4
$\begingroup$

Infrared lamps seem to be used in situations where you want to produce heat without producing (as much) light. But incandescent bulbs also produce heat. Why not just use incandescent bulbs?

My intuition is that when you use an infrared lamp, you're channeling (more of) the available energy into infrared radiation, whereas when you use an incandescent bulb you "waste" energy on producing visible light. Is this the case?

I'm also curious about infrared radiation generally -- it's just another type of electromagnetic radiation, right? Is it more efficient at transmitting heat than visible light? If so, what makes it more efficient?

$\endgroup$
5
$\begingroup$

Incandescent bulb are black body emitters. Basically, something is heated enough so that it radiates most of the power put into it. The black body radiation spectrum is well known, with it being a function of temperature. The lower the temperature, the more the bulk of the radiation shifts to lower wavelengths.

Normal incadescent bulbs used for lighting are actually horribly inefficient. We don't have a material that we can heat up so hot to maximize the amount of visible light in its black body radiation spectrum. Actually, we can heat material that high, but it won't last long. We don't have a material that won't evaporate or sublime slowly enough to make a useable lightbulb with its black body radiation optimized for visible light.

Real LEBs (Light Emitting Bulbs) are a tradeoff between efficiency and lifetime. The hotter you run the fillament, the more efficient but also the shorter it lasts. This effect is quite non-linear. A little hotter makes a big difference.

There is therefore another reason to have separate heat lamps from ordinary LEBs, which is to make a more economical IR emitting bulb. The fillament operating temperature is adjusted for good IR production and less light. This does increase efficiency a little, but a ordinary LEB still emits most of its radiation as IR already. However, running the fillament cooler greatly increases bulb life and forces less tradeoffs in the design otherwise. Not only does the fillament evaporate slower, but the wire can be thicker, for example.

So dedicated heat lamps are a little more efficient at their intended purpose, but also last a lot longer and are more rugged and durable than incandescent bulbs shifted more towards producing visible light.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Do you have a source to prove that the filament operating temperature for incandescent bulbs and heat lamp bulbs are different? $\endgroup$ – pentane Feb 11 '18 at 8:42
  • $\begingroup$ @pen: You can probably look at some datasheets. Or, just look at a visible bulb and "heat lamp" bulb run at the same power. The visible one will produce significantly more visible light. Going back to the physics of black body emissions, you know the one with more visible light is at a higher temperature. $\endgroup$ – Olin Lathrop Feb 11 '18 at 14:29
  • $\begingroup$ This is the only explanation on the internet that makes physics sense. Most others are some bs explanation. $\endgroup$ – Dan Z Dec 29 '19 at 14:54
4
$\begingroup$

You are right that IR lamps typically have the bulk of the energy shifted to longer wavelengths. Now it turns out that infrared radiation (IR) is more readily absorbed by water in the body - so when you use the same amount of power in an IR lamp, more of it will be "felt" as heat by the body (because less is scattered / reflected by the skin).

As for transmission of energy: one photon of IR carries less energy than one photon of visible light; but when you are talking about efficiency of transmission you really have to ask about the attenuation in air (or whatever medium is "transmitting") - and then it turns out that visible light (this is no coincidence) sits on a "window" in the atmosphere where little attenuation takes place.

Looking at this diagram (from http://www.rfcafe.com/references/electrical/ew-radar-handbook/images/imgh51.gif) illustrates the point:

enter image description here

As you can see, in the visible range the "transmission" curve is close to the "scattering losses" curve; when you get to the IR, there are regions of significant absorption. When you stay away from these (for example, around 10 um) you can indeed get very efficient transmission - but at other wavelengths, nothing gets through.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ So how does the plot work? $\endgroup$ – Dan Z Dec 29 '19 at 14:53
0
$\begingroup$

Are you asking about heat lamps, like those used to keep food warm or heat up bathrooms? Your intuition is correct-- those are normal incandescent bulbs (sometimes with red tinted glass). According to energystar.gov, only 10% of the energy from incandescent bulbs is visible light and the rest is infrared.$^\dagger$ Because our eyes don't see infrared, you would be completely unable to tell whether a true "infrared lamp" was on or off by looking at it. Lamps that produce only infrared for heating purposes do not exist to my knowledge.

The reason is that the mechanism used by heat lamps to produce heat does not permit much control of the wavelengths of light emitted. The spectrum is determined by the temperature of the filament via blackbody radiation. There are LEDs and lasers that produce only infrared light, but these are not used for heating.


$^\dagger$ This is actually the reason they are being phased out as lighting solutions (in favor of LEDs).

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.