Why the black objects absorb more light and heat from the other color objects?

[Physicsissuef]

Why the black objects absorb more light and heat from the other color objects?
What is the structure of the black color?

 

[pixel01]

Black objects absorb more lights than other colored so they appear black ! After all, things absorb more light and heat will appear more black.
Why they absorb more light and heat depends on the electron structure of the material.

 

[Andy Resnick]

Hang on... just because and object is black (highly absorptive) in the visible just not mean it is also black in the infrared. Snow is black in the IR region, for example.

How efficient an absorber a dielectric (or any material) is depends on both microscopic bulk properties (atomic/molecular absorption) but also on surface structure- textured surfaces can appear black under some conditions.

 

[DaveC426913]

Why the black objects absorb more light and heat from the other color objects?
What is the structure of the black color?

Without complicating the matter too much:

Black objects are black because almost all the light that falls on them is absorbed into the material. Little or no light is reflected back toward your eye, therefore you see black. That light that is absorbed ultmately becomes heat.

White objects are white because almost all the light that falls on them is reflected by the material. Because all the light is reflected back toward you eye, you see white. Little light is turned into heat.

Red objects absorb light at frequencies other than red and tend to reflect frequencies near red. They don't get as warm as objects that are black.


The nature of an object's colour has to do with the molecules and their electrons. Electrons in different states absorb different frequencies. When we build things - and especially when we paint them - we choose materials that contain molecules with very specific reflection frequencies. For example, Phthalo green - a paint colour - has very stable molecules in it that reflect light in a very specific, predictable band of green. Lamp black - another paint colour, has molecules that absorb ALL frequencies of light.

 

[TVP45]

Hang on... just because and object is black (highly absorptive) in the visible just not mean it is also black in the infrared. Snow is black in the IR region, for example.

How efficient an absorber a dielectric (or any material) is depends on both microscopic bulk properties (atomic/molecular absorption) but also on surface structure- textured surfaces can appear black under some conditions.

I used to design electrical enclosures and there was a white paint that had an IR emissivity above 0.9 but my boss always thought I was crazy since "white doesn't radiate".

 

[jobyts]

Without complicating the matter too much:

Black objects are black because almost all the light that falls on them is absorbed into the material. Little or no light is reflected back toward your eye, therefore you see black. That light that is absorbed ultmately becomes heat.

White objects are white because almost all the light that falls on them is reflected by the material. Because all the light is reflected back toward you eye, you see white. Little light is turned into heat.

Red objects absorb light at frequencies other than red and tend to reflect frequencies near red. They don't get as warm as objects that are black.


The nature of an object's colour has to do with the molecules and their electrons. Electrons in different states absorb different frequencies. When we build things - and especially when we paint them - we choose materials that contain molecules with very specific reflection frequencies. For example, Phthalo green - a paint colour - has very stable molecules in it that reflect light in a very specific, predictable band of green. Lamp black - another paint colour, has molecules that absorb ALL frequencies of light.

Is it correct to say, white objects do nothing with the incident EM waves, and black objects change the frequency (to IR range)?

 

[TVP45]

No, it's not an "all or nothing" case. White reflects maybe 60% to 90% of incident light depending on the actual pigment and the shininess of the surface. Black reflects maybe 5% to 30% for the same reasons. Light energy absorbed will be changed to IR and radiated according to the emissivity at that wavelength.

 

[Dale]

What always surprises me when I think about these kinds of things is shiny black objects! How weird. They reflect a lot of light (specular reflection), but they are black (diffuse absorption). Definitely a "surface structure" effect as mentioned above, but still interesting.

 

[DaveC426913]

Is it correct to say, white objects do nothing with the incident EM waves, and black objects change the frequency (to IR range)?

Too simplistic and misleading.

If a bond in a molecule absorbs a given photon of a given frequency, then that photon will be converted to kinetic energy of the bond - the bond will wiggle more. Kinetic energy = heat so the object is slightly warmer.

If the bond is impinged upon by a photon that is not of the right frequency, the bond will not absorb it. Depending on the structure, the photon (actually, the photon wavefront, but that's another story) will either pass straight through the material (meaning the material is transparent), or it will reflect off it (meaning the material is both opaque and reflective to that frequency of light.)

 

[dst]

What always surprises me when I think about these kinds of things is shiny black objects! How weird. They reflect a lot of light (specular reflection), but they are black (diffuse absorption). Definitely a "surface structure" effect as mentioned above, but still interesting.

Is that caused by a transparent & reflective layer on top of the black layer? An example would be anodized aluminium or powdercoating on, well anything. What raw material can actually do both without any change in composition?

 

[jobyts]

Kinetic energy = heat so the object is slightly warmer.

Why is kinetic energy == heat , for every substance?

 

[jobyts]

Hang on... just because and object is black (highly absorptive) in the visible just not mean it is also black in the infrared. Snow is black in the IR region, for example.

I was wondering how would any substance can be black for the infrared. If a substance reflects all the infrared rays, it should be white for the IR rays. If it absorbs the IR rays, it would cause the lattice to giggle, and emit heat, which is again in IR. So in any case, it should emit IR, right?

 

[DaveC426913]

Um. I am not sure exactly which frequency of light would cause the lattice to giggle.

 

[DaveC426913]

Why is kinetic energy == heat , for every substance?

That's what heat is.

 

[Andy Resnick]

I was wondering how would any substance can be black for the infrared. If a substance reflects all the infrared rays, it should be white for the IR rays. If it absorbs the IR rays, it would cause the lattice to giggle, and emit heat, which is again in IR. So in any case, it should emit IR, right?

Part of the conceptual problem may be with the concept of 'black'- it's typically associated with visible colors, not with thermal imaging.

Absoprtion = emission, by thermodynamics. A 'black body' is a perfect emitter in addition to being a perfect absorber. If the temperature of a black body is higher than the environment, it will be brighter than the environment. If it is colder, it will appear darker than the environment.

 

[Andy Resnick]

That's what heat is.

I disagree. Kinetic energy is a measure of the state of a system, is associated with how much work can be performed and is an exact differential. Heat, on the other hand, is the difference between the internal energy and the work, is not an exact differential, and is associated with a *process* rather than a state.

 

[meeeee5]

Heat

That's what heat is.

absolutely, heat is the process of energy transfer, and the kinetic energy is equal to 3/2*K*T.

 

[jobyts]

Let me be clear on some basics. Please correct if I am wrong.
Energy is a measure on the ability to do work on matter. Matter can do work on other matter, which is kinetic energy (eg; a moving object). EM wave can do work on matter, hence they have energy too. Heat is a just IR wave in the EM spectra.

In my understanding, the connection between EM wave to kinetic energy comes as follows:
An EM wave hits an atom; cause it to do work on the atom (causing the electrons to move; EM energy to kinetic energy conversion). Then the electron emits another EM energy which is Infrared ray, which we call heat.

My question is what property of the matter/electron caused it to emit waves in the IR frequency. Why not in some other frequency?

 

[Physicsissuef]

do the black body release the energy, that it absorbs?

 

[jambaugh]

I found it most useful to think of the surface color of an object as how its internal molecules couple to the electro-magnetic spectrum outside the object. "Blackness" is a kind of transparency of the surface to energy passing between the molecular vibrations (heat) and the electromagnetic vibrations we call light (and ir and radio etc.)

You see this deep inside a wood fire where the coals glow and the ashes are dark... shine a bright flashlight and it reverses with ashes white and coals dark. You can feel it when you pick up a teflon pan off the stove and hold it in front of your face. With the dark teflon side toward your face you feel the warmth, flip it around to the metal back and you fell much less heat.

From this mental picture you can then imagine bands of surface transparency effecting colors (and distinction between u.v., visible and i.r. reflectivity.)

do the black body release the energy, that it absorbs?

Yes but before it can be released it usually gets spread about in the kinetic energy spectrum (something called equipartition of energy) so that when visible light hits the black body it is re-released as infrared radiation in a thermal spectrum.

 

[Dale]

Is that caused by a transparent & reflective layer on top of the black layer? An example would be anodized aluminium or powdercoating on, well anything. What raw material can actually do both without any change in composition?

If you polish a black rock you can turn a dull black rock into a shiny black rock. I suppose that it is possible that the polishing process adds a "coating" but I thought it just rubbed pieces of rock off until the resulting stone was smooth.

 

[ok123jump]

Why the black objects absorb more light and heat from the other color objects?
What is the structure of the black color?

I agree with the various posters before me.

I would like to add that there is a slight confusion in terminology - which translates to some confusion in this initial question. To say that an object is black is really to say that it has absorbed the EM frequency which we are perceiving. I wouldn't say that colors truly exist as anything more than a residual effect of absorption/scattering mechanisms.

So, when you ask "Why the black objects absorb more light and heat from other objects?" - that is really a mistake in your thinking. I think that this question is equivalent to asking "Why do objects which absorb the frequency of EM which I perceive also absorb more heat as well?". To which the answer is - that is not necessarily the case, they don't always absorb more heat. They generate heat based on a conversion of energy from the incident light which gives the illusion that they absorb more heat - realistically, they are generating heat.

If you polish a black rock you can turn a dull black rock into a shiny black rock. I suppose that it is possible that the polishing process adds a "coating" but I thought it just rubbed pieces of rock off until the resulting stone was smooth.

The black rock is shiny because the process of polishing doesn't remove 100% of the debris. It may remove 99.9% of the debris, but that still implies that 0.1% gets smashed and compressed into an aggregate layer which ends up possessing some reflective properties.

This result relies on the fact that polishing generates a statistically broader distribution of incident angles of the light, relative to the direction in which the lattice structures of the rock powder point. While the majority of the light will pass through this layer to be absorbed by the black bulk below, a small percentage will interact chaotically to scatter the light (which is then picked up by our eyes).

 

[Physicsissuef]

Why Wt/At=Wt(black body)
Why At=1?
Stephan-Boltzmann law.

 

[Physicsissuef]

Yes but before it can be released it usually gets spread about in the kinetic energy spectrum (something called equipartition of energy) so that when visible light hits the black body it is re-released as infrared radiation in a thermal spectrum.

I think a very small portion of the absorbed energy is released.

 

[Andy Resnick]

Let me be clear on some basics. Please correct if I am wrong.
Energy is a measure on the ability to do work on matter. Matter can do work on other matter, which is kinetic energy (eg; a moving object). EM wave can do work on matter, hence they have energy too. Heat is a just IR wave in the EM spectra.

I'm not sure these statements are wrong, exactly, but they are not exactly correct, either. Energy is much more than work. Kinetic energy is just one form of work, and not the only form of energy that pertains to matter: latent heats, for example. It's better to simply think of material and non-material fluxes, flows, and diffusion. Energy and momentum can diffuse and flow just as smoke particles can.

In my understanding, the connection between EM wave to kinetic energy comes as follows:
An EM wave hits an atom; cause it to do work on the atom (causing the electrons to move; EM energy to kinetic energy conversion). Then the electron emits another EM energy which is Infrared ray, which we call heat.

My question is what property of the matter/electron caused it to emit waves in the IR frequency. Why not in some other frequency?

Again, the problem is that your model is too simple. Infrared transitions are usually vibronic and involve the entire molecule. Individual electrons have energy transitions in the visible, UV and higher energies.

So, your question is really, why does an excited electron not emit the same energy that it abosrbed? And the answer is, as you allude to, when a single electron absorbs a high energy photon and gets excited, that energy can flow into many excitation modes of the whole molecule via fast relaxation processes. Those modes are each of lower energy than the initial excitation, and so the visible energy cascades down into IR.

 

[jambaugh]

I think a very small portion of the absorbed energy is released.

It may be a matter of semantics. The object will warm until it reaches an equilibrium temperature at which the thermal emission equals the energy being absorbed. (Or it releases some of that energy convectively and diffusively).

 

[DaveC426913]

As the person who spurred the "what is heat" sub-discussion, I'd like to ask that we return to the OP's question. I think the answers are getting a little too academic to really be helping the OP. So:

"Why is kinetic energy == heat , for every substance?"

 

[Andy Resnick]

As the person who spurred the "what is heat" sub-discussion, I'd like to ask that we return to the OP's question. I think the answers are getting a little too academic to really be helping the OP. So:

"Why is kinetic energy == heat , for every substance?"

It's not. The confusion is between temperature and heat.

Temperature represents a thermodynamic state of a system, and in many simplified systems a unique temperature can be assigned based on the distribution of kinetic energy amongst the individual elements. Temperature is an equilibrium concept- a system not in thermodynamic equilibrium may not have a unique temperature.

Heat is associated with a thermodynamic process, the transition of a system from one state to another. The specific heat of a material represents the amount of additional energy required to raise the temperature by a certain amount.