Electronic – Coffee warmer capsule


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!!! 😀

THOUGH: In my test I had the resistor inside the water and it started "dissolving" so there is NO way you can put that in your drink.

I got fed up and just put 5 8.2Ohm resistors under my mug and got done with. It actually works…

Best Answer

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