Justify that a good absorber is also a good emitter for radiation heat transfer.
Answers
Step-by-step explanation:
A Good Absorber is a Good Emitter
According to the Stefan-Boltzmann law, the energy radiated by a blackbody radiator per second per unit area is proportional to the fourth power of the absolute temperature and is given by
For hot objects other than ideal radiators, the law is expressed in the form:
where e is the emissivity of the object (e = 1 for ideal radiator). If the hot object is radiating energy to its cooler surroundings at temperature Tc, the net radiation loss rate takes the form
In this relationship the term with Tc represents the energy absorbed from the environment. This expression explicitly assumes that the same coefficient e applies to both the emission into the environment and the absorption from the environment. That is, a good emitter is a good absorber and vice versa; the same coefficient can be used to characterize both processes. Why is that true?
Perhaps the most fundamental conceptual way to approach this question is to observe that a hot object placed in a room must ultimately come to thermal equilibrium with the room. The hot object will initially emit more energy into the room than it absorbs from the room, but that will cause the temperature of the room to rise and the temperature of the object to drop. But when they reach the same temperature, we can conclude that the amount of energy absorbed on average is exactly the same as the energy emitted. That is, the expression above for net energy radiated to the environment must give us zero when T=Tc.
The above argument is based upon the Second Law of Thermodynamics in the form that states that heat will not spontaneously flow from a cold object to a hot object. If the absorption coefficient were higher than the emission coefficient for the object, then it could absorb net energy from the room even when its temperature were higher than the room.
But suppose you wanted to argue that a good absorber must be a good emitter based on the microscopic processes involving the atoms in the surface of an object. Then it becomes quantum question and involves the following ideas:
1. All electromagnetic radiation can be considered to be quantized, existing as photons with energy given by the Planck hypothesis, E=hf.
2. In order for a solid (or any matter, but I am assuming we are talking about solids) to absorb a photon of given energy hf , it must have a pair of energy levels separated by that amount of energy hf, so that the photon elevates the system from the lower member of the pair to the upper.
3. For visible light or near visible, then energy level pairs involved in most absorption are electron energy levels, so that when you absorb a green photon of photon energy 2.2 eV, you are causing an electron very near the surface of the solid to jump upward 2.2 eV. It can't do it unless there is a level at 2.2 eV up to receive it.
Answer:
Prior to Kirchhoff's studies, it was known that for total heat radiation, the ratio of emissive power to absorptive ratio was the same for all bodies emitting and absorbing thermal radiation in thermodynamic equilibrium. This means that a good absorber is a good emitter. Naturally, a good reflector is a poor absorber
Step-by-step explanation:
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