Stephen-Boltzmann Law – Concept & Examples

Stephen-Boltzmann Law – Concept & Examples

ConceptExamples

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:

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)

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

 

 

 

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