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Good absorber = good emitter?

  1. Feb 23, 2015 #1
    Hello Forum,

    If a black object, being black, is a very good absorber or sunlight and at the same time a great emitter, why wouldn't it be cool? The total emission would seem to be the mechanism to remain cool. Maybe it works this way: if 100J of sunlight are absorbed, only a small % would be re-emitted? But then it would not be a good emitter...

    A white object rejects all energy at all wavelengths from the very beginning and remains cool.

    thanks,
    fog37
     
  2. jcsd
  3. Feb 23, 2015 #2

    Bystander

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    Think about the temperature dependence of the emission, T4, and balance power in and power out.
     
  4. Feb 23, 2015 #3
    Thanks.
    I see how after a transient phase, the objects reaches thermal steady state where the absorbed energy = emitted energy and the temperature stabilizes.

    Bystander, you bring the attention to the 4th power dependence of the temperature. So a white object stabilizes to its final temperature T_f earlier and T_f would be much lower than T_f for a black object. The black object takes longer to reach T_f and during that time it absorbs a lot of energy and warms up before it can emit an equal amount of energy out....

    Still, I am not sure how the 4th power dependency plays a role...any other hint?

    thanks,
    fog37
     
  5. Feb 23, 2015 #4

    Bystander

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    If both have the same heat capacity and the same areas exposed to radiative heat exchange? Maybe.

    Expose a couple flat rocks to the sun (~1 kW/m2). At steady state, each radiates at the same rate it absorbs energy, and that rate is a function of temperature (Stefan-Boltzmann), dq/dt = (5.67 x 10-8W/m2) εT4, where ε is a function of temperature and of the material (black rock, white rock for the case you've described).

    At what temperature shall we evaluate ε for our absorbing-radiating surfaces? We'll certainly use the temperatures of the surfaces for emission, and get ~ 450 W/m2 for black body (ε = 1) at 300 K, and expect the black rock to radiate more than the white rock (ε < 1).

    What about absorbtion? Source temperature (if it's a black body source). The sun is a nearly black-body, not perfect, and ε for most materials at 5600 K is approaching 1. "But the white rock is reflecting white light?" It's not reflecting all the white (visible) light. Does it have to get warmer than the black rock to emit the heat it absorbs at steady state?

    Bottom line? Black rock vs. white rock in the sun? Which is cooler? I have not done the experiment --- my curiosity is aroused.

    Reflecting surfaces? Burned your arm on chrome trim on automobiles? Literally, it will blister you. How? It's a very good reflector, blindingly bright reflections when driving in daytime. How does it get so hot? Bright metal surfaces have very low emissivities at ambient temperatures, ε = 0.01 to 0.1, ordinarily a few hundredths, and they simply do not radiate any of the energy they absorb, and there's plenty of energy in the solar spectrum outside the visible range for them to absorb. The only cooling automotive chrome trim gets is by conduction and convection to the rest of the auto body and to the air around it.

    Planck's radiation law, Wien displacement, and Stefan-Boltzmann are good places to start reading. Emissivities? Rohsenow & Hartnett for temperature dependent values for a variety of surfaces, Eshbach's Handbook and CRC Handbook, for tabulations, Perry's maybe. Emissivities are functions not only of temperature, but surface preparation and history, and you will find disparities in the tabulated values.

    Temperature dependence of emissivity is the biggest stumbling block to "blind" use of Stefan-Boltzmann for calculation of radiation from black, gray, and tattle-tale gray bodies; measured values of emissivities are the second largest problem.

    Sorry to be answering your question about radiative heat transfer with far bigger questions. Some of this should help you at least understand what's going on with some problems/systems.
     
  6. Mar 1, 2015 #5
    Thanks Bystander.

    Your answers are very helpful. I know that two different objects can have the same temperature T but, when touched, feel very different depending on their thermal conductivity. The classic example is the bathroom tiles feeling colder than the rest of the object in the bathroom regardless of all the objects having the same T. Thermal energy flows quickly away from the human body toward the tiles. That rapid heat subtraction makes us judge the tiles as "colder" than the other objects (regardless of them having the same T). I am rather clear on how thermal conductivity. Metals are not black, in general, but if we touch them after they have been left in sunlight for many hours, we would get burned (due to the metal high thermal conductivity).

    What I am still confused is the difference between a perfect black body (emissivity =1) and an ideal white body (emissivity=0). In the transient phase, both objects would have their temperature increase if left in sunlight. The temperature of the blackbody would increase fast and reach a larger final T in comparison to the ideal white body. At steady state, both objects will not experience any increase in T (energy in= energy out).

    Maybe, after a long enough time, both objects will reach the same temperature? Consider a piece of wood and piece of metal left in sunlight for 10hrs.

    I know that if the light source is removed, the blackbody will cool down much faster than the white body....

    Best,
    fog37
     
  7. Mar 1, 2015 #6

    Bystander

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    "White body" is going to change T only if it is in contact with some other body (earth, air, whatever). An isolated white body isn't going to exchange any radiation with anything.
    At equilibrium with the radiation source (in a closed system) they'll be at the temperature of the radiant source.
     
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