Conservation of Energy Equations

Conservation of energy implies that the amount of incident energy is equal to the sum of the absorbed, reflected, and transmitted energy.

(1) Incident Energy = Absorbed Energy + Transmitted Energy + Reflected Energy

Consider equation 1 for an object in a vacuum at a constant temperature. Because it is in a vacuum, there are no other sources of energy input to the object or output from the object. The absorbed energy by the object increases its thermal energy – the transmitted and reflected energy does not. In order for the temperature of the object to remain constant, the object must radiate the same amount of energy as it absorbs.

(2) Emitted Energy = Absorbed Energy

Therefore, objects that are good absorbers are good emitters and objects that are poor absorbers are poor emitters. Applying equation 2, Equation 1 can be restated as follows:

(3) Incident Energy = Emitted Energy + Transmitted Energy + Reflected Energy

Setting the incident energy equal to 100%, the equation 3 becomes:

(4) 100% = %Emitted Energy + %Transmitted Energy + %Reflected Energy

Because emissivity equals the efficiency with which a material radiates energy, equation 4 can be restated as follows:

(5) 100% = Emissivity + %Transmitted Energy + %Reflected Energy

Applying similar terms to %Transmitted Energy and %Reflected Energy,

(6) 100% = Emissivity + Transmissivity + Reflectivity

According to equation 6, there is a balance between emissivity, transmissivity, and reflectivity. Increasing the value of one of these parameters requires a decrease in the sum of the other two parameters. If the emissivity of an object increases, the sum of its transmissivity and reflectivity must decrease. Likewise, if the reflectivity of an object increases, the sum of its emissivity and transmissivity must decrease.

Most solid objects exhibit very low transmission of infrared energy – the majority of incident energy is either absorbed or reflected. By setting transmissivity equal to zero, equation 6 can be restated as follows:

(7) 100% = Emissivity + Reflectivity

For objects that do not transmit energy, there is a simple balance between emissivity and reflectivity. If emissivity increases, reflectivity must decrease. If reflectivity increases, emissivity must decrease. For example, a plastic material with emissivity = 0.92 has reflectivity = 0.08. A polished aluminum surface with emissivity = 0.12 has reflectivity = 0.88.

The emissive and reflective behavior of most materials is similar in the visible and infrared regions of the electromagnetic spectrum. Polished metals, for example, have low emissivity and high reflectivity in both the visible and infrared. It is important to understand, however, that some materials that are good absorbers, transmitters, or reflectors in the visible, may exhibit completely different characteristics in the infrared.