For a school project, I have to make a mug heater which draws its power from a computer's USB port, such as this one. I don't mean one that boils the tea, but one that keeps it warm-ish, at at least 40 degrees. Apparently, a USB port can give 500 mA at 5 Volts, which gives me at most 2.5 W. At first, I thought this wasn't nearly enough for a heater, but then I decided to experiment.

I decided to forget the rest of the circuit and pretended that the heating element could take all of the 500 mA. The best thing I could get my hands on was nichrome wire, and I worked out that to get a resistance of 10 Ω (to use all of the available power), I needed approximately 0.6m of swg 32. I experimented, and at 5 V and 500 mA, there was no heat produced whatsoever; as in, I could even touch the wire without feeling heat, let alone feel the heat above it.

How would I go about doing this? I know for a fact that this is possible (see the first link).

EDIT: Yes, there was current flowing (I used an ammeter), and I did coil the wire up.

  • Did you make sure that there actually was current flowing through the nichrome wire (i.e. did you put an amp-meter in series)? Also, did you coil up the wire a little? If you just have it as a straight piece of wire it can dissipate any heat produced quite quickly. – us2012 Oct 31 '13 at 19:21
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    What about a Peltier plate? With some additional circuitry, it could double as a cooler. – Matt Young Oct 31 '13 at 19:31
  • You may want to look here: electronics.stackexchange.com/questions/5498/… – Tut Oct 31 '13 at 19:32
  • @Tut "When a device is connected it goes through enumeration. This is not a trivial process and can be seen in detail on Jan Axelson's site. As you can see this is a long process, but a chip from a company like FTDI will handle the hard part for you." Remember, this is a school project... :) – Bluefire Oct 31 '13 at 19:47
  • @MattYoung can I run a Peltier Plate off 2.5W? – Bluefire Oct 31 '13 at 19:48

An electric kettle takes about a minute to boil a litre of water with 3,000 watts. if you've only got less than one-thousandth of that then in all honesty it would take forever but, assuming that heat was produced and kept in the kettle (mug or container) then it will start to warm stuff up.

Goin back to the 3 kW kettle, if it takes a minute to warm a litre of water from 20 ºC to 100 ºC that's an average of 1.333 ºC per second per litre. If it were more like 300 ml (mug size) then it would be about 4.5 ºC per second per litre. Now this has to be divided by about 1000 to represent your actual available power so maybe you could raise the temperature by 1 ºC in about 220 seconds (3 minutes and 40 seconds).

You then have to ask your self "how much does a mug of tea cool every second"? Do a test and you will probably find that it cools a bit quicker than you could warm it so then, you have to find a way of stopping a normal cup of tea from cooling so quickly.

Once you have got the mug (plus insulation and undoubtedly a lid) cooling less than the amount you can warm it with 2.5W you can devise any method you want of burning 2.5W in or around (close proximity to) the tea mug.

The most direct form of heating is better so maybe consider putting some form of heating element into the tea. It doesn't matter what sort of size or shape the resistor it is - if it only warms one small volume of the tea, heat conduction will cause this to equalize throughout the mug.

Also, remember that as the tea warms (hopefully), the constant power will keep it warming some more but, remember this is all a big waste of time without stopping the standard mug of tea losing its heat naturally - do tests and develop insulation techniques.

  • Way back in school science we did the "putting nichrome wire in a mug of water" experiment - no-one got very far as Teacher had failed to account for the conductivity of water. Or in other words, be careful what electronics you dip in your tea. – John U Nov 1 '13 at 9:35

2.5W is pretty small power to keep something warm. But the task is not impossible. The most important thing is to keep the heat where it is needed and to not allow it do dissipate in the air.

The rule is - on constant power, less surface means higher temperature. Big surface - lower temperature.

You will need to make a plate that to worm the cap. As long as the bottom of the cap is by porcelain, that is not very thermo conductive, you don't need to make the heating plate by metal. Use some thin glass plate with the diameter exactly as the bottom of the cap.

Put the heater under the plate and in thermal contact with the plate. Then under the plate (and heater) put some thermo insulation material - glass or mineral wool is a good material, but even cotton wool will do the job - your temperature should be lower than 100°C.

As a heater elements use some metal film resistors connecting them in parallel. You will need 10ohms resistance, so 10x100ohm resistors in parallel will do the job.

Use SMD resistors - they have flat surface that will provide better contact with the plate. Use some thermo paste in order to make the contact even better.

The main design rule - the heat must flow exactly to the bottom of the heated cup and to be insulated from all other ways.

There is one more way to increase the power if 2.5W is not enough. Use two USB ports to power your device. This way you will get 5W of power.

  • What if it's a tablet charger, which is >2000ma, not 500ma? – NoBugs Feb 15 '16 at 1:39

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