For example will an XM-L2 U2 at 3A put out the same amount of waste heat as an XM-L2 T6 at a given current or less waste heat?
I can see the T6 putting out the same heat because it may be putting out higher energy photons, or perhaps it put out more heat because more photons are blocked in the colour conversion.
Anyone have any data about this to share?
Yes, they have different heat output. But the difference is small compared to the total heat output.
The efficiency of the LED is a direct measure of how much heat and light it gives off. So a LED that gives off more visible light for a certain amount of energy will run cooler.
A XM-L2 U2 is more efficient than a XM-L2 T6, so it gives of more light and less heat. But the difference is small. Both still gives off much more energy in the form of heat, than they give off energy in the form of visible light.
The energy of the photons is the same as the color. The shorter the wavelength, the more energy in the photons. But more important is the fact that cool white LEDs typically are slightly more efficient at giving of visible light, and therefore becomes a little less hot.
So given the same energy a more efficient LED gives off a little less heat, and more light. But I don’t think the difference is enough that you can get away with significantly less cooling for a XM-L2 U2 than for a XM-L2 T6. There is a difference, but it is small.
A way to estimate is to use the output effect in lumen and subtract from the total effect. What is left over is heat. Losses. One lumen of (green) light effect represent about 1/683 W. http://en.wikipedia.org/wiki/Lumen\_(unit)
Green is the color that our eyes are most sensitive to, so one lumen of white light need slightly more power than green light. So we won’t get exact numbers, but close enough for good estimates and comparisons.
A 10 W flashlight that gives off 1200 lumen actually radiates less than 2 W visible light, and generates more than 8 W heat. (1200/683=1.76 W)
If you compare the output in lumen between two LEDs at the same total power usage, you can estimate the difference in heat generated.
difference in lumen / 683 = difference in heat output in Watts, assuming the same power is used to drive both flashlights.