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I am putting DC current through a wire to heat it. I would think the wire would heat up evenly but I have found that it is hotter the closer I get to the middle, or, respectively, colder the nearer to the clamps. Can anyone explain this?

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    \$\begingroup\$ Wire connectors on the sides act as heat sinks? \$\endgroup\$ – Nazar Aug 15 '16 at 13:57
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    \$\begingroup\$ Also, a high positive TCR can exacerbate the effect. \$\endgroup\$ – Dampmaskin Aug 15 '16 at 14:12
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    \$\begingroup\$ Post a picture of your setup and maybe put a rule or scale in so we can estimate dimensions. \$\endgroup\$ – Transistor Aug 15 '16 at 14:26
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    \$\begingroup\$ Can you confirm how your power is fed to the circuit? \$\endgroup\$ – Andy aka Aug 15 '16 at 19:42
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There are two effects going on. The heat sinking effect of the connections and the temperature coefficient on the wire.

Initially the wire is all at the same temperature.

You turn the power on and it starts to heat up.

The heating is determined by the electrical power dissipation in the wire, for any given section of the wire Power = Current * Voltage. All parts of the wire will have the same current. For a given length the Voltage = Current * Resistance giving Power = Current squared * Resistance.

Initially all the wire has the same resistance and so the heating is even along the length of the wire.

The heat flows from hotter to objects to cooler ones (this is the first law of thermodynamics). In this case the connection points are cooler and so heat flows from the ends of the wire to the connectors cooling the ends slightly. Since the very ends are cooler the bits of wire near them then cools a smaller amount and so on along the length of the wire. This results in a very small temperature gradient across the wire with the middle slightly warmer than the ends.

Copper has a positive temperature coefficient of about 0.4 percent per degree C. This means that the warmer the wire the higher the resistance.

The middle of the wire is hotter which means its resistance increases. From the above equations this means more power is dissipated in the middle of the wire than in the ends.

More power means more heating in the middle than the ends and you get a positive feedback effect. The middle is hotter which means it has a higher resistance and more power is dissipated there which means it gets hotter...

This continues until almost all the power is dissipated in the middle of the wire, you never get all of the power in a single point because the heat conduction along the wire means that the sections near the middle also have reasonably high resistance. Eventually you reach an equilibrium where the thermal conductivity spreads the energy enough to balance the positive feedback effect.

The best example of a positive temperature coefficient is an old style incandescent light bulb. If you measure the resistance when cold it will be a fraction of the value you would expect for its power rating, they operate at about 3000 degrees and so the cold resistance is about 1/10th of the normal operating resistance when on. They are made of tungsten not copper, copper would be a liquid at those temperatures, but the thermal coefficient is about the same.

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  • \$\begingroup\$ One other note - Wire has a low resistance which means the total power dissipated in a wire is generally not that high and so the effect isn't that large at normal currents. If you use a material with both a higher electrical and thermal resistance (e.g. the lead from a mechanical pencil works well for this) then the you can visibly see this effect as the middle slowly heats up to a few thousand degrees and starts to glow. The middle will burn/evaporate at those temperatures further increasing its resistance and increasing the effect until it fails. \$\endgroup\$ – Andrew Aug 15 '16 at 16:19
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    \$\begingroup\$ i'm sure you already know this, but for clarity's sake mechanical pencil "lead" is actually graphite, a form of carbon. lead has a low resistivity (hence its use in solder) \$\endgroup\$ – Steve Cox Aug 15 '16 at 20:22
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    \$\begingroup\$ Note that the OP never said that he was using copper (although it's probably a safe bet). Also, a good example of a material with a negative temperature coefficient of resistance is a carbon lamp filament, like those used in the very first bulbs. \$\endgroup\$ – Dave Tweed Aug 15 '16 at 21:37
  • \$\begingroup\$ @SteveCox: Yes, but metallic lead (and solder) still has roughly 10x the resistivity of copper. Which is why building up a PCB trace with solder to handle higher currents is less effective than you might think... \$\endgroup\$ – Dave Tweed Aug 15 '16 at 21:41
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    \$\begingroup\$ @DaveTweed just don't want someone to accidentally hook up "higher resistance" lead where they were expecting the resistance of graphite \$\endgroup\$ – Steve Cox Aug 15 '16 at 21:43
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Heat and temperature are two very different things. The equilibrium temperature occurs when the heat flow into a region equals the heat flow out.

In your case, the heat flow in per unit length of wire (the resistive heating) is essentially constant, as you surmise. However, the heat flow out — both along the wire itself and to the surrounding air — varies, mainly due to the proximity of whatever the ends of the wire are attached to, which acts as a heatsink.

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protected by Dave Tweed Aug 16 '16 at 11:45

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