Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Black body and white body

  1. Mar 26, 2016 #1
    Hello Forum,

    A black body is an object with emissivity=1 that absorbs as well as it emits. Does that mean that its temperature T stops rising after it reaches thermal equilibrium but it rises before then since absorption dominates over emission?

    What about a white body? That would be a perfect reflector. All the incident energy will be rejected and emitted (reflect). That sounds very similar, from a net energy standpoint, to what happens with a blackbody...

    so what is the difference? Does the white body reach this steady state (incident energy in= energy out) instantaneously? Theoretically, the white body temperature should be zero and never rise...

  2. jcsd
  3. Mar 26, 2016 #2
    A black body emits and absorbs according to the Stephan-Boltzmann law until (and after) equilibrium is reached. A whiter body is simply better insulated so that the emission and absorption of radiant energy is much less and equilibrium takes longer to reach.

    If it could be made perfectly white or reflecting (which would have to be not just in the visible but at all frequencies from microwave through UV), then I suppose it would be perfectly insulated to radiant energy (incident energy in = 0 = energy out) and could only reach equilibrium with its surroundings by other means such as by conduction and convection.

    What is interesting and not-so-intuitive is that if you coated the inside of a cave or cavity with this (almost) perfectly white reflecting surface, and then made a tiny hole to see what colour the inside the cavity looks like from the outside, then the tiny hole you made would appear to be a perfectly black spot!!!
  4. Mar 28, 2016 #3
    thanks jwinter.

    I am still not clear on what happens if we leave a black car and a white car under the sun for 10 hrs. The black should get much hotter and the white car remain cooler.

    In some situations, objects left in a room reach the same temperature, regardless of their color. Why? That is not what happens when they are in the sun, where they acquire different temperatures. Even if at the same room temperature, the object with larger thermal conductivity feels cooler because thermal energy from your body will quickly flow away from the touched area.
  5. Mar 28, 2016 #4
    A black body is an idealized physical body that absorbs all incidentelectromagnetic radiation, regardless of frequency or angle of incidence. A white body is one with a "rough surface [that] reflects all incident rays completely and uniformly in all directions."
    A black body in thermal equilibrium (that is, at a constant temperature) emits electromagnetic radiation called black-body radiation. The radiation is emitted according to Planck's law, meaning that it has a spectrum that is determined by the temperature alone, not by the body's shape or composition.

    A black body in thermal equilibrium has two notable properties:

    1. It is an ideal emitter: at every frequency, it emits as much energy as – or more energy – any other body at the same temperature.
    2. It is a diffuse emitter: the energy is radiated isotropically, independent of direction
    3. see for details https://en.wikipedia.org/wiki/Black_body
  6. Mar 28, 2016 #5
    Cars are not quite as black and white as you might imagine because the visible is only a very small segment of the EM spectrum. The vast majority of the radiant energy pouring down from the sun is infra-red and it is likely (but I don't know) that black and white paint has pretty much the same emissivity in the infra-red part of the spectrum.

    Black surfaces (high emissivity) absorb and emit radiant energy better than white surfaces (low emissivity), so a black object should heat up and cool down (by radiant pathways) quicker in a given situation that a white object. But because the temperature of earth objects is so low (in comparison to the sun), cooling down by radiating heat is very slow indeed and can probably be neglected in comparison with cooling down by other pathways such as conduction and convection.

    So yes a white car should stay cooler in the sun than a black car (but not by such a big factor because of the infra-red).
    This should be obvious - with no unbalanced radiant heat pathway there is no reason for an object to not be in equilibrium with its surroundings
    You are straying a long way from the original subject here! It only feels cooler when it is colder than your body. If it is just out of the oven then it will feel hotter!
  7. Mar 29, 2016 #6
    Thanks jwinter.

    Is it possible for a body to be a good absorber but a poor emitter or vice versa?

    Fundamentally, I guess I still don't fully grasp why a blackbody is, by definition, a good absorber and automatically a good emitter....
  8. Mar 30, 2016 #7
    No it is not possible. If it was you could defeat the second law of thermodynamics and make a perpetual motion machine (of the second kind).
    A black body is a good absorber, because that is how it is defined. It is something that absorbs all radiation that shines on it so that none is reflected (emmissivity = 0).

    If you try to imagine the blackest possible surface and then think how can you tell there is even a surface there. If it was truly black you would not be able to tell that it was there. Shine the brightest light on it and it should still look completely black. So black in fact that you cannot tell if there is a surface there or simply a hole opening into an infinite black chasm beyond. So you should have no trouble grasping why a "black body" is a good absorber - it simply absorbs all the radiation you can shine on it.

    Since nature really doesn't want you to get around the second law of thermodynamics, it must ensure that emission is identical with absorption, and so nature makes such a surface (or opening) the best possible emitter also!
  9. Mar 31, 2016 #8
    Ok, thanks jwinter.

    I continue with my quest. Why is heat (thermal energy) defined as that portion of the EM spectrum that resides in the infrared? IS it because the resonance frequencies of most common materials are in that region of the spectrum?
  10. Apr 1, 2016 #9
    My guess is that there is no better description for it. All radiation ends up as heat, but some of it we can see with our eyes (just as it is turning into heat) - and that makes it much more interesting to us. Frequencies above the visible region tend to be more destructive or penetrating and that also makes them interesting. But frequencies below the visible really don't do much at all except get things warmer.
    I don't know whether that is true or not. Anything that is dark or opaque to visible must be absorbing or reflecting those frequencies by something akin to resonances I suppose. There are very few crystal clear substances, and substances which are clear in the visible may not be clear in the infra-red and vice-versa.
  11. Aug 25, 2016 #10
    A lot of people here are confusing specific heat with temperature. If you leave a white car and a black car in the summer on a 90 degree day and check them in a couple hours they will both have a temperature of 90 degrees! However, they will each have a different specific heat. Confusing heat and temperature will really make it hard to do thermodynamics.

    Thermal energy is defined as the infrared because it is a property of mater that all bodies above 0 Kelvin give off infrared radiation and absorb it. This is caused by heat being converted to EM radiation according to Planks Equation.
  12. Aug 25, 2016 #11

    Charles Link

    User Avatar
    Homework Helper
    Gold Member

    The car with absorbing paint when sitting in the sun will get much hotter than a car with reflective paint=white is basically a diffuse reflection while a shiny mirrored surface would be specular reflections. In any case, the sun will not heat up the highly reflective surfaces any significant amount (if they are reflective throughout the entire spectrum including the IR), while the absorbent (black) paint can get very hot (temperature of surface climbs) in the sun before it radiates enough to reach dynamic equilibrium. (Radiation (in this case mostly infrared) per unit area is proportional to T^4 where T is in degrees kelvin. It is in dynamic equilibrium when heat absorbed =heat radiated.) The highly reflective surfaces will be at the temperature of the ambient air, basically by conductive heating, etc. while the absorbing surface(absorbing the direct sunlight) can get much hotter than the ambient air and the ambient air can even supply some conductive cooling to the very hot absorbing surface.
    Last edited: Aug 25, 2016
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted