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House Color and Heat

  1. Jun 5, 2008 #1
    It is often claimed that in a hot climate, it is better to paint one's house white because it will reflect more radiation, while black paint will absorb more radiation. It seems clear that other things equal, this is correct. But I have two questions regarding this.

    1) Which is more important in heating a house through radiation: the visible properties (tendency to reflect or absorb visible light from the hot sun) or IR properties (tendency to reflect or absorb infrared from surrounding objects at 300 K or so) of the surface?

    2) Is there a strong connection between an object's visible properties and its IR properties? In other words, could something mostly reflect visible while absorbing IR or the reverse? Can we reasonably infer an object's IR properties by looking at it?
     
  2. jcsd
  3. Jun 5, 2008 #2

    mgb_phys

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    The peak of the suns power output is in the visible so it's important to reflect visible.

    It's very common and useful to have a strong difference between the visible and infrared reflectivity. Most glass used for houses and offices is coated to transmit visible, so you can see through it, but reflect infrared, to keep the heat in the room.
    A lot of dyes are designed to absorb visible light, so they appear black, but reflect UV and infrared, so they don't absorb the extra energy and break down.
     
  4. Jun 5, 2008 #3

    DaveC426913

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    Well, that would be a big mistake to make when observing a stove element.
     
  5. Jun 5, 2008 #4

    mgb_phys

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    Only with your puny human eyes - to say, a pit viper it would be obvious!
     
  6. Jun 5, 2008 #5
    Thanks much. So your answer to my second question is no, we can infer little about an object's IR properties from its visible properties.

    But I'm not sure if you answered my first question. Radiation needn't come directly from the sun, so how can we be sure that most of the energy is visible?
     
  7. Jun 5, 2008 #6

    mgb_phys

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    Only for a certain type of surface (called a black body) can you determine it's output at different wavelengths from just it's temperature. The sun is a very good blackbody and so we can calculate it's power output at different wavelengths from only knowing it's temperature.

    You can calculate the power incident on the house from the sun and the surrounding houses but since the power output increases with Temperature^4, the sun tends to dominate!
     
  8. Jun 6, 2008 #7
    Temperature Doesn't Explain Everything

    The fact that the sun is hot is only part of the solution. There are stars much hotter than the sun, but we don't get much energy from them because they are so far away. I've read somewhere that the mean intensity of sunlight at the surface of the earth is in the neighborhood of 100 W/m^2, most of which would be in the visible range. But what is the approximate mean intensity of IR light?
     
  9. Jun 6, 2008 #8

    Borek

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    Image from wiki solar radiance article:

    [​IMG]

    Can't include image in the post, so you have to click the link. Looks like at the sea level about 1/3 of energy is in the visible range (area was eye-integrated, so don't quote me). That means around 2/3 in the IR - but different coatings will be able to reflect IR in in different ranges, so things get more complicated.
     
  10. Jun 6, 2008 #9

    DaveC426913

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    :rolleyes:

    Yeah, but my point is made. Objects can look very different in visible light than in IR.

    Of course, I cheated. My example is that of a body that is emitting EM. The poster was asking about objects that reflect EM.
     
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