$Q$ 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 $$?
Is it the temperature difference? And if so, would the equation look like this?
$$
\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)
 A: 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.
All these topics are discussed in detail in introductory heat transfer textbooks, e.g., Incropera & DeWitt's Fundamentals of Heat and Mass Transfer.
