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Use spectral radiant exitance per unit wavelength. The Planck law isA hotter blackbody radiates more at every wavelength, its wavelength peak moves to shorter wavelengths according to Wien's displacement law, and its integrated flux rises as $\sigma T^4$ by the Stefan–Boltzmann law. The wavelength convention matters: a spectrum plotted per unit frequency has a different peak transformation.

For planetary equilibrium temperature, assume an isotropic blackbody star, a circular orbit, negligible internal heating, unit surface thermal emissivity and efficient global heat redistribution so the surface has one temperature. The received stellar irradiance is $F_{\ast }=\sigma T_{\ast }^4(R_{\ast }/a{)}^2$. A planet of radius $R_p$ intercepts $\pi R_p^2{F}_{\ast }$ but radiates over $4\pi R_p^2$, giving the projected-area factor in planetary energy balanceWithout the uniform-temperature assumption, this calculation gives $⟨T^4{⟩}^{1/4}$, not a unique arithmetic mean of surface temperature.

With Bond albedo $A=0.05$, the absorbed fraction is $1-A$. Hencea decrease of approximately $3.87\mathrm{K}$. This albedo weights incident stellar light and does not imply thermal emissivity $0.95$.

Define the planetary effective emissivity by outgoing top-of-atmosphere flux $F_{\mathrm{out}}={ϵ}_{\mathrm{eff}}\sigma T_s^4$. Maintaining the same absorbed stellar power with $T_s=312\mathrm{K}$ givesThe greenhouse effect allows thermal radiation to escape from colder layers, so the outgoing flux is less than a black surface at $T_s$ would emit. Downward atmospheric emission contributes to the surface energy balance; this does not violate the top-of-atmosphere energy balance.^[ref]

A single-layer greenhouse model makes the distinction quantitative. Take a shortwave-transparent, isothermal layer with infrared absorptivity and emissivity ${ϵ}_a$, above a surface black in the thermal infrared. Writing $Q=(1-A)F_{\ast }/4$, its surface and atmospheric balances areFor ${ϵ}_a>0$, $T_a^4=T_s^4/2$. Radiation escaping to space is $(1-{ϵ}_a)\sigma T_s^4+{ϵ}_a\sigma T_a^4$, soThus $0.849$ is the planet's effective emissivity; $0.301$ is the layer's infrared emissivity within this particular model. They are not interchangeable. Real atmospheres require wavelength-dependent absorption, a vertical temperature profile and nonradiative heat transport; the measured temperatures alone do not uniquely retrieve an atmospheric optical profile.