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A diffuse thermal emitter has emitted radiance independent of outgoing direction over the outward hemisphere. Its emitted radiant intensity from a planar patch still varies with viewing angle through projected area, as described by Lambert's cosine law. Diffuse emission does not mean equal power per solid angle from the patch. This approximation does not by itself impose a grey body spectrum or specify how incident radiation is reflected.

Directional emissivity is the ratio of emitted radiance to blackbody radiance at the same temperature in a specified outgoing direction. It can be total over wavelength or additionally spectrally resolved. Normal-direction emissivity is not automatically total hemispherical emissivity: the latter integrates all directions with the projected area weighting. A diffuse thermal emitter removes this directional dependence of emitted radiance.

Emissivity compares thermal emission from a surface with emission from an ideal blackbody at the same absolute temperature, using the same wavelength and angular averaging convention. For ordinary passive surfaces in the far field at local thermal equilibrium, the ratio lies between zero and one. A blackbody has emissivity one. Unless a spectral or directional qualification is given, the $ϵ$ in $M=ϵ\sigma T^4$ is total hemispherical emissivity: a ratio of emitted radiant exitances, not of all outgoing radiant flux including reflected radiation.^[ref]

Emissivity belongs to a specified surface state, not just a bulk material name. Coating, oxidation and surface finish can change it; wavelength, viewing direction and temperature also matter. A surface that looks white in visible light need not have low infrared emissivity. This distinction is useful when choosing a radiator coating or interpreting a radiation-based temperature measurement: a narrow-band detector needs the appropriate spectral emissivity, and its signal can include reflected surrounding radiation. The grey body and diffuse thermal emitter assumptions are separate simplifications, addressing wavelength and direction respectively.

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 isAssume an opaque grey body that is a diffuse thermal emitter, with the same emissivity $ϵ$ 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 ${\dot{Q}}_{\mathrm{rad}}$ means net loss by the surface; negative means net gain. This expression counts thermal radiation alone, not all heat transfer.^[ref]

The reasoning is a two-way energy balance. Black walls supply irradiance $E=\sigma T_s^4$, so the surface absorbs $P_{\mathrm{abs}}=ϵAE$ and emits $P_{\mathrm{emit}}=ϵA\sigma T^4$. Subtract these to obtain ${\dot{Q}}_{\mathrm{rad}}$. For an opaque surface the remaining fraction $1-ϵ$ 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 $ϵ=0.8$, $A=0.01{\mathrm{m}}^2$, $T=500\mathrm{K}$ and $T_s=300\mathrm{K}$. These are stipulated model inputs, not measured material properties. Using $\sigma \approx 5.670374419\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$ givesThe incoming irradiance is $459.300\mathrm{W}{\mathrm{m}}^{-2}$, 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.

Near equal temperatures, write $T=T_s+\Delta T$. Taylor's theorem givesFor constant emissivity, the radiative heat transfer coefficient $h_{\mathrm{rad}}$ is approximately $4.8992\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ at $T_s=300\mathrm{K}$ and $ϵ=0.8$. 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,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.