MadSci Network: Physics

Re: Is a black object hotter then a white object even in a dark room.

Date: Fri Nov 10 10:04:58 2000
Posted By: Todd Jamison, Staff, Image Science, Observera, Inc.
Area of science: Physics
ID: 973719029.Ph

Believe it or not, the answer depends on a lot of little details in how you 
run the "experiment".  I would have to say that in the most general 
interpretation of the experiment you are describing, the two objects will 
end up at the same temperature.  Here is the idealized experiment to confirm 

Two objects of exactly the same material, one coated white and one coated 
black, are placed into a room that is completely shielded from ALL 
radiation.  Hypothesis:  Both plates will eventually reach the ambiant 
temperature of the room and stay that way.  

WHY?  The mechanism by which dark objects get warmer is a function of the 
amount of radiation that is absorbed by them.  In fact, objects are dark in 
color, BECAUSE they absorb radiation in the visible band (i.e., the bands 
that we can see with our eyes).  It also happens that many dark materials 
that absorb visible band radiation also absorb radiation in other bands (for 
example infrared, uv, xrays, microwaves).  The absorbed radiation is turned 
into thermal energy by "exciting" the atoms in the absorbing material, thus 
heating the object.  Lighter objects reflect visible radiation making them 
appear bright.  Many light objects also happen to reflect many other bands 
of radiation.  Because the radiation is reflected away, the material stays 
cool.  Almost all materials are a combination of "light" and "dark", which 
means that they both absorb some radiation in some bands and reflect some 
radiation in other bands.  

If you remove all sources of radiation, then eventually both objects will 
reach the ambient temperature of the room through conduction between the 
objects and the air.  Conduction is another form of heating that occurs when 
one molecule touches another molecule at a different energy level and they 
exchange some of the energy, like a pan on a heating element.  In this case, 
like an oven or a refrigerator, the air molecules at one temperature bump 
into the "plate" molecules at a different temperature causing them to 
eventually get to the same temperature if you wait long enough. 

THE OTHER SIDE OF THE STORY:  By coating the objects, you have changed the 
material makeup of the objects, so they are NOT THE SAME, and it is 
particularly true of the critical outer surface, which is where most 
absorption of radiation will occur.  Furthermore, it is difficult, or 
perhaps impossible to create a room completely devoid of radiation, since 
all objects that exist above absolute zero give off some radiation (black 
body radiation), including the air in the room and the plates.    

WILL THE DARK PLATE BE WARMEST?  Not necessarily.  First, the radiation 
bands that are absorbed depend upon the type of coating.  Thus a plate with 
a dark coating that absorbs mostly in the visible bands (making it dark to 
our eyes), but does not absorb well in other bands, may well absorb less 
energy and be cooler than a coating that reflects visible light, but absorbs 
microwave and infrared radiation.  Complicating this, there is the issue of 
"emission".  So called "black bodies" that absorb radiation well also give 
up their energy in the form of radiation better than "white bodies".  Thus, 
the "black" plate may cool (give up its thermal energy) faster than the 
"white" plate, since it is more like an ideal "black body".  Since cooling 
is an "infinitely" long process, the black plate MAY ALWAYS BE COOLER than 
the white one in the dark room.  

SO, WHATS THE ANSWER?  For most purposes in the real world, the "black" and 
"white" plates in the "dark" room will converge to the ambient temperature 
of the room.  From a strictly physics-based point of view, however, it is 
more complicated.  It would be difficult (or perhaps impossible) to create 
the idealized conditions needed to prove this and the results depend heavily 
upon the makeup of the objects and the surrounding room. 

Let me know if this answers your question.  I am curious to hear what you 
decide is "your" answer after reading this.

Yours in science, 
Todd Jamison

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