Does a frying pan use radiation?
You may estimate the contribution of thermal radiation by using the Stefan–Boltzmann law, which describes the power $P$ radiated from a black body in terms of its temperature $T$:

$$P=\sigma\cdot A\cdot T^4$$

where $\sigma=5.670\,367(13)\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}$ is the Stefan–Boltzmann constant[reference] and $A$ is the surface area.

The formula can be corrected for real materials by introducing the emissivity $\varepsilon$:

$$P=\varepsilon\cdot\sigma\cdot A\cdot T^4$$

For an ideal black body, the emissivity is $\varepsilon=1$. For polished metal surfaces, the values are significantly lower.

For example, for a hot black frying pan with $\varepsilon\approx1$, $A\approx600\ \mathrm{cm^2}=0.06\ \mathrm{m^2}$, and $T\approx230\ \mathrm{^\circ C}\approx500\ \mathrm K$, the power may be estimated as

$$\begin{align}
P&amp;=\varepsilon\cdot\sigma\cdot A\cdot T^4\\
&amp;=1\times5.670\,367\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}\times0.06\ \mathrm{m^2}\times\left(500\ \mathrm K\right)^4\\
&amp;\approx200\ \mathrm W
\end{align}$$

However, in addition to thermal radiation, heat transfer also occurs via thermal convection and thermal conduction. You could exclude these modes by placing the frying pan in a vacuum.

You could also estimate the total heat transfer due to all three modes from the electrical power of the stove that is required to keep the frying pan at a constant temperature.