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I want to make a small capsule that you can throw in your coffee and keep it warm (not hot, lets say about 40-50C).

In the first place, I don't care about power supply - I will have 2 wire going into my coffee. However I do care about it being water-resistant

I found wire-wound resistor, will it do the job?

Next, is there any chance to pack the "thing" (resistor) and a battery in cr2032-sized package? I need 1-2 hour lasting battery.

NOTE: I do not want to "heat up" my coffee (from 20C up to 50) but to slow down the heat loss. So lets say my coffee is ~60C and I need to be @~50C 1 hour later.

Final test: I actually ordered 5W 8.2Ohm resistors and in about 30-45' it was able to WARM UP a ceramic cup of water which was about 15C to a warm-like temperature 25-30C which is way more than I expected (I just wanted to slow down the "getting cold" time of my coffee) @5V drawing min 480mA and max 540mA. Temperature measurements done with my finger but that's actually the kind of precision I need (the amp draw was done with an actual precise amp meter and the power supply was a 300W PC PSU so the voltage was very steady 5-5.1 Volts). KUDOS TO THE ANSWER!!! :D THOUGH: The resistor started "dissolving" so there is NO way you can put that in your drink, a simple heat shrink should do the job...

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For your second question: no. :) – pipe Jan 3 at 12:07
My guess is that you'd need at least 2-3 Watts to keep a coffee mug warm. You want the battery to last one hour, so the maths is simple. Just google for "3Wh battery" and see how small you can get. You'll probably be disappointed. – Armandas Jan 3 at 12:16
Start with a fully insulated coffee mug so that it loses almost zero heat to the surroundings. However opening a lid every time to take a sip seems a big pain to me. Another idea is to use a bigger rechargeable battery like lipo. Put the battery outside the mug (maybe stick to one side) and just dip the resistor inside which can be insulated well enough to prevent you from adverse health effects. – Whiskeyjack Jan 3 at 12:58
Don't forget to spec the battery for 80°C+. – transistor Jan 3 at 13:40
I don't care about "mug solutions" (insulated etc.) – dzervas Jan 3 at 15:42
up vote 22 down vote accepted

Newtons Law of Cooling - Scala graduum caloris

the heat which hot iron, in a determinate time, communicates to cold bodies near it, that is, the heat which the iron loses in a certain time is as the whole heat of the iron; and therefore, if equal time of cooling be taken, the degrees of heat will be in geometrical proportion

\$ \frac{dQ}{dt} = h\cdot A \cdot \Delta T(t)\$

Q = Thermal Energy (cooling rate)

h = Heat transfer coef - Taking a minimum of 3Wm-2K-1 [1]

A = Heat transfer area - \$\pi r^2\$, take a 8cm diameter mug = 0.02m2

\$\Delta T\$ = Temperature of the object - Ambient temperature = 50 - 22 = 28

Thus the cooling rate is: 3 * 0.02 * 28 = 1.68W

This is what you need to counter. So you need a resistor to transfer 1.68Watts.

Take a typical AA battery: 1.5V @ 3.9Wh (typical Alkaline). This potentially could source the needed energy for your required 1-2h (2.32h).

From \$P = \frac{V^2}{R}\$ R would therefore need to be: \$1.339\Omega\$, but this equates to 1.12A CONTINUOUSLY from a AA, which it will not do (50mA is a typical drain)

This should show you the methodology needed & its a simple case of finding a suitable battery, suitable resistor, for the given environment.

[1] http://www.engineeringtoolbox.com/convective-heat-transfer-d_430.html

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What about the heat transfer through the mug? – Johannes Jan 3 at 12:56
Then things get more complicated as it becomes a two part equation & is dependent on how long the mug was allowed to cool: one for the top of the fluid and one for the sides. A ceramic cup however does act as an insulator to some degree but is still a source of heatloss. – JonRB Jan 3 at 13:04
Well, I don't need a "floating point precision temperature" for my coffee. I just need it to be "warm" which means 40-60C (I guess...) So a 2W 1.4kΩ resistor will do the job, right? – dzervas Jan 3 at 15:45
You lost me at Watts/second... – tomnexus Jan 3 at 16:39
This completely ignores the BIGGEST source of cooling in an open mug, which is evaporative cooling. – Dave Jan 4 at 23:13

You need to measure the heat loss in your mug around 50°C.

  • Measure the mug capacity.
  • Fill it with hot water or coffee.
  • Insert a temperature probe and close the lid as much as possible. Use tape to stop any heat leaks.
  • Record the temperature drop over time in the zone of interest.

Power loss (and power required to maintain temperature) will be given by

$$ P = \frac{ΔT·m·SHC}{t}$$

Where ΔT = temperature drop (°C), m = mass (kg), SHC = specific heat of water (4200 J·kg-1·K-1) and t = time (s).

I ran a test with 400 ml coffee (0.4 kg) and it took 21 minutes to cool from 53.5 to 52°C. Popping these into the formula we get

$$ P = \frac{1.5·0.4·4200}{21·60} = 2 W$$

This is the continuous power needed to supply to maintain 50°C in that much coffee in my cup. For two hours heating you will need a 4 Wh battery.

Thermos lab

The laboratory kitchen setup.

I was agreeably surprised at how good the insulation was on the cup.

One thing that hasn't been covered is that you really want to suppress the heating until the temperature drops to 50°C. That implies some electronics or a thermostat in your slug.

Phase change heating

Note this section does not provide an electrical solution to the original question but the question made me look up the information and I offer it as an alternative.

During change of phase from liquid to solid a material gives off its latent heat. The temperature remains constant until the phase transition is complete. I did a quick web search for phase change materials with a phase-change temperature around 50°C and found an interesting article on Better Pizza with Phase Change Materials in which the author describes a student project to keep pizzas at eating temperature for an extended time.

Phase change temperature plot

Temperature versus time during phase-change cooling.

This article lead me to savEnerg which lists their PCM-OM55P as having a phase change temperature of 55°C which is almost perfect for this application. The latent heat is given as 210 kJ/kg. Time for some numbers!

Lets say we could tolerate 100 g of this material in our cup. (Density is 0.84 kg/litre so its volume would be \$\frac{100}{0.84} = 120 ml\$). If we heat it up and convert it to liquid then on cooling down it would give off \$210,000 J/kg \times 0.1 kg = 21,000 J\$.

Since a watt is one joule per second and we require 2 W to counter heat loss at 50°C then the time to make the transition is \$ t = {21,000\over2} = 10,500 s = 2.9 hours\$. This, I suggest, meets the OP's requirement.

There are a few practical considerations.

  • The phase change material needs to be heated. This shouldn't be a problem if there is enough energy to make a mug of coffee.
  • The PCM-OM55P max operating temperature is 80°C. I'll leave the reader to figure out how not to overheat the phase-change material.
  • I have no idea what format the material is available in and how it would be packaged for this application.

On the positive side there are no electrics and it should have a long life. The best solution is the simplest one that works!

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For 1 hour heating I need 2Wh, right? – dzervas Jan 3 at 15:50
Oops! Yes, and for two hours a 4 Wh battery. I'll fix. @JonRB's calculation is close to my experimental result. – transistor Jan 3 at 15:50
How was the "taste" of 52C? Is the temperature JUST below the "I burned my tongue" temperature or less? Is 60C drinkable? – dzervas Jan 3 at 15:53
@ttouch: I'm not a hot-drinks guy. I didn't test it. I think it really depends on how much slurp-noise you think your company can tolerate. The slurp draws in air to rapidly cool the drink. – transistor Jan 3 at 15:58
+1 for actually doing the test! – Dave Tweed Jan 3 at 16:25

For low voltages, you can get away without insulating the element as the water will have a higher resistance and you won't get much electrolytic current. For higher voltages, insulate to prevent electrocution. You also might want to insulate anyway to prevent electrolysis in case the element breaks. Also, it is highly unlikely that the manufacturers of the resistor took care to make it food safe, so you are probably better off putting a clean, insulating coating on it.

A CR2032 is close to 1ml in volume. To warm 250ml of coffee from ambient 10°C to 50°C is 4181 J/L° * 0.25L * 40° ~= 42kJ. This means you need an energy density of 42MJ/L.

You have no chance of doing this with a battery. Jet fuel comes close but would also need an oxidiser and something to regulate the reaction, so might be possible in 3ml. But if you really want something that small to generate enough heat to warm a cup of coffee you need thorium pellets.

The off-the shelf coffee warmers are either catalytic or gel solutions which give off heat when crystallizing, and are much larger than CR2032 cells.

Also, if you put something that small in your coffee you might swallow it by mistake.

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He wants to keep it warm, not warming a cup of coffee – JonRB Jan 3 at 12:19
yes I need to slow down the heat loss, not heat up... "jet fuel", that would be COOL :D – dzervas Jan 3 at 15:48
"might swallow it by mistake" ... mission double accomplished, the coffee keeps you warm too :) – rackandboneman Jan 3 at 16:03
Then you'll have fever :P – dzervas Jan 3 at 16:36
+1 for thorium pellets option. "If the coffee don't kill you ..." – transistor Jan 3 at 23:26

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