You must have a short somewhere.
24AWG is 0.511 mm in diameter and has a resistance of 84 mOhm/m. That means that a 1 A current will generate 84 mW in heat loss per meter.
Copper has a specific heat of 0.385 J/g, this means you have to 0.385 Watt-second to heat 1 g of copper 1 °C.
Specific mass of copper is 8.93 g/cm^3, or 1 m of 24AWG weighs 1.83 g. So you need 0.705 Watt-second to heat 1 m of 24AWG 1°C, provided it can't loose this heat. Now we're generating 84 mW, this means that having 1 A for 1 second will heat -- well, warm -- your cable up by 0.12 °C.
In 1 minute your cable will be 7 °C warmer, in practice less, because the cable can conduct/convect/radiate part of this heat.
The insulation is HDPE (High Density PolyEthylene), which has a melting point around 120 °C.
In conclusion: even 1 A will not melt your cable, and a LED uses far less than that.
My proposed answer is as follows, based on this FAQ plus some physical realities.
The Working state is as described above, as is the Broken Connection state, but the Shorted state is not as has been presented on various over-clocking websites in most heating applications.
Firstly, it appears that peltiers often slowly decline according to the referenced article, with increased resistance and decreased heat pump capacity.
Obviously an all-at-once failure and increase of resistance to the maximum capacity of the device can happen, but that does not mean that the heat produced by a failed 100W peltier in a heating application would be the same as if it were replaced by a 100W resistor as typically used:
Since the peltier will be installed as a heatpump, not a heater, and therefore will have a heatsink or some other ability to exchange temperature on both sides, the total heating capacity will be halved, assuming there were no other variables involved and the installation setup was the same on both the cold and hot side (same heat sink, same fan). But, in reality, since the cold-side will represent an already-colder environment and the hot-side will represent an already-hotter environment it will dissipate heat more quickly and more effectively on the cold side, at least until the two sides eventually equalized in temperature (if that ever occurred).
Therefore, my proposition is that in a typical heating application the heat produced by a perfeclty-shorted 100W peltier would be the equivalent of a 50W, poorly back-insulted, resistance element.
Best Answer
The part you are missing is the thermal resistance from the wire to ambient air (or whatever medium you have it in). This is commonly specified in °C/W for semiconductor devices. For wires, of course the total power depends on the length, so you are looking for figure like °C/(W m).
For example, let's say you want the wire temperature to be 100 °C and that you've found its thermal resistance to ambient air is 10 °C/(W m). Ambient air in this example is at 20 °C, so the temperature rise required is 80 °C. That means that you need to dump 8 W into every meter of wire. If your wire segment is 200 mm long, then you need to dump 1.6 W into it.
Once you know the power required, you can use the electrical resistance to compute the voltage and current that will attain that power.
Keep in mind the resistivity of many substances changes significantly over temperature. Old LEBs (light emitting bulbs) were a good example. The filament gets so hot it gives off significant light. It is several times more resistive at that temperature than at room temperature. LEBs therefore had large currents for a short time after being switched on until they got to glowing temperature.
One way to deal with varying resistivity is to measure both voltage and current in a microcontroller, do the multiply to find the power, and adjust the power supply accordingly to maintain the desired power. The power supply is likely a switcher controlled by a micro anyway, so this doesn't add much complexity.
A even better way is to regulate the temperature directly. This scheme makes use of the fact that the wire resistance changes with temperature. You measure voltage and current, but this time compute the resistance. The power supply is then regulated to keep this resistance constant. This method is tricky if the material doesn't have a lot of resistivity change per temperature at the operating point you care about.