After reading this question asking why kettles aren't insulated, (a question I had myself) I questioned, "What about people who boil the kettle 2~5 times a day?".

So, my question is, if we were to create a thermally insulated Kettle, and it's qualities in thermal retention were similar to that of a flask, how long could we expect the water to stay hot in the kettle?

Assumedly, any heat conserved by the time we boil again is energy saved. That said lets define hot to be more or less ~40% of boiling temperature, which is therefore around 40 degrees C, right?

So, the question is then, how long does it take a thermal flask of 100 degree C water to drop below 40 degrees C? I know that's not a strait answer, with many variables involved, but there should be a ball park figure right?

I am going to do my own experiment this afternoon too.

  • $\begingroup$ How much do you think an item like that would cost compared to a plain old kettle, and do you think that most people would be willing to pay the difference in price? Also, the kettle could not be insulated on the bottom where heat is added, and heat would be lost out the bottom, which comprises a significant fraction of the kettle surface area. $\endgroup$ Sep 24 '16 at 12:13
  • $\begingroup$ @ChesterMiller, I think there are fairly simple ways of getting around most of the issues. Both kettles and insulated flasks are fairly cheap. The biggest issue I forsee is that of a decent lid, since sealing a kettle closed is a bad idea, and not having a sealed flask is a waste of heat... the element could be embedded inside the flask, or you could go an induction route, ah the options. At the end of the day, bulk production is the only thing that will create an affordable product, in this case. But yes, I would pay a fair amount more for a kettle that saved on electricity. $\endgroup$ Sep 24 '16 at 20:00
  • $\begingroup$ It's difficult to transfer heat into a thermally insulated kettle, which should be designed to boil water. In other words, there are mutually exclusive constraints involved in the design of a well insulated kettle. $\endgroup$ Jul 20 '20 at 3:33

After a simple experiment, here are my results.

Note: Tap water temperature: ~ 16°C.

  • 12:02 Midday, 95.6°C right after boiling, and pouring water into flask. The flask was nearly full.
  • 3:07 pm, 85.8°C
  • 5:32 pm, 75.6°C, before this measurement I forgot the top open for a while, maybe 20 min?
  • 7:41 pm, 71.7°C

That's a good 7~8 hours and the water is still pretty hot.

So, I think I can conclude that a simple store bought vacuum flash should be fairly good at keeping in the heat when compared to a kettle, which cools in a matter of minutes. The kettle is one of those glass see-through ones though, so it's not winning any prizes for thermal insulation.

For myself, and anyone interested, I have resolved to try do the following:

  1. Boil only as much as I need (and convince others in the family to do the same), and
  2. Keep a flask next to the kettle, for those days when I plan on being in the house for the whole day. This is to put the left over water in when I forget to measure the right amount, then boil that the next time round. :-)

As you say, there are many variables involved. You have not specified the amount of water in the flask, which is probably the most important variable. Rate of heat loss depends on surface area, whereas heat content depends on volume, so having a large volume will maximise the time for the temperature to drop a certain amount.

The Ultimate Thermos Flask can retain heat for around 24 hours. 'Everyday' flasks retain heat for around 12 hours. For cheaper models it is probably around 6-8 hours. (There is no indication on the manufacturer's website what the parameters for these tests were.)

From my own experience, using whatever is handy at home (cotton towels, woollen clothing) to insulate your kettle, you might achieve 1-2 hours. Using polystyrene - eg from the original packaging of the kettle - you might achieve 4-8 hours.

The usual advice is to boil only as much water as you need, making sure the element is covered (in an electric kettle). In this case there is little or no water stored between use, and little need for insulation. (As the linked answer to Why aren't domestic kettles insulated? points out, the boiling time is quite short - around 15 to 60s - so there is little time in which to lose much heat during use.)

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    $\begingroup$ I think this answer is beautifully concise and points out an issue in my thinking, the simplest way to preserve energy is to not use it to begin with. For these reasons, I marked this answer as correct. I will still post my simple experiment, and suggestion for anyone else who may want to preserve what energy they can rather than measure out the water needed. $\endgroup$ Sep 24 '16 at 18:00

The full details based on Wikipedia Heat Loss From Vacuum Flask

This is a pretty comprehensive math description, but from other sources, and when you do your experiment, you can check, it should take about 4/5 hours to drop in temperature.

The rate of heat (energy) loss through a vacuum flask can be analyzed thermodynamically, starting from the second $TdS$ relation

$${\displaystyle TdS=dH-VdP}$$

$${\displaystyle T_{surr}\Delta S=mc_{p}\delta T-Vdp}$$

Assuming constant pressure throughout the process,

$${\displaystyle T_{surr}\Delta S=c_{p}\left(T_{b'}-T_{c}\right)}$$

Rearranging the equation in terms of the temperature of the outside surface of the vacuum flask's inner wall,

$${\displaystyle T_{b'}=T_{c}+{\frac {T_{surr}\Delta S}{c_{p}}}}$$


$T_{surr}$ is the temperature of the surrounding air

$∆S$ is the change in specific entropy of stainless steel

$c_p$ is the specific heat capacity of stainless steel

$T_c$ is the temperature of the liquid contained within the flask

$T_b'$ is the temperature of the outside surface of the vacuum flask's inner wall

Now consider the general expression for heat loss due to radiation:

$${\displaystyle Q'_{0}=A\epsilon \sigma \left(T^{4}-T_{0}^{4}\right)}$$

In the case of the vacuum flask,

$${\displaystyle Q'_{0}=A_{in}\epsilon _{s.s.}\sigma \left(T_{b}'^{4}-T_{surr}^{4}\right)}$$

Substituting our earlier expression for $T_b'$,

$${\displaystyle Q'_{0}=A_{in}\epsilon _{s.s.}\sigma \left[\left(T_{c}+{\frac {T_{surr}\Delta S}{c_{p}}}\right)^{4}-T_{surr}^{4}\right]}$$


$Q'_0$ is the rate of heat transfer by radiation through the vacuum portion of the flask.

$A_{in}$ is the surface area of the outside of the inner wall of the flask

$ε_{s.s}$. is the emissivity of stainless steel

$σ$ is the Stefan–Boltzmann constant

Assuming that the outer surface of the inner wall and the inner surface of the outer wall of the vacuum flask are coated with polished silver to minimize heat loss due to radiation, we can say that the rate of heat absorption by the inner surface of the outer wall is equal to the absorptivity of polished silver times the heat radiated by the outer surface of the inner wall,

$${\displaystyle \alpha Q'_{0}=Q'_{in}}$$

In order for energy balance to be maintained, the heat lost through the outer surface of the outer wall must be equal to the heat absorbed by the inner surface of the outer wall,

$${\displaystyle Q'_{in}=Q'_{out}}$$

Since the absorptivity of polished silver is the same as its emissivity, we can write

$${\displaystyle Q'_{out}=\epsilon _{s.s.}Q'_{0}}$$

We must also consider the rate of heat loss through the lid of the vacuum flask (assuming it is made of polypropylene, a common plastic) where there is no vacuum inside the material. In this area, the three heat transfer modes of conduction, convection, and radiation are present. Therefore, the rate of heat loss through the lid is,

$${\displaystyle Q'_{lid}=Q'_{cond}+Q'_{conv}+Q'_{rad}}$$

$${\displaystyle Q'_{lid}=kA_{lid}\left({\frac {T_{b}-T_{surr}}{\Delta x}}\right)+hA_{lid}\left(T_{b}-T_{surr}\right)+A_{lid}\epsilon _{p.p.}\sigma \left[\left(T_{c}+{\frac {T_{surr}\Delta S_{p.p.}}{c_{p}^{p.p.}}}\right)^{4}-T_{surr}^{4}\right]}$$


$k$ is the thermal conductivity of air

$h$ is the convective heat transfer coefficient of free air

$ε_{p.p}$ is the emissivity of polypropylene

$A_{lid}$ is the outer surface area of the lid

$c_{pp.p}$. is the specific heat capacity of polypropylene

$∆S_{p.p}$. is the specific entropy of polypropylene

$∆x$ is the distance over which conduction across the temperature gradient takes place

Now we have an expression for the total rate of heat loss, which is the sum of the rate of heat loss through the walls of the vacuum flask and the rate of heat loss through the lid,

$${\displaystyle Q'_{total}=Q'_{out}+Q'_{lid}}$$

where we substitute each of the expressions for each component into the equation.

  • $\begingroup$ I tried to upvote both you and Sammy, but I have not the rep :-(, also, I didn't do the calculations, I have proved enough for myself to try follow through with my conclusions. Thank you all the same. $\endgroup$ Sep 24 '16 at 19:54
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    $\begingroup$ Totally agree. ::thumbs up:: Just got up-vote privilege! $\endgroup$ Sep 24 '16 at 20:04

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