0
$\begingroup$

Infrared waves are widely used for quick heating objects, but a quick look at electromagnetic spectrum astonished me and raised a question that how come an infrared wave with lower energy compared to visible light is able to heat objects.Is there anything here related to resonance ...??

$\endgroup$
5
  • $\begingroup$ "" how come an infrared wave with lower energy compared to visible light is able to heat objects"" I can't deduct this from looking at a EM spectrum $\endgroup$
    – Georg
    Jan 26, 2015 at 18:19
  • 1
    $\begingroup$ Infrared waves are used in many heating applications because they are easier to generate than visible light waves. That is, they are produced at a lower temperature, and therefore require less exotic materials (like tungsten) that are used in incandescent light bulbs. If you generate 100 Watts of IR, and 100 Watts of visible light, how much an object heats up depends on how much it absorbs at those frequencies. The IR source won't necessarily work better. $\endgroup$
    – David Rose
    Jan 26, 2015 at 18:59
  • $\begingroup$ Heating means energy transfer to the piece of material to be warmed up. Often its molecules have vibrational states with (quantum) level differences in the infrared. Due to their close packing in the material these levels are broadened, so much of the energy in a broad band IR source will be absorbed. $\endgroup$
    – Urgje
    Jan 26, 2015 at 20:18
  • $\begingroup$ All the close votes, and I'm not sure why. Two comments so far answer the question, so clearly they understand what is being asked (and should endeavor to make complete answers...). $\endgroup$
    – user10851
    Jan 27, 2015 at 0:26
  • $\begingroup$ Short answer: visible light can also heat objects and you can easily test it with a powerful LED light. $\endgroup$
    – gigacyan
    Jan 28, 2015 at 12:23

1 Answer 1

2
$\begingroup$

When you say:

an infrared wave with lower energy compared to visible light

I'm guessing you're referring to the photon energy:

$$ E = h\nu $$

Visible light has a higher frequency than IR light, so the energy per photon of visible light is higher than IR light. However this is not related to the energy carried by the wave. The power transmitted by a wave is the energy per photon multipled by the number of photons per second, and we can adjust the number of photons per second simply by adjusting the intensity of the radiation. So a high intensity IR wave can transmit more power than a low intensity visible wave.

We tend to use IR for heating things simply because it's easier to generate high intensity IR than high intensity light. Many of the radiation sources we use approximate black bodies, and the peak frequency of black body radiation is proportional to the black body temperature. So to generate visible light requires a considerably higher temperature than to generate IR.

$\endgroup$

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