With conductive and convective heat transfer I have talked about heat transfer that has occurred through a material medium. Radiative heat transfer however does not require a material medium to occur, and is the only type of heat transfer that can occur through a vacuum. It does this by transferring heat through the electromagnetic spectrum. The amount of heat energy that an object can emit is represented by the following equation.

(Eq-1)  $q_{emitted} = σAT^4$

σ = 5.669 X 10-8 W/m2K4 Stefan Boltzman Constant

A = Surface Area

T = Absolute Temperature

For the above equation the temperature must be in an absolute temperature scale. For standard international that is Kelvin.

## Example

There is sphere with a surface area of .166m2 and a current temperature of 432 degrees C. How much energy does the sphere emit assuming it is a blackbody?

(Eq 2) $q_{emitted} = 5.669X10^-8(.166)(432+273)^4 = 2.1kW$

Now, the above example only represents the amount of heat that an object will emit if it were a black body. A black body is an object that has an emissivity of 1. The emissivity can range from 0-1. The equation below takes emissivity into account.

(Eq 3) $q_{emitted} = ε F_GσAT^4$

ε = Emissivity Function

FG = Shape Factor

Now let’s say that instead of the above example being a black body it instead has an emissivity of .83, ignoring any shape factors, the energy emitted would change to the following.

(Eq 4) $q_{emitted} = 0.83(5.669X10^-8)(.166)(432+273)^4 = 1.74 kW$

(Eq 5) $q=ε_1σA_1(T_1^4-T_2^4)$