# A cheap experiment to demonstrate the relationship between resistivity and heat output

I'm trying to design a cheap experiment that will demonstrate relationship between resistivity and temperature change under current. Optimally I would

• Find some relatively resistive substance
• Run some current through it.
• See how much it heated up

Any suggestions from either the practical angle (choice of material, range of current) or the math (temp change <>watts) would be a big big help

Joe

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There is no relationship between resistivity and heat. There is a relationship between resistivity, and the ease with which you can, with common tools, cause something to produce a lot of heat. However, this relationship is purely one based in the limitations of common lab tools. With the right equipment, anything with any resistance at all can be made to produce any amount of heat (well, until it self-destructs, anyways). –  Connor Wolf Jan 25 '12 at 1:41
Agreed and understood. Sorry I misexpressed myself. –  Joe Stavitsky Jan 25 '12 at 1:48
As a side note, what you are making sounds a lot like a simple Calorimeter. The wikipedia Calorimeter page (as linked earlier) may be useful to you in your pursuits of this project. –  Connor Wolf Jan 25 '12 at 1:51
@FakeName actually there is: in fact I think practically any material, conductor or semiconductor, increase its resistivity with temperature; I know it's not what Joe was asking, but just to make it clear :) –  clabacchio Jan 25 '12 at 8:33
@clabacchio - but a change in heat output does not necessarily result in a change in temperature or resistance. Certain materials may have temperature coefficients, but that is a relationship between temperature and resistance. If you have a material that is perfectly heatsinked (e.g. idealized), the heat output can vary infinitely without affecting the resistivity at all. Everything else is materials or implementation specific. –  Connor Wolf Jan 25 '12 at 10:10
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Strap a piece of this stuff across the leads of a 6-volt lantern battery and you can melt thread, fishing line, Styrofoam, etc. It gets really hot - sometimes visibly red hot, depending on the battery. Check this out: http://www.amazon.com/Nickel-Chromium-Resistance-Bright-Diameter/dp/B000FMUF4E/ref=pd_sbs_indust_1

You might also be able to find it at your local hardware store.

Or, for an even cheaper (though much less awesome) experiment, use a thermistor: http://en.wikipedia.org/wiki/Thermistor. I've gotten them on Mouser for a few dollars each (of course, I don't buy in bulk).

They are very cheap, and the temperature rise is easy to calculate (calculations might fall under the category of "self-heating" in datasheets, and on Wikipedia).

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This: (because I've done it, almost exactly as follows)

1. Get a chassis-mount resistor. They're fairly cheap; you can get a 50W resistor for under \$4 in single quantities from Digikey. Pick a resistor value that works for your voltage source, whether it's a power supply or AC (WARNING: do not wire to the AC mains unless you know what you're doing!)

2. Get a block of aluminum or steel or something (important: make sure you know what kind of material it is), with a flat surface, from a scrap yard or a machine shop. Machine shops are probably better, as they can help you with the next step, and they'll probably have a small block of scrap metal they wouldn't mind giving away.

3. Tap machine screw holes in the metal block, placed so that you can screw the resistor down to the block of metal.

4. Screw the chassis mount resistor onto the block of metal. You probably don't even have to worry about thermal grease to ensure good conduction between these parts, because of the way this experiment will work.

5. Wire the chassis-mount resistor to your voltage source.

6. Attach your favorite temperature sensor to the block of metal, not too close to the chassis mount resistor.

7. Put the whole thing into an insulating box. Styrofoam is best, but an ordinary cardboard box is probably fine.

8. Apply voltage to the resistor (WARNING: see above warning about AC mains), and record the voltage and temperature over time.

What you expect to happen is that the temperature of the block will rise linearly. The rate of temperature rise will be proportional to the amount of power being dissipated in the resistor: dT/dt = P / C, where C is the heat capacity. C will be equal to the specific heat capacity (joules per degree C per gram) times the mass of the block, in grams.... and you may have to take into account the heat capacity of the chassis-mount resistor, but if you choose a block that is larger than the resistor, this will be negligible.

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This is indeed excellent as an experimental demonstration, but unfortunately unsuitable for my purposes because we need each student to perform the experiment. –  Joe Stavitsky Jan 25 '12 at 4:12
Use an incandescent light bulb. :) Or an electric boiler plate, the kind where you can see the filament and do boil some water.