First step is to understand that the heat in the wire comes from the resistance of the wire, which is like friction in a mechanical system. The amount of heat peer second that is generated is given by Q = I^2R, where Q = rate of heat (in joules/second), I is current going down the wire in Amps, and R is the resistance of the wire in ohms. The resistance R depends on the wire's cross-sectional area and length, following the equation R = pl/A, where p = the resistivity of the conductor in ohm-m. Hence the heat generated in the wire is I^2 pl/A. The resistivity of copper is 1.7x10^-8 ohm-m, but a more common material that might be used in a heating element would have a higher resistivity: one common alloy for heating elemsnt si cupronickel,which has p = 2x10^-7 ohm-m.

Now as the wire heats up it begins to shed some of that heat to the surrounding air, so that the rate of temp rise slows. The hotter it gets the quicker it sheds heat, until ultimately the rate at which it sheds heat equals the rate at which I^2R heating creates heat. That's the equilibrium temperature. The problem is that calculating how quickly a wire sheds heat is not easy, because convection is difficult to model. It really depends on the geometry of the wire and the way in which air is allowed to flow over the wire. If the air is stagnant the wire will get hotter than if the air is moving. The equation to use is:

Q = hS (Delta T)

where h is a constant with units of J/s/m^2/K whose value depends on the geometry of the wire and the manner of air flow around it, S is the surface area of the wire, and Delta T is the temp difference between the surface of the wire and the air. The hardest part is determining the value for h. It can be done experimentally, though there are some empirical models that may give reasonable results. Here's a model for horizontal cylinder, like a wire:

http://en.wikipedia.org/wiki/Heat_tr...ontal_cylinder
In general for free convectio in air the value of h is typically between 5 and 25 W/m^2-K.

Putting it all together: the temperature rise of the wire above ambient air temperaturein Kelvin is:

Delta T = I^2R/hS = I^2 pl/AhS

Recognizing that the cross-sectional area A andthe surface area S are functions of the wire's radius r and length, we can replace A with pi r^2 and S with 2 pi rl, and so l/AS = 1/(2 pi^2 r^3). So we get finally:

Delta T = I^2 p/(h 2 pi^2 r^3)

Phew! Now if you plug in numbers of I= 10 Amps, p = 2x10^=7 ohm-m, h = 10 W/m^2-K, and r = 0.002 m you get Delta T = 100K, or 180 degrees F, which seems pretty reasonable. Class dismissed!