Why do dark objects feel hotter than reflective objects?

[fog37]

Dear Forum,

An object, like a metal, that is highly reflecting (white body) is very different from a very emissive object (black body).
An object that closely behaves like a black body absorbs all the incident radiation. But it also "emit" all the absorbed radiation (over a range of wavelengths). Emission implies that the object has first absorbed the radiation. An ideal white body reflects instead all the incident radiation without absorbing it. Is this correct? Reflection does not imply that the process of absorption takes place.

At the end of the day, it seems that if 200J of energy are incident on a perfect black body or a perfect white body, all that energy will eventually be returned to the environment. That means that an object that does not have an internal source of energy eventually gives back all the energy it receives (with one mechanism or the other), correct?

I am sure there are some differences. MAybe in the case of a perfectly reflecting object all the reflected energy has exactly the same spectrum as the incident energy. Maybe that does not happen in the case of a black body. However, I think I read that the absorption spectrum of a black body has the same shape as the emission spectrum of the blackbody...

So why does a dark object feel hotter than a reflective object? Via conduction, we receive energy from the hot dark object. But we should receive energy also from the reflective object and feel it hot...why not? Maybe the feeling of hot and cold only depends on touching (conduction) and reflection does not involve conduction. Reflection is only about radiation.

To keep a liquid hot inside a bottle, should the bottle inside walls be dark or light in color? Eventually, all the heat emitted by the liquid will be either reflected or emitted back into the liquid...

thanks,
fog37

 

[DrClaude]

Emission implies that the object has first absorbed the radiation.

Not necessarily. The emission of a blackbody depends on temperature, and there is no reason to think that the heating of a blackbody can only be achieved through radiation.

At the end of the day, it seems that if 200J of energy are incident on a perfect black body or a perfect white body, all that energy will eventually be returned to the environment. That means that an object that does not have an internal source of energy eventually gives back all the energy it receives (with one mechanism or the other), correct?

I'm not sure what you mean by "all that energy will eventually be returned to the environment." The final situation will always be thermal equilibrium.

I am sure there are some differences. MAybe in the case of a perfectly reflecting object all the reflected energy has exactly the same spectrum as the incident energy. Maybe that does not happen in the case of a black body. However, I think I read that the absorption spectrum of a black body has the same shape as the emission spectrum of the blackbody...

A blackbody has an identical spectrum in absorption and emission. This is known as Kirchhoff's law.

So why does a dark object feel hotter than a reflective object?

I guess you're talking about an object that has been left in the sun. The answer to your question is that because it is hotter. This is not an equilibrium situation.

To keep a liquid hot inside a bottle, should the bottle inside walls be dark or light in color?

They should be as reflective as possible: http://en.wikipedia.org/wiki/Vacuum_flask

Eventually, all the heat emitted by the liquid will be either reflected or emitted back into the liquid...

What about the other side of the container?

 

[fog37]

Thanks DrClaude.

I am considering when the only energy that gets absorbed is radiation.

I am considering the situation of two objects with identical shape: a piece of metal and a piece of wood exposed to the sun for a long enough amount of time. I think they will reach a very similar temperature which is also very similar to the air temperature. However, when we touch the metal it feel hotter than the wood even if they have the same T: different heat conductivity so heat flows faster to the human body than in the case of wood.

If the metal was instead compared to a dark piece of plastic, the plastic would be more like a blackbody and absor more energy than the metal which would instead reflect it. Regardless, both the plastic and the metal would reject the the same amount of incident energy into the air.

thanks,
fog37

 

[dauto]

Yes, they will both eventually reach thermal equilibrium at which point the amount of energy coming in matches the amount of energy going out (whether by reflection or emission). But their temperatures will be different.

 

[DrClaude]

Yes, they will both eventually reach thermal equilibrium at which point the amount of energy coming in matches the amount of energy going out (whether by reflection or emission). But their temperatures will be different.

Sorry, but this is a contradiction in terms. Thermal equilibrium means that the objects will have the same temperature, which will not be the case here. It is a steady state situation, not equilibrium.

I am considering the situation of two objects with identical shape: a piece of metal and a piece of wood exposed to the sun for a long enough amount of time. I think they will reach a very similar temperature which is also very similar to the air temperature.

As I just mentionned, the temperature will not be the same, and far from ambient temperature. The system is not isolated, but receives energy from an external source, namely the sun. Each object will reach a steady state, where the energy gained from the sun will be exactly compensated by radiative losses and heat transfer to the air, with a different temperature for each.

However, when we touch the metal it feel hotter than the wood even if they have the same T: different heat conductivity so heat flows faster to the human body than in the case of wood.

There is an additional complexity here due to the difference in materials. Indeed, for the same temperature, a piece of wood and a piece of metal will not feel the same. But that's a human experience. Better stick to what is measured by a thermometer.

A better system to think about would be two objects made of the same material, one a blackbody, the other painted in white.

If the metal was instead compared to a dark piece of plastic, the plastic would be more like a blackbody and absor more energy than the metal which would instead reflect it. Regardless, both the plastic and the metal would reject the the same amount of incident energy into the air.

It depends what you mean by "rejecting energy." My guess is that you are thinking about the case where you have two objects of identical shape, one a blackbody and the other not, illuminated by the same source (the sun). In that sense, at steady state, the amount of radiative energy per unit of time hitting both is the same, and thefore by conservation of energy that radiative energy must be equal to the energy emitted by each object + the heat transferred to the air + the radiative energy reflected (in the case of the object that is not a blackbody).

 

[dauto]

Potatoes Potahtoes. A steady state is a type of equilibrium.

 

[fog37]

Sorry, but this is a contradiction in terms. Thermal equilibrium means that the objects will have the same temperature, which will not be the case here. It is a steady state situation, not equilibrium.


As I just mentionned, the temperature will not be the same, and far from ambient temperature. The system is not isolated, but receives energy from an external source, namely the sun. Each object will reach a steady state, where the energy gained from the sun will be exactly compensated by radiative losses and heat transfer to the air, with a different temperature for each.


There is an additional complexity here due to the difference in materials. Indeed, for the same temperature, a piece of wood and a piece of metal will not feel the same. But that's a human experience. Better stick to what is measured by a thermometer.

A better system to think about would be two objects made of the same material, one a blackbody, the other painted in white.


It depends what you mean by "rejecting energy." My guess is that you are thinking about the case where you have two objects of identical shape, one a blackbody and the other not, illuminated by the same source (the sun). In that sense, at steady state, the amount of radiative energy per unit of time hitting both is the same, and thefore by conservation of energy that radiative energy must be equal to the energy emitted by each object + the heat transferred to the air + the radiative energy reflected (in the case of the object that is not a blackbody).

the heat transferred to the air is also emitted energy. It only refers to EM radiation of very long wavelength, correct? That is all heat is.

Ok, so, if a metal and a piece of wood stay in the sun for 10 hrs, you say they will not reach the same (or close) temperature. Why? Is it because they have a different specific heat with the metal having a lower specific heat?

I found this information interesting: http://boards.straightdope.com/sdmb/showthread.php?t=143186

 

[DrClaude]

Potatoes Potahtoes. A steady state is a type of equilibrium.

Thermal equilibrium has a very specific meaning, and can be seen as the fundamental principle that allows us to have thermometers, and therefore talk about temperature (0th law of thermodynamics). Thermal equilibrium between two objects requires them to have the same temperature, otherwise we couldn't even define what temperature is.

I do have a tendency to be pedantic, but this goes beyond that. We are talking here about terms with specific scientific meaning, and I think that proper thinking only come through proper usage.

 

[DrClaude]

the heat transferred to the air is also emitted energy. It only refers to EM radiation of very long wavelength, correct? That is all heat is.

No. There are three ways in which heat can be transferred: conduction, convection, and radiation. Emission of IR radiation is thus only one of the possibilities.

Ok, so, if a metal and a piece of wood stay in the sun for 10 hrs, you say they will not reach the same (or close) temperature. Why? Is it because they have a different specific heat with the metal having a lower specific heat?

No. It's mostly due to the rate of absorption of heat from the sun. Again, forget about different materials, but htink of identical objects differing inly in their color. Fun experiment: on a sunny day, go to a parking lot, or better yet a car dealership. But your hand on similar cars but one light in color, the other dark. You'll feel the difference!

 

[fog37]

Thanks DrClaude. I have done that experiment and the dark car is surely hotter. The dark car has "absorbed" a lot of the incident EM radiation while the light car has mostly reflected the incident IR radiation.

In a room, during the day, away from direct sun exposure, two identically shaped objects, one made of wood and one made of metal will "feel" different. Their temperature will be the same but the higher heat conductivity of the metal will make it feel cooler even if they have the same T. This is because the flow of energy from our hand to the metal is large and we perceive that flow as the sensation of coolness...

 

[DrClaude]

You can check out this post

https://www.physicsforums.com/showpost.php?p=4752798&postcount=17

and the corresponding thread for more information about the human sensation of heat.