# Why does a good thermos keep a cold drink to temp twice as long as a hot drink?

I've owned a number of insulating vessels. It seems the best they can do for keeping a hot drink hot is about 5 hours.

The same vessels will keep cold drinks with ice intact overnight, sometimes longer.

Why does it not work the same for either extreme?

As far as starting temperature, I am using fresh brewed coffee, about 200 F. For cold beverages, something along the lines of ice water, which is close to 32 F. I would never use too much ice, maybe 5 solid cubes, max.

Ambient temperature in either case is room temperature, about 68 F.

• It would be useful to know 1) how hot the hot drink is, 2) the ambient temperature, and 3) the amount of ice in the cold drinks.
– Bio
Commented Jan 6, 2022 at 9:17
• @Bio answered in question Commented Jan 6, 2022 at 9:31

Black body radiation formalizes this, the Stefan Boltzmann law:

For the same surrounding temperature there is a power to the fourth dependence on temperature.

A thermos has a bottle within a bottle with vacuum in between to stop conductivity loss of heat, but black body radiation happens anyway. As the formula above shows, the higher temperature of the object will radiate energy away through the vacuum between the bottles much faster .

• Anna, would not the main mechanism of heat transfer between the thermos contents and the thermos external surface be conduction, and then from the external surface to the air be convection? I thought that radiation from the external surface of the thermos would be less than convection since the external surface temperature of the thermos wouldn't be high enough for radiation to be the primary heat transfer mechanism. What's wrong with my thinking? Commented Jan 8, 2022 at 3:20
• @BobD the way I see it it is: take the inner bottle to outer space , its surface chosen so that it is a bad conductor of heat. It will certainly radiate a black body radiation and cool to the space temperature eventually. The hotter the content the faster the cooling according to the SB law. What happens when a second bottle is introduced keeping the vacuum will affect, because some of the inner radiation will be absorbed on the surface of the second bottle but the main difference in heat lost must be the SB law. Commented Jan 8, 2022 at 6:21
• I heard thermos are also mirror coated to minimize radiation exchange Commented Jan 11, 2022 at 21:20

Temperature difference
The cooling/heating of a thermos is mainly due to thermal conduction through the material that it is made of. The heat flow is thus dependent on the difference of temperatures, in the first approximation: $$\frac{dT_{internal}}{dt}=\alpha (T_{external} - T_{internal})\Rightarrow T_{internal}=T_{external} + (T_{internal}-T_{external})e^{-\alpha t}$$ The difference between a typical temperature that you consider cold and the ambient temperature is smaller than the difference between the hot temperature and the ambient temperature. Thes, over the same amount of time the temperature drop in a thermos xwith hot drink is greater (in degrees) than the temperature increase in the one containing a cold drink.

Thermal conductance
The proportionality coefficient $$\alpha$$ is also not the same, if we are talking about a hot drink and ice cubes. While the former is in direct contact with the container walls, the latter are largely surrounded by air, which is itself a bad heat conductor.

• How do we know that the cooling/heating of a thermos is mainly due to thermal conduction and not to radiation? Commented Jan 6, 2022 at 16:08
• @Usually an effort is made to reduce all the heat exchange: convection, radiation, and thermal conductance. The latter is most difficult to exclude, since the content is necessarily in contact with the container. Radiation, on the other hand, can be controlled by making the inside of the container reflecting - which is often the case. Commented Jan 6, 2022 at 16:16

Whether we think a thermos is better at keeping things hot or cold is probably to some degree influenced by our perception of temperature differences. In general we may be more aware of changes in the temperature of hot liquids than cold liquids.

That said, the rate of heat transfer by conduction and convection is proportional to temperature difference, all else being equal. Therefore the greater the temperature difference between the contents of the thermos and room ambient, and contact surfaces at room temperature, the greater the heat transfer rate (the quicker the contents will warm or cool), all else being equal.

So given a difference in temperature between hot coffee, at say 80 C. and room temperature, at say 20 C, of 60C, compared to the difference between water at 0 C and room temperature of 20 C, the initial rate of cooling of the coffee would be three times greater than the rate of warming of the cold water. If you had a mixture of ice and water it would warm even more slowly because of the latent heat of fusion of the ice.

However, as time goes on and the temperature difference between the contents and the room narrow, so will the differences in heat transfer rates narrow.

Hope this helps.

The rate at which your thermo bottle loses heat is described by $$dh/dt = R (T_o -T_i) \,.$$ Whether the temperature difference is positive or negative makes no difference in the linear approximation. However, with the cold drink you throw in a couple of ice cubes. That does the trick as to melt 1 gram of ice costs 70 calories.

I assumed that the heat exchange is not dominated by radiation.

You wouldn't expect significantly better insulation against heating than against cooling under normal circumstance. What you might be missing in your comparison is the room temperature that each scenario stabilises at.

Room temperature is $$\sim 68 \,\mathrm{^\circ F}$$ ($$\sim 20 \,\mathrm{^\circ C}$$).

• Starting from $$\sim 200 \,\mathrm{^\circ F}$$ gives a larger temperature difference, $$\Delta T=-132\,\mathrm{^\circ F}$$.
• Starting from $$\sim 32 \,\mathrm{^\circ F}$$ implies a smaller temperature difference, $$\Delta T=36\,\mathrm{^\circ F}$$.

The energy exchange from the thermo would presumably mainly be due to convection from the flask surface to the surrounding air or conduction to a table surface or similar. Both convection and conduction can be expressed as:

$$\dot Q=hA\Delta T,$$

where $$\dot Q$$ is the energy absorbed or lost per second (rate of energy transfer), $$h$$ is a heat coefficient matching the transfer type and the scenario (for conduction this mainly covers the material conductivity, whereas for convection it might vary a lot from scenario to scenario) and $$A$$ is exposed or contact area.

So here you can see that the larger the temperature difference, the larger the rate of heat absorbed or lost and thus the faster will the temperature change.