# $Q$ Transfer via Radiation Formula

According to the formula: $$\frac{\Delta Q}{\Delta t}=\sigma\epsilon A T^4$$ What does $$T$$ refer to in a situation where I am modelling the power of radiation from air of temperature to surface of emissivity $$𝜖$$?

Is it the temperature difference? And if so, would the equation look like this? $$\frac{\Delta Q}{\Delta t}=\sigma\epsilon A (T_{2} - T_{1})^4 \hspace{0.5in}or\hspace{0.2in} \frac{\Delta Q}{\Delta t}=\sigma\epsilon A (T_{2}^4 - T_{1}^4)$$

Further, what would the equation look like if the surface had greater temperature than the air?

Edit: the source of the formula is from the IBO Exam Data Booklet (https://ibphysics.org/wp-content/uploads/2016/01/annotated-physics-data-booklet-2016.pdf)

• You should cite the source you're getting this formula from, for clarity. Nov 21, 2022 at 19:20
• Added. Thanks for reminding me Nov 22, 2022 at 14:21

The term $$\sigma\varepsilon A T^4$$ is used to model the graybody output power—i.e., the outgoing radiative heat flux from some surface with emissivity $$\varepsilon$$, area $$A$$, and temperature $$T$$, without considering any radiative input.
If that surface entirely faces an environment at temperature $$T_\mathrm{env}$$, then the net output is typically modeled as $$\sigma\varepsilon A(T^4-T_\mathrm{env}^4)$$ because that environment itself radiates heat toward the surface. (Alternatively, the net rate of heat gain at the surface can be modeled as $$\sigma\varepsilon A(T_\mathrm{env}^4-T^4)$$.)
Special care may be needed to treat a surrounding environment of gas only (as in our atmosphere), as some wavelengths may be unabsorbed and some radiation passing essentially transparently to outer space ($$T\approx 0\,\mathrm{K}$$). This can cause the effective temperature of the atmosphere for radiative heat transfer calculations (or the so-called "sky temperature") to differ from the actual temperature.