2$ for $\epsilon=0.1$ and 416.8736 W/m$^2$ for $\epsilon=0.9$, a factor 10 difference.

2

IR wavelengths range from 700 nanometers to 1 millimeter, which is narrower than the insulative region of the cup.

If the hidden zone of steel is shiny and reflective to the thermal radiation which is emitted by the beverage, then it can bounce a lot of the radiation back towards the inside of the cup.

In industry, they use aluminium foil and bubble wrap fairly similar to the design of the mug, and it's designed to reflect IR radiation.

Steel varies for insulative properties, for example automotive brake rotors have grains of carbon in the structure which are excellent for thermal conductivity, whereas some alloys are less conductive.

The plastic is 20 times less conductive, although it requires a far bigger diameter to achieve the same strength.

enter image description here

bandybabboon
  • 1,479
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You say that steel conducts heat about 10 times better than plastic, but that number applies to a cross section. If you take a look at the connection between "inner" and "outer" cup, the connection of the steel mug relevant for direct conduction is a very thin rim. Touching your way around it will make you realise that there is a considerable gradient near that rim. So the rim is rather relevant for the separation of heat. Outside of the rim, transfer involves two steel-vacuum barriers, and steel is shiny, so it neither emits nor absorbs much heat radiation.

The plastic does not reflect significantly. While its visible-light behavior is not necessarily the same as its infrared behavior, it will be translucent and/or dark to infrared, either of which will result in significant heat transfer even when discounting that you can't evacuate the space between cups.