I am trying to figure out if I can melt Aluminum cans using a heater coil from a toaster oven.

I've already used q=mc∆T and q=m*(Latent heat of fusion) to figure out how much energy is required to melt a mass m of aluminum, considering no heat loss to surroundings.

What I'm having a hard time with is that I know the coil needs to be at least 660°C in order to melt the aluminum. So, for example if at full draw the coil's temperature is only 500°C, once the surroundings reach 500°C, what happens? Obviously the aluminum isn't going to melt, but does the coil just stop drawing current since it isn't able to transfer the energy?


  • \$\begingroup\$ "Scotty, We need more power!" Do you have any fire brick around your heater elements? \$\endgroup\$ – George Herold Aug 29 '14 at 1:26
  • 2
    \$\begingroup\$ You're confusing 'heat' and 'temperature'. Just because the temperature doesn't rise, doesn't mean there is no heat transfer anymore. Basically, the temperature is constant because the amount of heat added (by converting electrical energy into heat) is the same as the amount of heat lost due to radiation/convection to the surroundings. \$\endgroup\$ – RJR Aug 29 '14 at 5:11

The heater will never stop drawing current, the current will be quite constant because it is made of material with low thermal coefficients of resistance.

The heating element loses heat by three principles: conduction (contact with cold air), convection (wind) and radiation (far-IR)

Conservation of energy works also in this case. When heating up, consumed power is used to heat the element, and cover heat losses.

The hotter it gets, the higher percentage of input power it dissipate into air.

When it reaches its maximum temperature, it can not heat itself more because all input power is lost in air.

Side note: some heating elements can get hotter then their own melting point (e.g. clothes iron) and can not work without temperature controller.


The coil will keep drawing power an dissipating energy as heat.
By itself, a coil (resistive heater) doesn't have a negative feedback mechanism.

As the material around the coil is heating up, the coil will keep drawing electric energy and dissipating it, so it will keep heating the target material. The gradient between the coil and the material it's heating can be bigger or smaller. In a steady state a powered heater will be at a higher temperature than the material that's being heated.

Assumptions: steady state operation (the transient response has already decayed), ambient is colder than the coil an material, catastrophic failures in the heater are not considered.


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