We have a body suspended in a vacuum at temperature T (purple circle).

We have a single heat source of incident radiation, A, shown by red arrow.

Heat is shown radiated away as blue arrow, B. There is no single heat-sink, so this would be in many directions.

The system is in equilibrium, and temperature T is stable. The purple body is not a heat source.

We want to change or control temperature T.

We cannot change A or B in any way.

We cannot change the mass of the purple body; we can only change its chemical make-up.

What kind of things could we do to the purple body to control temperature T ?

CLARIFICATION: I want to control or determine the temperature at the surface of the body, by constructing the body from different materials / structures.

My first thought is to try to change the reflectiveness of the surface. If the surface was made shiny would that cause T to drop, because heat A bounces off?

What else could we do to change temperature T?

Also interested if the size of the body would affect how this operates.

Body in vacuum

  • $\begingroup$ Are you allowed to do work on the body? Are you allowed to have a chemical reaction within the body? $\endgroup$ – Chet Miller Sep 23 at 12:27
  • $\begingroup$ @ChetMiller Well, yes, you could have internal structure to the body, and so have chemicals mixing inside. But I'm mainly interested from the viewpoint of how the body responds or reacts to heat source A. If there was a chemical reaction inside the body, consider it temporary so that I'm interested in T after the reaction is spent, and equilibrium has been re-established. $\endgroup$ – Stewart Sep 23 at 13:08
  • $\begingroup$ Are you saying there is a single temperature T associated with the purple body? I would think there would be a temperature gradient from the side facing A to the side facing B, or perhaps I misunderstand the set up. $\endgroup$ – Bob D Sep 25 at 13:24
  • $\begingroup$ Isn't there an inconsistency in your question?: you mention changing the reflectiveness of the body's surface to change A; however, you also say that 'A and B cannot be changed in any way'. $\endgroup$ – Time4Tea Sep 25 at 13:28
  • 1
    $\begingroup$ @Time4Tea I mean incident radiation. Thanks for helping me with that. Making edit. $\endgroup$ – Stewart Sep 25 at 14:03

To enable a mathematical analysis, let us simplify the problem to an infinite plane with radiation incident from one side, and uniform in space:

enter image description here

Energy enters the body via the incident radiation; since it is a vacuum the only way for it to leave is through emitted radiation.

The fraction of the incident radiation which it absorbs is given by the emissivity $\varepsilon$. If the body is opaque, the remainder of the incident radiation is reflected. The amount of emitted radiation is determined by the Stefan-Boltzman law. So if the power per unit area of the incident radiation is $J$, then we have

$$\frac{\mathrm{d}E}{\mathrm{d}t} = A \varepsilon \left( J - \sigma T_H^4 - \sigma T_C^4 \right),$$

where $T_R$ is the temperature of the hot side of the plane (left in the figure) and $T_C$ is the temperature of the cold side. I am ignoring any possible wavelength-dependence of the emissivity.

To get equillibrium, set the time derivative of the energy to zero (i.e. make the rate of emission equal to the rate of absorption). Thus:

$$T_H^4 + T_C^4 = \frac{J}{\sigma}.$$

So far, the properties of the material haven't seemed to make a difference! ($\sigma$ is a universal constant, and the emissivity dropped out of the equation!) But where they enter is in determining the relationship between $T_C$ and $T_H$.

To determine the equillibrium relationship between these two temperatures, we must balance conduction from the hot side to the cold side against radiation from the cold side. Conduction in the body is described by the heat equation, which at equillibrium inside the slab gives a linear temperature profile, with the temperature decreasing linearly from $T_H$ to $T_C$.

The rate of heat conduction (per unit area) in this quasi-1d problem is

$$J_{\mathrm{conduction}} = \kappa \frac{\partial T}{\partial x} = \kappa \frac{T_H - T_C}{h},$$

where $\kappa$ is the thermal conductivity and $h$ is the thickness of the slab. At equillibrium, this flux must balance the outgoing radiation on the cold side of the slab, so

$$ \frac{\kappa}{h \varepsilon \sigma} \left(T_H - T_C \right) = T_C^4.$$

Notice that in this simplified geometry, we only have one control parameter, the dimensionless ratio which I will call

$$ \phi \equiv \frac{\kappa}{h \varepsilon \sigma}.$$

We can make $\phi$ very large either by making the material conduct heat very well, making it very thin, or making it very reflective. Any of the above will allow the body to equillibrate internally, and give $T_H = T_C$. On the other hand, if we have a slab which conducts very poorly or is very very thick, then we will have $T_H \gg T_C$.

In principle we can completely solve the problem by solving the above equation for $T_C$ as a function of $T_H$, then plugging it back into the equation containing $J$. But let's just solve the two extreme limits, and we'll basically understand everything. In the $\phi \to \infty$ limit there is one temperature, $T$, and

$$T = \left( \frac{J}{2 \sigma} \right)^{1/4}.$$

In the limit $\phi \to 0$, $T_C = 0$ from the second equation, and therefore $T_H = \left( \frac{J}{\sigma} \right)^{1/4}$.

So in summary:

  • We only have one control paramter, which is basically (conduction rate) / (radiation absorptivity).
  • If this number is high, the entire body will come to a single, medium-ish temperature.
  • If the number gets lower, the "sunward side" will get hotter, but the "dark side" will get colder.

A more realistic geometry will complicate this analysis, but I think the conclusions should be similar. I don't know if a very "weird" $\varepsilon{\left(\lambda\right)}$, produced by some sort of metamaterial, would have an interesting effect.

  • $\begingroup$ Outstanding! I am super-impressed by your answer being both thorough, in-depth, and yet also fairly easy to understand! $\endgroup$ – Stewart Sep 26 at 7:14

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