Convert your temperatures to Kelvin (K) if the problem you're solving doesn't provide them as such. Add 273.13 to your Celsius (C) figures to convert them to K. For example, if your problem asks you to calculate to radiant heat a 246 degrees C cooking surface loses as it emits heat toward a 24 degrees C wall, your "K" values would be 246 + 273.15 (cooking surface) and 24 + 273.15 (wall), or 519.15 K and 297.15 K, respectively.
Calculate the object's surface area in meters squared. For example, if the cooking surface is flat and round with a radius of 12 cm, convert the radius to meters; 1 m = 100 cm. So the example radius would be .12 m. Express its area (πr^2, where "π" is pi, approximately 3.14159, "r" is the radius and "^2" means "squared" or "to the second power") as follows: π x .12^2 = π x .0144 = .045 m^2.
Solve the "Radiant Heat Loss" equation, Pnet = S.B. x Ah x Eh x (Th-Ts)^4, where "S.B." is the Stephan-Boltzmann constant (5.67 x 10^-8 K^-1), "Ah" is the surface area of the heating surface, Eh is its emissivity and Th-Ts is the different in Kelvin temperatures of the heating surface and the non-heating surface. Assume that the example cooking surface has an emissivity of .96 watts per meter squared (W/m^2) and solve the equation as follows: Pnet = (5.67 x 10^-8) x .045 x .96 x (519.15 - 297.15)^4 = .0000000567 x .045 x .96 x 222^4 = 5.95 Watts.