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I Specific Heat of a Blackbody

  1. Jun 4, 2018 #1

    I was thinking about how a blackbody (and any other type of body) eventually reaches a steady-state, constant and finite temperature once the absorbed energy is equal to the emitted energy. The specific heat of a substance indicates the temperature change causes by the absorption/emission of a certain amount of energy.

    That said, what would the specific heat of an ideal blackbody be?

  2. jcsd
  3. Jun 4, 2018 #2


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    Can be anything ! Depends on the material.
  4. Jun 5, 2018 #3
    I see. Thanks.

    I just see how certain materials, like a black painted object and a white painted object, reach their steady state temperatures, when left in sunlight, at different times. I think the black painted object reaches a higher steady state temperature faster and sooner than the white painted object (whose steady-state temperature is also lower). I thought that that speed would be somehow related to some specific heat property of the objects...
  5. Jun 5, 2018 #4


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    Blackbody has a specific meaning in physics (see below). It sounds like you are asking about objects painted black; not a blackbody.

    Such an object will reach an equilibrium temperature, depending on its size, shape, material, orientation toward the sun, and what it is in contact with. Materials with higher specific heat change temperature more slowly, but many factors are involved.

  6. Jun 5, 2018 #5
    Search "thermal diffusivity" and Biot number.
  7. Jun 5, 2018 #6


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    It is, but that isn't what you asked. You asked to equate specific heat and radiation (emissivity) when as you correctly note here, they are two separate variables in this problem.
  8. Jun 5, 2018 #7

    Well, materials having constant emissivity, absorptivity and reflectivity across the entire spectrum don't exist. All real objects are selective radiators/absorbers.

    I am trying to figure out how to determine which, between two real objects, will remain cooler under sunlight. I think it will first depend on the reflectivity of each object over the incident solar spectrum (300nm-2500nm). The object will the highest reflectivity in this spectral range will absorb very little solar radiation. But that is not enough.

    We need to consider the emissivity of each object in the far infrared (FIR) and the object with the largest emissivity in the FIR will be cooler and have a lower steady-state temperature.

    So, the coolest object would have a high reflectivity in the VIS and a high emissivity in the FIR. An object with a a high VIS reflectivity and low FIR emissivity will potentially get very hot. I am still not sure why a high emissivity in the FIR would lead to a lower final steady-state temperature than an object with low emissivity. Emissivity correlates on how much energy per unit time and unit area the object emits. I guess an object with high emissivity will reach steady-state sooner and at a lower temperature....
  9. Jun 6, 2018 #8


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    The radiation an object emits will ideally be a black body spectrum. The peak of that spectrum will depend on temperature and, for an object sitting out in the sun, the resulting temperature will give a peak in the FIR. If such an object has a high emissitivity in the FIR, it will radiate more strongly than an object with a low emissivity. It will lose heat faster. If it loses heat faster than it gains heat, its temperature will fall.

    All other things being equal, that means that an object with a high emissivity will equilibriate at a lower temperature where the rate of heat loss is reduced to become the same as the rate of heat gain.
  10. Jun 6, 2018 #9
    Thank you jbriggs444.

    I am think you about your statement "All other things being equal, that means that an object with a high emissivity will equilibriate at a lower temperature where the rate of heat loss is reduced to become the same as the rate of heat gain."

    So, the object gets progressively hotter as time goes by and eventually reaches a steady-state temperature. But your point is that if the emissivity is high that stable temperature will be lower. could you add another clarification about this point? I know that at steady state, energy in = energy out but I cannot articulate well why that condition would be reached at a lower temperature if the emissivity is high.

    So, while high emissivity implies high absorptivity in the same spectral region (the FIR in the case), the high absorptivity in the FIR does not imply the object will become hot since the solar spectrum does not have FIR radiation. What matters is the absorptivity in the spectral region of the incident solar spectrum.

    For an object to be cool in the sun, high reflectivity in the VIS and high emissivity in the FIR would be the requirements. Other combinations (low reflectivity in the VIS and high emissivity in the FIR, etc.) would not be as good at keeping the object cool.

  11. Jun 6, 2018 #10


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    In general, an object will warm up until it emits at the same rate that it absorbs. The hotter it gets, the more it emits until eventually an equilibrium is reached.

    If the object already emits enough at a low temperature to match what it absorbs then it will never get to a higher temperature.

    Or, looking at it the opposite way, if an object with a high emissivity is at a high temperature, it will be emitting too much. It will cool down toward a lower temperature equilibrium condition.
  12. Jun 6, 2018 #11
    Thank you! That was a great explanation.

    So, for comparison purposes, I would argue that an ideal blackbody, which has maximum emissivity of 1 at all wavelengths and max absorptivity of 1 at all wavelengths, absorbs maximally and emits maximally and would become hotter than a body which is a selective radiator/absorber and has low absorptivity in the VIS and emissivity 1 in the FIR. The selective radiator would be much cooler than the ideal blackbody.

    A metal object, which has high reflectivity (and low absorptivity) in the VIS but low emissivity in the FIR, would get much hotter than an ideal blackbody.

    A black painted object: black paint has high absorptivity in the VIS and high emissivity in the FIR and white paint has low absorptivity in the VIS and high emissivity in the FIR. Hence a white painted object will be cooler than a black painted object. What about a metal object compared to a black painted object? Which one should get hotter? I think the metal one would get hotter
  13. Jan 1, 2019 #12


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    This seems like an old blog
    but I would propose that in equilibrium all bodies (black or gray) in a constant radiation flux would reach the same temperature
    The black ones heat and cool at the highest possible rate, and the gray ones heat and cool at a lower rate, but the emissivity is on both sides of the heating and cooling equation so it cancels out. Temperature is only defendant on the radiation flux.
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