Electronic – difference in heat produced on a conductor between AC and DC at the same amperage and voltage

The best way to do this would be like a real multimeter/ammeter measures current. Put a very low value shunt resistor (<1 ohm) in series with the DC power supply (cut the connection and connect it back together through a resistor). The only thing you have to watch out for in this scenario is that the higher your load, the greater the voltage drop across the shunt.

So, using Ohm's law:

V = IR

Assuming a 20 mA load through a 1 ohm shunt:

V = .02A * 1 ohm = .02mV

So, with a 20 mA load, the shunt drops 20mV. As you can see, using a 1 ohm shunt is particularly convenient because the voltage drop equals the current (20mA results in 20mV drop).

The Arduino has a voltage regulator on board, so as long as the voltage at the Arduino stays above about 6 volts everything should be fine. Keep in mind that this method measures ALL of the current used by the Arduino, including inefficiencies of the linear regulators and the current usage of the AVRs. However, for a rough figure of current consumption this is fine. Remember that power consumption will be the current through the shunt times the wall supply voltage (minus the voltage drop across the shunt, but that will probably be negligible). Since your wall supply voltage is known, you only need a lead showing current.

The fact that you're trying to measure this with an Arduino complicates the issue. You need a common ground between the DC power supply and your measuring Arduino. Depending on the Arduino's power source (if you connect both Arduinos to USB), that means you cannot just choose the more negative lead on the shunt to be ground (doing so would cause a short and probably some smoke). This means you need to design a differential amplifier using an Op-Amp.

This is fairly simple, but I highly recommend you simply use a multimeter to measure the voltage drop across the shunt. This will be both easier to implement and more accurate.

If your soldering iron heater is intended for 240v, then giving it 12v would result in a very low heat output indeed. Feeding it with 1/20th of the voltage would result in 1/400th of the rated power (\$power=\frac{v^2}{R}\$), you might feel the chill taken off it, but no heat to speak of.

Before you look at 24v irons made by Weller, they run off 24v AC. I doubt the thermostatic switch is rated for DC, so you could expect the switch to fail in short order if you tried to run on off 24v DC.