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excuse me if this is too basic question. why is it that black materials BOTH "absorb" and "radiate" more heat ( energy ) and white materials do it less. i mean is there a rule that any material absorbs and radiate heat in the same amount and can't absorb more (like black materials) and radiate less ( like white materials) at the same time. i think i have seen such thing when i was in college (8 years ago and i don't have major in physics).

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collinsmark
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You might want to ask yourself these questions:

• Absorption: What property of a black material makes it black? (In terms of its interaction with incident light.)
• Radiation: Given the above, can you explain how a black material might tend to be at a higher temperature than a non-black (e.g., white) material, all else being the same? [Hint: assume that both materials are in a state of equilibrium.]
Edit: And regarding your second statement, if a material's temperature is constant (i.e., not changing), what does that tell you about the net heat flow into or out of the material, given conservation of energy?

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black material is black becuase it absorbs more energy and reflect less.
what i think is :
the amount of radiation depends on temperature and material.
the amount of absorption depends on the amount of radiation the object is exposed to and the material
if we have two objects white(W) and black(B) with same temperature not exposed to any radiation. B's temperature would fall more quickly because it radiates more energy. (compared to W).
but if they are exposed to radiative energy, B would absorb more energy (so higher
temperature).
but my question is can a material radiate less( like W in the first situation) and absorb more ( like B in the second situation)? i think it can't, but is there a general law here that makes it obvious?
where am i wrong?

collinsmark
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black material is black becuase it absorbs more energy and reflect less.
'Sounds reasonable. :)
what i think is :
the amount of radiation depends on temperature and material.
Correct.
the amount of absorption depends on the amount of radiation the object is exposed to and the material
Correct, if I'm interpreting that right.
if we have two objects white(W) and black(B) with same temperature not exposed to any radiation. B's temperature would fall more quickly because it radiates more energy. (compared to W).
You might want to rethink that. Assuming that both materials are each isolated in darkness (and I mean an absolute-zero darkness in a vacuum), and they have identical heat capacities, identical heat conduction properties, and are both at the same temperature, then I don't think that statement is true.
but if they are exposed to radiative energy, B would absorb more energy (so higher
temperature).
That part I agree with. :)
but my question is can a material radiate less( like W in the first situation) and absorb more ( like B in the second situation)?
It can if its temperature is increasing.
i think it can't, but is there a general law here that makes it obvious?
where am i wrong?
You might look up Fourier's law (although that applies more to heat conduction in terms of space rather than time). I think the application of conservation of energy (first law of thermodynamics) is sufficient though.

(Edit: I think Stefan–Boltzmann law might be what you are looking for, if you want to get more detailed than just conservation of energy.)

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collinsmark
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;)By the way, radiated light not only involves energy, but also imparts momentum. This can be exploited in what is called a Crookes radiometer.

We've established that black objects will get hotter than white objects (all else being equal) when exposed to light. So when exposed to light, we can conclude that the black side of the Crookes radiometer vane is slightly hotter than the white side.

Now, if we put the thing in the darkness, what happens? ;)

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jbriggs444
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Sina89 is correct. A black object will at a given temperature will tend to radiate more strongly than a white object at that same temperature.

The second law of thermodynamics demands this. Suppose that you have a black body and a white body each in equilibrium with a shared environment that is, itself, in equilibrium. If the two bodies are not at the same temperature then you have a second law violation.

Both bodies are radiating and absorbing energy continuously. The black body, being black, absorbs more strongly than the white. If it fails to radiate more strongly then it will not be at equilibrium with its environment. It follows that it must radiate more strongly than the white.

jbriggs444
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;)By the way, radiated light not only involves energy, but also imparts momentum. This can be exploited in what is called a Crookes radiometer.
Note that the toy in your animation rotates backwards from the direction one would expect based on the momentum of light.

[And note that the environment in which a Crookes radiometer is placed must be decidedly non-uniform in order for it to function]

collinsmark
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Sina89 is correct. A black object will at a given temperature will tend to radiate more strongly than a white object at that same temperature.
No, the black object will radiate more only because it is at a higher temperature.

Consider initially that both objects are in darkness and are at equal temperature. At some point they are exposed to light/radiation. The black object absorbs more radiation than the white object, thus its temperature increases to a greater extent. Eventually each object will reach an equilibrium with the environment. Each object will radiate the exact amount of energy that it absorbs (the black object radiating/absorbing more than the white object, of course -- my point being that a given object radiates the same amount that that given object absorbs). But the black object remains hotter than the white object because it has absorbed more energy when it was initially exposed to the light/radiation. [Edit: and thus the black object radiates more power than the white object because it is hotter.]

The second law of thermodynamics demands this. Suppose that you have a black body and a white body each in equilibrium with a shared environment that is, itself, in equilibrium. If the two bodies are not at the same temperature then you have a second law violation.
That's the case only when the radiation/light source is removed.

But when the radiation/light source is present, heat transfer is taking place, and when in equilibrium (with the light/radiation source present), the black object is hotter.

I'm sure you know that wearing black clothes on a hot summer day in the sunshine makes you hotter than a white clothes. It's the same idea here.

Both bodies are radiating and absorbing energy continuously. The black body, being black, absorbs more strongly than the white. If it fails to radiate more strongly then it will not be at equilibrium with its environment. It follows that it must radiate more strongly than the white.
That's absolutely correct. When in equilibrium each object radiates exactly the same amount of energy that it absorbs. But the black object is still hotter when the light/radiation source is present.

Note that the toy in your animation rotates backwards from the direction one would expect based on the momentum of light.
There was a decade of debate on this after the invention of the device. It depends on how you approach the problem, I suppose.

[And note that the environment in which a Crookes radiometer is placed must be decidedly non-uniform in order for it to function]
Could you elaborate on that?

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jbriggs444
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No, the black object will radiate more only because it is at a higher temperature.
That assertion is incorrect.

It is true that a black object will equilibriate with its environment more rapidly than a white and because of this it will initially be at a higher temperature when exposed to light. But it will attain the same temperature in the long run (assuming a uniform environment). At that equilibrium temperature, it will be radiating more than the white object because of its color, not because of its temperature.

Note that the toy in your animation rotates backwards from the direction one would expect based on the momentum of light.
There was a decade of debate on this after the invention of the device. It depends on how you approach the problem, I suppose.
There has been debate on the details of the mechanism, but there is no debate on the fact that a typical toy radiometer does not rotate in the "right" direction. It is not the momentum of the light, but the temperature difference between the two faces of each vane that is relevant.

[And note that the environment in which a Crookes radiometer is placed must be decidedly non-uniform in order for it to function]
Could you elaborate on that?
If the radiometer is exposed to sunlight from all directions and if it is at the associated equilibrium temperature (the temperature of the surface of the sun) it will fail to operate.

No, the black object will radiate more only because it is at a higher temperature.

Consider initially that both objects are in darkness and are at equal temperature. At some point they are exposed to light/radiation. The black object absorbs more radiation than the white object, thus its temperature increases to a greater extent. Eventually each object will reach an equilibrium with the environment. Each object will radiate the exact amount of energy that it absorbs (the black object radiating/absorbing more than the white object, of course -- my point being that a given object radiates the same amount that that given object absorbs). But the black object remains hotter than the white object because it has absorbed more energy when it was initially exposed to the light/radiation. [Edit: and thus the black object radiates more power than the white object because it is hotter.]
I have to clear something here that
in the second scenario when talking about radiating, im supposing that the temperature of two objects are equal.
actually i remember some sentence that i have heared in school "Black objects both absorb and radiate heat more (stronger) that white objects "
and i'm not talking about conservation of energy in two sequential experiments ( absorption then radiation),
but i'm talking about "rate" of radiative transfer in two seperate experiments.
1. energy absorption of W vs B : B absorbs more energy in a certain time interval
2. energy radiation of W vs B (everything including temperature is equal) : B radiates more energy in a certain time interval
thank you.

collinsmark
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That assertion is incorrect.

It is true that a black object will equilibriate with its environment more rapidly than a white and because of this it will initially be at a higher temperature when exposed to light. But it will attain the same temperature in the long run (assuming a uniform environment). At that equilibrium temperature, it will be radiating more than the white object because of its color, not because of its temperature.
[Edit: Okay, I think I see what you mean now. What you say is true in situations lacking in some sort of cooling mechanism, such as the objects transferring heat to the surrounding air (like why we are not at the temperature of the sun just because we take an afternoon walk). But in the presence of such mechanisms, black objects become hotter when when exposed to light/radiation such as sunlight.]

An energy source (the light/radiation source) is present here, and that changes things.

The filament of incandescent light source requires an energy source for it to work. After the light bulb has been turned on for some time, its temperature stabilizes and the light bulb emits as much energy as is given to it (overall). It is, in a sense, in a state of equilibrium. Its temperature is constant.

The room in which the light bulb is in might initially heat up slightly, but assuming the room is not perfectly insulated, the ambient temperature of the room will eventually reach a state of equilibrium too, and the ambient temperature of the room will remain constant. So both the light bulb and the room are in a state of equilibrium.

But it's altogether different to claim that the temperature of the light bulb filament (when still turned on) is equal to the ambient temperature of the room. I claim that the filament is hotter.

----

There exist solar water heaters that operate by piping water through pipes exposed to the sunshine. The color of the pipes matter. Given the choice between black and white pipes, the black pipes heat the water more.

There's plenty of experimental evidence for this: paint two objects, one black and one white (all else being equal except for the paint color). Put them both in the bright sunshine for awhile and measure their temperature. The black object will be hotter.

There has been debate on the details of the mechanism, but there is no debate on the fact that a typical toy radiometer does not rotate in the "right" direction. It is not the momentum of the light, but the temperature difference between the two faces of each vane that is relevant.
It depends on what you mean by "right." Approaching the problem with the wrong principles of how light imparts momentum (i.e., wrong choice of "right") can lead to conclusions that do not conform to nature and experiment.
If the radiometer is exposed to sunlight from all directions and if it is at the associated equilibrium temperature (the temperature of the surface of the sun) it will fail to operate.
Can you verify that with a credible reference?

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collinsmark
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I have to clear something here that
in the second scenario when talking about radiating, im supposing that the temperature of two objects are equal.
actually i remember some sentence that i have heared in school "Black objects both absorb and radiate heat more (stronger) that white objects "
and i'm not talking about conservation of energy in two sequential experiments ( absorption then radiation),
but i'm talking about "rate" of radiative transfer in two seperate experiments.
1. energy absorption of W vs B : B absorbs more energy in a certain time interval
2. energy radiation of W vs B (everything including temperature is equal) : B radiates more energy in a certain time interval
thank you.
That fine, the rate of change of energy is called "power." I'll try to keep things in terms of power when appropriate. :)

When the black object is at the same temperature as the white object, they radiate the same power. It's as simple as that. [Edit: nevermind, that's not right. Sorry.]

When exposed to a light source, the black object will absorb more power (and integrating that, more energy initially) than the white object, causing it to initially increase its temperature, and then will radiate more power than the white object because it has become hotter.

By the way, there is an interesting and related phenomenon, that if you allow the two objects to cool in the absence of a light source (put them in the freezer with the door almost closed so that the freezer light doesn't turn on), the black object will cool quicker. And that's the reverse situation that demonstrates that the black object not only absorbs radiated power more easily, it also emits radiated power more easily. [After some times passes, both objects will eventually reach the same temperature though, so long as the freezer remains dark.]

In summary, when in the dark, in the absence of a radiated power source (and after allowing them to reach equilibrium), black and white objects have the same temperature, and they both radiate the same power. But expose them to a radiated power source, and the black object will absorb more of this power, increase its temperature to a greater extent, and radiate more power as a consequence (after reaching equilibrium).

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jbriggs444
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That fine, the rate of change of energy is called "power." I'll try to keep things in terms of power when appropriate. :)

When the black object is at the same temperature as the white object, they radiate the same power. It's as simple as that.
That statement is utterly incorrect.

By the way, there is an interesting and related phenomenon, that if you allow the two objects to cool in the absence of a light source (put them in the freezer with the door almost closed so that the freezer light doesn't turn on), the black object will cool quicker.
BINGO!! You have just admitted that the black object radiates more power.

The second law of thermodynamics demands this. Suppose that you have a black body and a white body each in equilibrium with a shared environment that is, itself, in equilibrium. If the two bodies are not at the same temperature then you have a second law violation.

Both bodies are radiating and absorbing energy continuously. The black body, being black, absorbs more strongly than the white. If it fails to radiate more strongly then it will not be at equilibrium with its environment. It follows that it must radiate more strongly than the white.
:w yes this is second law of thermodynamics
wow i didn't have that in mind
second law is not always straightforward

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Bandersnatch
@collinsmark: the wiki page on Crookes radiometer provides ample discussion on the correct and incorrect reasons for the motion. It's got very little to do with actual radiation itself, and is all about the (radiation-induced, imparted through the vanes) temperature differences in the low pressure gas in the bulb. For one, the rotation stops if the bulb is evacuated well enough.

Here:
http://math.ucr.edu/home/baez/physics/General/LightMill/light-mill.html
you can find a non-rigorous (about wiki-level) analysis, with sources listed at the bottom.

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collinsmark
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@collinsmark: the wiki page on Crookes radiometer provides ample discussion on the correct and incorrect reasons for the motion. It's got very little to do with actual radiation itself, and is all about the (radiation-induced, imparted through the vanes) temperature differences in the low pressure gas in the bulb. For one, the rotation stops if the bulb is evacuated well enough.

Here:
http://math.ucr.edu/home/baez/physics/General/LightMill/light-mill.html
you can find a non-rigorous (about wiki-level) analysis, with sources listed at the bottom.
I stand corrected.

[Forgive me for the confusion.]

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collinsmark
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That statement is utterly incorrect.
Yes, that was my mistake. You are correct.

thanks all.