I am an electrician at a commercial electrical company. There is an in-house training program that posits the datum that when talking about the designations of color temperature for lights (for example an LED light with a 5000 Kelvin color temperature) that the designation means that the light is that temperature, thermodynamically....but that would mean that the light is 8540.33 degrees Fahrenheit. Yet LED lights are not only known for running much cooler than incandescent, but I mean, the idea that the light has over 8,000 degrees of heat seems illogical. My position on it is that when using K when talking about color temperature is that we are strictly talking about color, exclusive of heat or lack thereof. I am not sure and need help sorting this out.


Your intuition is correct. For a normal filament type bulb we run current through a narrow wire and the current heats it up due to ohmic resistance and it achieves something like thermal equilibrium - and the physical temperature and the color temperature are the same. This means that there is a lot of infra-red which is perceived as being hot. This is a real waste of power unless you are using the bulb in an "easy bake oven".

With the LED bulbs the LED's emit only a few wavelenghts of light and are definitely not in thermal equilibrium. Semiconductor junctions are used to emit just certain visible frequencies. Then, usually, the LED's are surrounded by phosphorescent material that absorbs the LED light and "smooths it out" by re-emitting over a range of visible frequencies. When assigning a "color" to these LED bulbs the visible section of a thermal radiation curve (a.k.a. black body radiation) is compared to the light of the LED and a color temperature is assigned. However, this assignment is partially art and partially science (and largely regulation) since the LED's are not thermal emitters.

  • $\begingroup$ Wow that is very helpful thank you! I did some more internet research in the meantime, but if you are still available for more info: Just to clarify, could one say that specifically with a black body radiator, like an incandescent light filament, or graphite (another article said graphite is pretty close to the theoretically perfect black body radiator) that as you heat it, it's color temperature in Kelvin, and its thermal temperature in Kelvin, would be about the same? Like the nearer something is to a perfect BBR, the more equilibrium there would be between these two qualities? $\endgroup$ – ElectricianPat Jan 7 at 16:48
  • $\begingroup$ Again, you are correct. Filaments are reasonably to being black body radiators and they get very hot! A typical filament heats up to about 2700K. This is why they are made of tungsten whose melting point is 3700K. The glass bulb surrounding it does two things - 1) it keeps oxygen out so the metal doesn't burn in the extreme heat, and 2) it keeps anything from getting too close to that super hot filament. Some glass bulbs are very small and come close to the filament - but you are warned not to ever handle them bare handed lest oil from your skin be left on the bulb and burn and break it. $\endgroup$ – Paul Young Jan 7 at 17:10
  • $\begingroup$ I want to add this: 5000K is closer to daylight but people are more used to yellow-ish light bulbs. Why is that? Because tungsten melts at 3700K and you don't want to get anywhere near that melting point - it shortens the lifetime of the filament. So the public got used to 2700K. So, why not use a metal with a higher melting point? Answer: none exists. $\endgroup$ – Paul Young Jan 7 at 17:14
  • $\begingroup$ Good to know! Gosh this is great. Thank you so much for your answers. I have one last one: If we go down deep into the LED where the semiconductor is (just to play "devil's advocate" a little bit) when the semiconductor fires off, and the electrons are flowing briefly, I know it's not quite an arc, but is that little bit of conductive material even if very briefly, in equilibrium? I don't see how it could be otherwise the LEDs would melt from all that heat, but it seems to make sense. Considering how "hot" an electrical arc is. Last question, promise, but your insights are very appreciated. $\endgroup$ – ElectricianPat Jan 7 at 17:19
  • $\begingroup$ I cannot answer with certainty, but I will give an educated guess. I think the answer is "no" - the LED's crystal is not in thermal equilibrium with the visible radiation. Electrons and holes combine at a p-n junction via specific electronic transitions emitting photons. If the junction is sufficiently close to the surface of the crystal the photons escape before they can be absorbed and heat up the crystal to the point of thermal equilbrium. I think most of the heat from an LED bulb actually comes from rectifying the power to the right D.C. voltage for the diode. $\endgroup$ – Paul Young Jan 7 at 17:31

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