What is a blackbody? How do real bodies differ from blackbodies? Consider heat t

Question

What is a blackbody? How do real bodies differ from blackbodies? Consider heat transfer between two identical hot solid bodies and their environments. The first solid is immersed in a large container filled with water, while the second one is allowed to cool naturally in the air. For which solid is the lumped system analysis more likely to be applicable? Explain The outer surface of a spacecraft in space has an emissivity of 0.8 and a solar absorptivity of 0.3. If solar radiation is incident on the spacecraft at a rate of 950 W/m^2, determine the surface temperature of the spacecraft's affected area when the radiation emitted equals the solar energy absorbed. The roof of a house consists of a 3-cm-thick concrete slab (k = 2 W/m. degree C) that is 15 m wide and 20 m long. The convection heat transfer coefficients on the inner and outer surfaces of the roof are 5 and 12 W/m^2. degree C, respectively. On a winter night, the ambient air is reported to be at 10 degree C. The indoor temperature is maintained at 20 degree C. Determine the rate of heat transfer through the roof, and the inner surface temperature of the roof. If the house is heated by a furnace burning natural gas with an efficiency of 80 percent, and the price of natural gas is $0.60/therm (1 therm = 105500 kJ of energy content), determine the money lost through the roof that night during a 12-h period.

Explanation / Answer

Question 1 (a)

An object is a "Blackbody" if the radiation it emits into space originates completely from its temperature. This means the radiation produced by the object comes from light waves mixing it up with the jiggling motions of all the innumerable atoms that make up the object. Inside a blackbody, radiation can not travel very far before it is absorbed by a jiggling atom.  It is then quickly re-emitted, travels a short distance and then gets absorbed again by another atom.  This happens zillions of times so there is a constant interplay between the matter and the radiation bouncing around in a blackbody. The one-to-one relation between the amount of jiggling heat motion of atoms and the spectral signature they produce make blackbodies unique, distinctive and of primary importance. Blackbody spectra do not depend on an object's chemical composition, its size or its age.

Keep in mind, Blackbodies do not have to be black. They can be blue or red or yellow. Actually, anything coloured black absorbs all the wavelengths (colors) of light that fall onto it. Blackbodies do this as well and that is why physicists came up with the name "blackbody".

Anything which has heat and is dense enough will emit as a blackbody.  That means you, the chair you’re sitting on and the Earth on which the chair rests are all blackbodies.

Every blackbody emits light with an easily identified pattern, its "spectral" signature (also called a spectral energy distribution or more specifically the blackbody curve). The blackbody curve is the particular way the total light emitted by a blackbody varies with its frequency. The number of red photons, the number of green photons, the number of infrared and ultraviolet photons are all exactly specified by the blackbody curve. Now here is the killer point - the exact form of the curve depends only on the object's temperature. Every blackbody at 2000 degrees emits light with exactly the same curve. The spectral signature of a 2000 degree iron bar in a blast furnace is identical to a 2000 degree star a trillion, trillion miles away in deep space. That is what makes blackbodies so useful and that is why the color of a star is also a measure of its temperature.

Now, Difference between Black bodies and Real Bodies;

A perfect blackbody absorbs all wavelengths of incoming electromagnetic radiation perfectly and none of it is reflected. A perfect blackbody at a given temperature emits radiation in accordance with Planck's Law.

Any real blackbody is imperfect in both of these respects; it does not absorb all incoming radiation, and it does not have an emission spectrum that perfectly matches Planck's law.

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