# What material can achieve the highest temperature before failing when being heated by electricity?

Just as a qualifier, I barely made it through physical science in college, so... Please be detailed in your answers and assume I know nothing.

Let's say I wanted to make an electrical coil for the simple purpose of heating a volume of air (like an oven does, for example.) I'm wondering what material, or class of materials, can produce the most heat before failing?

To try and keep this simple, let's stick with the oven example, because I assume size is going to make a difference. And, also, since this is theoretical, cost of said material is no concern.

For bonus points, I'd love for someone to explain to me how to calculate out how well different materials will perform in this capacity, especially if I were to decide to change the parameters to a much smaller application...

(And yes, this questions is poorly put together, please suggest all the edits you want. I'm more than happy to improve the quality of my post if it means getting a better answer.)

• By produce the most heat before failing do you mean reach the highest temperature before failing? Commented Feb 5, 2015 at 17:56
• @JohnRennie Yes, I'll edit it in a moment. Commented Feb 5, 2015 at 17:56
• @JohnRennie - Though what I really should be asking is this: are those terms interchangeable? Can some materials produce more heat than others as they become thick or thinner, for example? Or is the heat given off always something like (maximum temp * amount of material)? Commented Feb 5, 2015 at 18:28
• You might find the Wikipedia article on refractory metals interesting. There are other materials that can withstand higher temperatures (like ceramics) but most of them aren't conductive enough to be heated effectively by a current through them. Commented Feb 5, 2015 at 18:36
• Also: Tungsten carbide. Commented Feb 5, 2015 at 18:42

Specifically, you can ask any wire to get equally "hot", but since different materials have different melting points, some are better-suited for heating elements. In electronics, heat is a by-product and is unavoidable. Most components generate a more-or-less fixed amount of heat "per second" in the silicon. The plastic that microchips are encased in is responsible for pulling heat off of the silicon. A heatsink attached to the black plastic is responsible for pulling heat off of the plastic. A fan attached to the heatsink is responsible for moving cooler air over the metal fins. Larger microchips are usually equipped with a large pad on their bottom side which mates with an equally large pad on the circuit board. Those pads are designed and physically constructed to transfer a large amount of heat directly to the PCB. The large thermal pad is sometimes also electrically grounded. All other microchip pins also conduct heat away from the silicon and into the PCB.

Basically, as long as you can dissipate the heat faster than it is generated, nothing melts. So, more important than how you generate the heat, or which material will survive the most heat (unassisted), it is a question of how you will dissipate that heat away from the actual conductor.

Consider shorting the terminals of a 12V battery with a piece of wire. The power dissipated by the wire is determined by Ohm's law.

E = I * R
Where:
E = voltage(volts)
I = current(amps)
R = resistance(ohms)

The wire, being the sole "component" (modeled as a resistor), drops the entire voltage along it's length. That is to say that the voltage on one end is 12V, while on the other end, the voltage is "ground" or 0V. That means the wire has consumed those 12V (into heat) all along it's length. Wire of a particular gauge has a rated DC resistance per length at a specific temperature. If one foot of wire measures 1 ohm, then:

E = I * R so:
I = E / R
I = 12V / 1ohm = 12Amp (assuming the power supply can actually push this much)

Then:
P = I * E
Where:
P = power(watts)
E = voltage(volts)
I = current(amps)
P = 12V * 12A = 144Watts (compared to about 2,000W for an oven at 350 degrees)

Working the oven backwards:
P = 2000W (google any oven)
E = 240VAC RMS (US standard)
P = I * E so:
I = P / E
I = 2000W / 240V = 8.3Amp
E = I * R so:
R = E / I
R = 240V / 8.3Amp = 28.9ohm (expected approximate resistance of heating element)

1BTU is approximately 0.293071 watt-hours so, if the wire does not self-destruct, it will produce about 491BTU per hour.

To keep the device (wire) from destroying itself, you simply need to either:
1. Limit the current through it or,
2. Heat-sink it like any other component

A few materials commonly used for heating elements: resistance alloys(nickel chrome, iron-chrome aluminum, molybdenum, tungsten), stainless, copper, aluminum, and nickel alloys. Some are coated, some are not. Coatings (heat-sink) also vary but ceramics are common.

Typical current ratings and resistances of multiple gauges of copper:
http://hyperphysics.phy-astr.gsu.edu/hbase/tables/wirega.html

I couldn't find an actual photographic cross-section of an element but:

Given that:
E = I * R and
P = I * E
Then:
P = I * I * R (substitute for E)
P = E * E / R (substitute for I)

• Please remember that I'm very uneducated on this subject. Could you possible label the variables in the equations and add a bit more explanation? I think this is what I'm looking for but...having trouble following. Commented Feb 6, 2015 at 4:39
• expanded on units and formulas
– Jon
Commented Feb 6, 2015 at 5:04
• Consider that a resistive, counter-top, stove burner, left on high, with no pan on it, will destroy itself. Likewise, an empty pan left on the same burner will be destroyed. If you will be generating a significant amount of heat, please design your coil and enclosure as if the fan distributing the heat has already failed. The element you create should not require a fan to not catch on fire. :)
– Jon
Commented Feb 6, 2015 at 5:23