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Please consider this thermal radiation problem.

Preliminary/Background: A spherical black body B1, like a star, is in a setting with no other thermally active objects nearby. Space is at a temperature of 0 K. The body has internal reactions (nuclear, say) which cause its surface temperature to be 1000 K when at steady state in this setting. A similar (same radius, mass, thermal diffusivity) spherical black body B2, in the same setting, has nuclear reactions that cause its surface temperature to be only 900 K.

The problem: Body B1 is now brought close enough to B2 (say their surfaces are separated by a distance of 2x the radius) to cause a new steady state condition to be established. Ignore gravity.

How would I go about calculating the new temperatures of the bodies after they come to thermally interact? As in, what other information is needed? It is intuitive that the temperatures of both would increase from the case where they were each in isolation because they went from thermally interacting with an environment at 0 K to an environment on average above 0 K (since the environment of each now includes the other). Assume the nuclear reactions inside each are not impacted by the presence of the other. I am sure that more information is needed to calculate the new steady state temperature of each. What information would that be? If we assume near infinite thermal conductivity such that each body is at a uniform temperature then that will make the problem easier. It seems clear we would need the heat capacity as well. Any ideas about what other variables are needed and the governing equations to solve?

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  • $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Commenters, please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    Aug 18, 2020 at 7:22

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Assume two spherical blackbodies at temperatures $T_1$ and $T_2$ with constant radii $r_1$ and $r_2$ and infinite thermal conductivity. The two objects are initially individually radiating into empty space at temperature $T_{\mathrm{inf}}=0\,\mathrm{K}$. Assuming steady state, the corresponding heat generation must be $$Q_i=4\pi r_i^2\sigma T_i^4$$ (corresponding to volumetric heat generation of $3\sigma T_i^4/r_i$), where $\sigma$ is the Stefan-Boltzmann constant.

Assuming the two objects are placed in the same region at a center-to-center distance $d>>r$, each object $i$ now receives an additional incoming flux of approximately $a_{ij}\sigma T_j^4$ from a solid angle of $a_{ij}=A_j/4\pi d^2=r_j^2/4 d^2$, where $A_j$ is the cross-sectional area of object $j$. The new energy balance is therefore now $$4\pi r_i^2\sigma T_i^{\prime 4}= 4\pi r_i^2\sigma T_i^4+ r_i^2r_j^2 \sigma T_j^{\prime 4}/d^2,$$

where the new equilibrium temperatures $T_i^{\prime}$ and $T_j^{\prime}$ can be found iteratively, for example.

The case of $d$ comparable to $r$ requires a more complex view factor, generally obtained from a table of values or an empirical fit, as discussed here.

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  • $\begingroup$ Where did you get that view factor equation for when d is comparable to r? If I take d = 4r, such that the separation between surfaces is 2r, then the calculated view factor is 0.0557 for r1=r2. This does not match up with the results from this calculator (0.015877). thermalradiation.net/calc/sectionc/C-137.html $\endgroup$
    – Karlton
    Aug 17, 2020 at 23:07
  • $\begingroup$ That's a great site; I edited to include it. The other formula, now removed, was a cheat using two coaxial parallel disks. $\endgroup$ Aug 18, 2020 at 0:30

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