0
$\begingroup$

How does a blackbody actually heat up and change temperature if it emits all the radiation that it absorbs by definition? And by extension, if the absorptivity and emissivity of a body are equal, then how is the temperature actually changing?

$\endgroup$
1
  • 1
    $\begingroup$ > "it emits all the radiation that it absorbs by definition?" This is not so, and also not in the actual definition. Where did you find such claim? $\endgroup$ Nov 21, 2023 at 0:25

3 Answers 3

3
$\begingroup$

To begin with, in real life the absorptivity and emissivity are not exactly the same as both of these things are functions of wavelength. This means it is possible for an object sitting in sunlight to absorb well across a range of wavelengths but if it emits poorly in the infrared then it will eventually become hot enough to burn your hand if you touch it.

Furthermore, the whole blackbody derivation relies on the black body being at the same temperature as its surroundings: this represents a state of equilibrium in which the heat going in equals the heat going out and temperature therefore remains constant. While a cold body is being heated, it own blackbody emission spectrum does not match the blackbody emission spectrum of its environment and heat transfer is biased in the direction of the cold body getting hotter.

$\endgroup$
4
  • $\begingroup$ So a body cannot be modelled as a blackbody whilst it's temperature is increasing? $\endgroup$ Nov 20, 2023 at 17:18
  • 2
    $\begingroup$ while its temperature is increasing, it emits a blackbody spectrum at every moment corresponding to its temperature at that moment but that spectrum changes shape at every moment as the object heats up. $\endgroup$ Nov 20, 2023 at 17:26
  • $\begingroup$ Oh Oh Oh I see what you mean (Lightbulb moment)! The amount its absorbing and emitting is only equal when we have equilibrium conditions so though it is a perfect emitter based on it's temperature this quantity is still smaller than the amount of radiation it's receiving (assuming net heat transfer to the blackbody) until thermal equilibrium is reached and by definition this becomes the same. Thank you so much!!! $\endgroup$ Nov 20, 2023 at 17:30
  • $\begingroup$ You got it man!!! $\endgroup$ Nov 20, 2023 at 22:50
1
$\begingroup$

A blackbody is in thermodynamic equilibrium. It is not getting hotter or colder.

Having realised this, then it is more straightforward to understand why a true blackbody emits as much radiation as it absorbs.

If an object is being heated, in the sense that its temperature is increasing, its spectrum may still approximate to a blackbody if the timescale on which the temperature is increasing is much longer than the timescale for it to achieve a thermodynamic equilibrium - i.e. if the temperature increases slowly enough to model it as the evolution through a sequence of quasi-static temperatures.

$\endgroup$
6
  • $\begingroup$ Black body need not be in thermal equilibrium even though it is always assumed to be in local thermodynamic equilibrium (with thermal emission independent of incoming radiation). E.g. Sun photosphere is approximately a black body, almost in LTE, but it is not in thermal or thermodynamic equilibrium, because there is substantial radial flux of energy and loss of energy to the outer space. $\endgroup$ Nov 21, 2023 at 0:53
  • $\begingroup$ The Sun approximates to a black body (but clearly isn't one). An actual blackbody must be in thermal equilibrium, as I said. $\endgroup$
    – ProfRob
    Nov 21, 2023 at 7:39
  • $\begingroup$ Must be in LTE. In LTE is not the same thing as in thermal equilibrium. $\endgroup$ Nov 21, 2023 at 9:43
  • $\begingroup$ The definition of a black body implies/requires LTE, but does not require it to be in thermal equilibrium with other outer bodies. A black body solid sphere at non-zero temperature in vacuum, losing energy via radiation, is a valid case of a black body. $\endgroup$ Nov 21, 2023 at 9:48
  • $\begingroup$ Thermodynamic equilibrium is the unique stable stationary state that is approached or eventually reached as the system interacts with its surroundings over a long time. The temperature cannot be changing. @JánLalinský Perhaps I see what you mean though. $\endgroup$
    – ProfRob
    Nov 21, 2023 at 10:43
1
$\begingroup$

it emits all the radiation that it absorbs by definition?

This is not the correct definition of a black body. The Wikipedia definition of a black body says:

A black body or blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.

In other words a black body is defined as one that absorbs all incoming radiation. Not as one that emits all the radiation that it absorbs.

Indeed, any object at thermal equilibrium in purely radiative heat exchange will emit all of the radiation that it absorbs, whether it is a black body, gray body, or white body. Your definition is a definition about thermal equilibrium, not about the type of object that is in thermal equilibrium.

While a black body is colder than its environment it is absorbing all of the incoming radiation. It is emitting a characteristic black body spectrum, which contains less total power than the absorbed power. Thus it gains thermal energy until it reaches the equilibrium point and behaves as described above.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.