Bottom Science
According to Stefan Boltzmann Law,
The energy of thermal radiation emitted per unit time by a body of surface area A and temperature T is given by:
$\frac{E}{t}$ = P = e σ A $T^{4}$
where,
P = Power (Energy/Time)
t = Time
T = Temperature (in Kelvin)
σ = called Stefan Boltzmann’s constant = 5.67 x $10^{-8}$ W $m^{-2}$ $K^{-4}$
e = emissivity of the surface = 0 to 1 (e=1 for black body)
Examples
Example 1: The power per unit area emitted by a surface of a blackbody at temperature 230 K is about?
Solution:
We know that,
E= e σ A $T^{4}$
So, the power per unit area emitted by a surface of a black body will be:
P/A = e σ $T^{4}$
Substituting, T=230 K, e=1 (for the black body), A=unit and σ in the above equation,
We get,
P = 1 x 5.67 x $10^{-8}$ x $(230)^{4}$
So,
P = 158.67 W$m^{-2}$
Example 2: A blackbody of surface area 10 cm$^{2}$ is heated to temperature 127 $^{o}C$ and is subtended in a room at temperature 27$^{o}C.$ Calculate the initial rate of loss of heat from the body to the room?
Solution:
Given,
room temperature = T1 = 27 $^{o}C$ = 300 K
blackbody temperature = T2 = 127 $^{o}C$ = 400 K
Again, as we know that,
The energy of thermal emission of a body is given by:
E= e σ A $T^{4}$
In the case of a blackbody, e=1
So,
E=σ A $T^{4}$
So, at room temperature T1,
E1=σ A $T1^{4}$
and, at room temperature T2,
E2=σ A $T2^{4}$
Hence, net emission of energy (in the form of heat),
E1-E2 = σ A ($T2^{4} – T1^{4}$)
Substituting, T1=300 K, T2=400 K, A=10x$10^{-4}$ m$^{2}$ and σ in the above equation,
We get,
E1-E2 = 5.67 x $10^{-8}$ x 10x$10^{-4}$ m$^{2}$$(400^{4}-300^{4})$
= 0.99 W
