The Stefan’s law of radiation is applicable to both ordinary bodies and also to perfectly blackbody.
Statement:
The Stefan’s law of radiation states that the quantity of radiant energy emitted per unit time per unit surface area is directly proportional to the fourth power of the absolute temperature of the perfectly blackbody.
Explanation:
- If Q is the quantity of the radiant energy emitted by the perfectly blackbody
- A is the area of the perfectly blackbody
- Let t be the time for which perfectly blackbody emits the radiant energy
- T be the absolute temperature of the perfectly black body
- Then according to Stefan’s law of radiation,
Q/At α T4
Or Ebα T4
- Thus, Eb = σT4
- Where σ is the constant called as Stefan’s constant.
- Its value is σ = 5.67*10-8 J/m2sK4
- The dimensions of the Stefan’s constant σ are [M1 L0 T-3 K-4]
Emissive power:
- The emissive power of the body is defined as it is the radiant energy emitted by the body per unit time per unit surface area of the body at a given temperature.
- Hence, emissive power can be given as
E = Q/At
- The SI unit of emissive power is J/m2s or W/m2.
- And the dimensions of emissive power will be [M1 L0 T-3].
- The emissive power of the body depends on mainly following factors:
- temperature of the given body
- nature of the given body
- nature of the surroundings
- surface area of the body
- The emissive power of the perfectly blackbody is greater than any other bodies always.
For example:
Lampblack is having emissive power as 98%.
Example:
If the perfectly blackbody is radiating energy at rate of 22.68*108 watt/m2 then what is the temperature of the perfectly blackbody.
Solution:
Given that,
E = 22.68*108 watt/M2
σ = 5.67*10-8 watt/m2K4
T =?
We have, according to Stefan’s law of radiation,
E = σ T4
T4 = E/σ = 22.68*108/ 5.67*10-8
T4 = 4*1016
T = (4)1/4*104
T = √2*104
Thus, T = 1.4142*104
T = 141.42*102 K