= Net radiative heat exchange
{title2=$\dot Q_{\rm rad}=\epsilon\sigma A(T^4-T_s^4)$}
For a small, uniform surface at <temperature> $T$, enclosed by much larger black walls all at <temperature> $T_s$, the outward net <heat transfer> by <thermal radiation> is
$$
\dot Q_{\rm rad}=\epsilon\sigma A(T^4-T_s^4).
$$
Assume an opaque <grey body> that is a <diffuse thermal emitter>, with the same <emissivity> $\epsilon$ and <radiometric absorptance> at the surface's current <temperature>. Its emitting <area> $A$ sees only the enclosure: the <view factor> to the walls is one, with no self-viewing surface, opening or intervening obstruction. The intervening medium must neither absorb nor emit appreciable <thermal radiation>; a <vacuum> supplies a useful idealization. The enclosure is maintained at uniform $T_s$. The small-body assumption makes that reservoir description physically reasonable; genuinely black isothermal walls supply the incoming <irradiance> even without an infinite area ratio. Positive $\dot Q_{\rm rad}$ means net loss by the surface; negative means net gain. This expression counts <thermal radiation> alone, not all <heat transfer>.<Thermodynamics and Propulsion>[§19.1, cavity irradiance; §19.2, grey-body absorptance/emittance; §19.4, enclosed-body view factor one]{ref}
The reasoning is a two-way <energy> balance. Black walls supply <irradiance> $E=\sigma T_s^4$, so the surface absorbs $P_{\rm abs}=\epsilon AE$ and emits $P_{\rm emit}=\epsilon A\sigma T^4$. Subtract these to obtain $\dot Q_{\rm rad}$. For an opaque surface the remaining fraction $1-\epsilon$ of incoming <radiant flux> is reflected; counting emitted plus reflected outgoing <radiant flux> minus total incoming <radiant flux> gives the same result. <Kirchhoff's law of thermal radiation> justifies equal spectral absorption and emission factors; the <grey body> assumption lets one factor be taken outside both spectral <integrals>.
Choose $\epsilon=0.8$, $A=0.01\,\mathrm{m^2}$, $T=500\,\mathrm K$ and $T_s=300\,\mathrm K$. These are stipulated model inputs, not measured material properties. Using $\sigma\approx5.670374419\times10^{-8}\,\mathrm{W\,m^{-2}\,K^{-4}}$ gives
$$
P_{\rm emit}\approx28.3519\,\mathrm W,\qquad
P_{\rm abs}\approx3.6744\,\mathrm W,\qquad
\dot Q_{\rm rad}\approx24.6775\,\mathrm W.
$$
The incoming <irradiance> is $459.300\,\mathrm{W\,m^{-2}}$, corresponding to $4.5930\,\mathrm W$ incident on the surface; only $80\%$ is absorbed. At $T=T_s=300\,\mathrm K$, emitted and absorbed <powers> are both $3.6744\,\mathrm W$, so <radiative equilibrium> has zero net exchange despite continuing emission. Changing only <emissivity> to $0.2$ reduces both one-way <powers> and the net loss by a factor of four, giving $6.1694\,\mathrm W$ net. A low-<emissivity> coating therefore suppresses both radiative cooling to colder walls and radiative heating from hotter walls.
\Image[/grey-surface-black-enclosure-exchange.png]
{title=Two-way exchange in a black enclosure and the effect of <emissivity>}
{description=Original schematic, not an experimental apparatus. The central convex body has total emitting <area> $0.01\,\mathrm{m^2}$, <temperature> $500\,\mathrm K$ and <view factor> one to black walls at $300\,\mathrm K$. Arrows show gross emission, absorbed enclosure radiation and reflected enclosure radiation for <emissivity> $0.8$; widths are illustrative. The graph compares stipulated <emissivities> under the same geometry and <temperatures>.}
Near equal <temperatures>, write $T=T_s+\Delta T$. <Taylor's theorem> gives
$$
\dot Q_{\rm rad}=4\epsilon\sigma AT_s^3\Delta T+O((\Delta T)^2),
\qquad h_{\rm rad}=4\epsilon\sigma T_s^3.
$$
For constant <emissivity>, the radiative <heat transfer coefficient> $h_{\rm rad}$ is approximately $4.8992\,\mathrm{W\,m^{-2}\,K^{-1}}$ at $T_s=300\,\mathrm K$ and $\epsilon=0.8$. This local linear approximation is useful for small <temperature> differences; the $200\,\mathrm K$ difference in the worked example needs the full fourth-power expression.
A cooling experiment requires a wider <first law of thermodynamics> balance. In a uniform-<temperature> model without a heater,
$$
C(T)\frac{dT}{dt}=-\dot Q_{\rm rad}-\dot Q_{\rm conv}-\dot Q_{\rm cond},
$$
where $C(T)$ is the sample's <heat capacity> and the last two terms are signed outward <convection> and <thermal conduction> losses. A <vacuum> suppresses <convection> but does not eliminate <thermal conduction> through supports or sensor leads. A cooling slope therefore cannot establish <emissivity> unless these paths and <heat capacity> are independently constrained. Multiple wall <temperatures>, incomplete <view factors>, wavelength-selective materials or appreciable intervening absorption require a more complete <radiative transfer> calculation.
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