I am studying the thermal radiation (Stefan–Boltzmann law) by myself

$$P = \epsilon \sigma A T^4$$

here $\epsilon$ is the emissivity, $\sigma$ is Stefan-Boltzmann constant, $A$ is the surface area of the radiating object, $T$ is the temperature of the radiating object. As my understanding, $P$ is the power radiating out of the source. But to figure out the power on the object away from the source due to the radiation, I find the calculation from the book

$$P_\text{desc} = \epsilon \sigma A (T_\text{src}^4 - T_\text{desc}^4)$$

again, $\epsilon$ and $A$ are the parameters for the radiating source. My question is why the power on the destination object only depends on the temperature^4 difference? So the parameters of the destination object (like it's surface area, heat capacity, etc.) have nothing to do with that power? I don't understand this from the physical point of view.

So if that's true, to calculate the power absorbed by the earth due to sun, should the following calculation sufficient?

$$P_\text{earth} = \sigma A_\text{sun}(5778^4 - 287^4) $$ where $5778K$ is the temperature of the sun surface, $287$ is the average temperature of the earth surface.


1 Answer 1


The wiki article has a lucid calculation of what you are asking.

Notice that this is the incident power, and somewhwere the book must say per meter square on the target.

Incident and absorbed are two different concepts.

The incident/incoming radiation is computed at the location of the earth, so it depends only on the parameters of the sun and the distance from it. After it arrives on the earth we talk of energy per meter square, and then come the calculations dependent on measurements, like cloud albedo (reflectivity from clouds). In terms of heat capacity it is complicated by air absorption, water absorption/ reflectivity land absorption/reflectivity.

In terms of climate studies all these are included in the so called energy budget, which mixes up a lot of concepts as the earth is not a simple thermodynamic object to be just described by a heat capacity.

  • $\begingroup$ Thanks. I read the wiki also. But I still don't understand how to estimate the radiating energy reach an object. There is a question in the text read: what's the energy from run and reach an ice (with emissivity about 0.050) on earth? This is pretty confusing because I think the energy from sun only depends on the emissivity of the sun and the temperature but in this question, it looks like I have to use the emissivity of the ice to calculate that energy, why? $\endgroup$ Commented Apr 28, 2013 at 5:13
  • $\begingroup$ you have a spelling mistake and i do not understand your "what's the energy from run and reach an ice (with emissivity about 0.050) on earth". generally the impinging radiation energy falls as 1/r^2 with distance from the source, as all incoherent radiation. Once you have the energy per meter square falling on some substance to find how much of this enerhy can be absorbed you need the reflectivity: a perfect mirror would absorb nothing . emissivity is the coefficient for the black body radiation at the temperature reached. $\endgroup$
    – anna v
    Commented Apr 28, 2013 at 6:18
  • $\begingroup$ Sorry for the typo. It should be: what's the energy transferred from the sun and reach an ice with area A and height H on the ground of the earth. So to my understanding your reply, if the energy per meter^2 from the sun is E, the actual energy research the ice should be E*A*0.05 (0.05 is the emissivity same as absorptivity), is that correct? $\endgroup$ Commented Apr 28, 2013 at 6:46
  • $\begingroup$ the energy reaching the ice will be E*A. have a look nsidc.org/cryosphere/seaice/processes/thermodynamic_melt.html . i think it should be .5, the reflectivity not the emissivity. so the factor should be 0 .5 not 0.05 . emissivity is not the same as absorptivity. of that 50% energy not reflected, look at ice solarmirror.com/fom/fom-serve/cache/43.html . $\endgroup$
    – anna v
    Commented Apr 28, 2013 at 8:30
  • $\begingroup$ It is confusing about the emissivity because my text "Physics for scientists and engineers" state that "the emissivity is equal to the absorptivity" $\endgroup$ Commented Apr 28, 2013 at 16:24

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