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Why do you feel warmer in a dark T shirt on a hot day than a white T shirt?

  1. Dec 5, 2008 #1
    People often explain this by saying that dark colours are good absorbers of radiation (compared to white). But in itself this explanation is not sufficient; a good absorber of radiation is (by definition) also a good emitter of radiation!

    Does the dark T shirt enable your body to reach thermal equilibrium with the warm surroundings faster that a white T shirt? If this is the case, then if you're outside for a while (long enough for the white T-shirt to reach thermal equilibrium with its surroundings) wouldn't a white T-shirt be just as warm as a dark T-shirt?

    Or is it that a dark T-shirt reaches thermal equilibrium at a higher temperature than a white T-shirt? (If so please explain. This seems to contradict my previous understanding of thermal equilibrium)
  2. jcsd
  3. Dec 5, 2008 #2


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    But what TYPE of radiation it absorbs and what TYPE does it radiates?

    There are many objects that absorb, say, UV radiation and then either gets warm or radiate that has IR, which bodies pick up as heat. A black object can do just that, where it absorbs visible light, but then due to the molecular vibrations, radiate that back as IR/heat. Your dark t-shirt absorbs more visible/UV than lighter ones, and it gets hot and also radiate IR back out. That's why it feels hotter.

  4. Dec 6, 2008 #3
    I haven't thought about this for more than a minute or two, but I believe that your assumption that good absorbers are good radiators is mistaken. Such a statement is true only in limited contexts -- such as, perhaps, looking at thermal absorption in metals.

    If you look outside of a very narrow context, than this clearly is not true. Water in a microwave absorbs small (micro) wavelength radio waves, but it does not re-radiate any of these wavelengths (I.E. it is not a good radiator at all; just a good absorber of these wavelengths). Instead, it re-"radiates" the energy as thermal-kinetic energy (energy of the more rapidly moving H2O molecules). Thus, it has not only changed forms of energy, it has also changed from radiation type energy to conductive (remember: radiative, convective and conductive are the three types of thermal transfer -- convective is really a special case of conductive, but for practical purposes such as in engineering, it is practical to consider it a third type).

    It is in fact not even necessarily true that the water will re-radiate or pass on *any* energy. Put an ice cube in a microwave, heat it to room temperature, and it it will have absorbed energy, but it will not heat up your hand, or heat up the cup you put it in.

    Consider also endothermic (heat absorbing) chemical reactions: a vat of liquid will absorb heat, the chemical reaction happens, molecules are formed -- and the heat energy stays in the vat. It is not re-"radiated". It has been absorbed into the new molecular bonds. In actuality, the vat may turn colder, or may turn warmer, depending on the particular endothermic reaction -- but the energy put in will not equal the energy that will be "radiated" (i.e. conducted and convected) back out into the room that the vat is in.

    For the T-shirt, most of the energy from the sun's ray's probably turns into molecular kinetic heat energy, transferred through contact to your body, and (much less efficiently) to the air.
  5. Dec 6, 2008 #4


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    Note that the absoption or emission of EM energy is the same for a material at a particular frequency. But that isn't what determines what is going on here: black body radiation is temperature dependent, so the radiated EM will be of a lower frequency than the absorbed EM.
  6. Dec 7, 2008 #5
    Hey thanks for your replies.

    What you are saying makes sense and it is what I overlooked. The claim that "a good emitter is an equally good absorber" is something I have come across in a conceptual physics course that I am teaching. Russ; you also concede this statement is correct (for a particular frequency)

    This still bothers me. If this is the case, why doesn't a black object appear as white. Black objects are black because they absorb the full range of visible EM frequencies. However if they were equally good emitters of these frequencies- they would also emit the full range of visble frequencies. Wouldn't the black object then appear the same as a white object that reflects all frequencies of visible radiation?

    What am I overlooking now?
  7. Dec 7, 2008 #6


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    You're overlooking what I said in the second sentence:
    The surface of the sun is 11,000 F. The surface of your black shirt on a hot day is, perhaps 110F. So your shirt radiates at a different frequency than the sun. Heat your shirt up to 11,000 F and if it survives, it will radiate white light.
  8. Dec 7, 2008 #7
    The often-quoted principle that "a good absorber is also a good emitter" is true for the Einstein coefficients for photon absorption and emission. Perfect for atomic physics. But when you consider bulk matter, not atoms, you find that even if an individual atom re-emits the light, it could just as easily emit it into the substance as opposed to out of the substance. Then it could become absorbed again, emitted again, and so on. From this point on, it becomes thermodynamics.

    Big example: the sun's interior, at millions of degrees, would probably be classified as a good emitter. But the individual photons can take millions of years to diffuse out to the surface because of absorption / emission / reabsorption in such a high density medium.
  9. Dec 7, 2008 #8


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    A photon emitted into an object must eventually make its way out of the object, otherwise the interior would "fill up" with photons. Eventually the object reaches an equilibirum. As you say, it's thermodynamics. So I don't see how that changes anything.
  10. Dec 7, 2008 #9
    Precisely. It wasn't meant to.
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