thermodynamics - How to calculate temperature of an incandescent bulb filament?

Suppose we have a light bulb, for which we know its power rating, like voltage of $12\mathrm V$, and power consumption of $10\mathrm W$. We also know it's a halogen bulb with a tungsten filament inside. Suppose we also know temperature of the surrounding air.

Is this data enough to compute temperature of the filament? If not, what should also be included? And anyway, how do we find the temperature?

Answer

You really are asking two questions.

First - how do we calculate the temperature:

At the typical temperatures of a halogen bulb, the large majority of heat loss is due to thermal radiation (although there is some conductive loss in a halogen bulb as the bulb is not evacuated). Because of this, the most important factor is the "apparent size" of the filament. I say "apparent" because when you have a tightly wound coil, the parts of the coil facing other parts of the coil don't contribute to a net heat loss.

If you took for example a 5 mm long, tightly wound filament with a mean diameter (after winding) of 0.5 mm, you would have a surface area of approximately 5 * π * 0.5 ~ 8 mm². If you had 10 W of emission, you would use the Stefan Boltzmann law to get the power per unit area:

$$I = \sigma T^4$$

from which we get a temperature of

$$T = \sqrt[4]{\frac{10}{8\cdot 10^{-6}\cdot5.67\cdot 10^{-8}}} \approx 2100 K$$

Getting more accurate numbers is quite hard - there are lots of subtle effects (conduction down the support wires, heat lost to the filler gas, and "true effective area" to name just three).

Second - how do we measure the temperature. For such high temperatures, the disappearing filament pyrometer is very effective: you send a current through a calibrated filament, and compare its color against the color of the object of interest. When the filament "disappears" against the background, the temperatures are matched. Often, filters are used to do the comparison in a narrower range of wavelengths; the result can give resolution down to 10C according to the above linked article. There are obvious problems with this is the emissivity of the object of interest is very different than that of the filament - but if you are trying to determine the temperature of a filament that is not likely to be a problem.