Stefen's law|stefen's ka niyam tharmodynamics|blackbodyradiation|stefen's formula|science|physics

Stefen's law 

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According to stefen's law "the total amount of heat rediation by a perfectly black body per second per unit area is directly proportional to the fourth Power of its absolute temparature"

                              E∝T⁴

                                Or 

                           E=σT⁴

Where σ is a constant called the stifen's constant.

This law is sometimes also called as stefen's fourth Power law.

In the above form this law refer only the emmiton and not to the net loss. If a black body A at absolute temparature T is surrounded by another blackbody B at absolute temparature T' then

Amount of heat lost in blackbody A=  σ T⁴

 amount of heat absorbed by black body A from black body B = σ T`⁴

so net amount of heat lost by A per second per unit area = σ(T⁴ - T`⁴)

This is also known as Stefan Boltzmann lau Thermodynamic Proof: Suppose the radiation is enclosed in an evacuated cylinder with perfectly reflecting walls and perfectly reflecting moving piston. The object of assuming perfectly

reflecting walls is to avoid heat exchange between the walls and the radiation otherwise the thermal capacity of walls will come in the calculation. Let u energy density (energy per unit volume) of the radiation inside the cylinder

V = volume of the cylinder

P = pressure of radiations

Total energy of radiation-energy per unit volume:-

➣U = u×V। .....(1)

volume According To Maxwell's electromagnetic theory of radiation, the pressure p exerted by:

➣p = 1/3( u)। .....(2)

Let us suppose that a small amount of heat dQ is brought into the cylinder and at the same time the volume is changed by an amount dV. If dU is the change in the internal energy of radiation and d'W the external work done, then according to first law of thermodynamics,

➣ dQ = dU+dW 

➤dU+pdV। ......(3)

From equations (1) and (2) substituting the values of U and p in equation (3), we have.

➤dQ = d(uV) + 1/3 (udV)

➤u dV+V du+ 1/3 (u dV)

➤ 4/3 u dV+V du.

According to second law of thermodynamic, we get

➤dQ = TdS get

➤TdS = 4/3 udV + Vdu

➣ds = 4/3 u/T dV+ V/T du. ....(4)

Considering S as a function of (V; u) i.e.,S=/(V,u) we have

➣dS=(∂S/∂u )dV+( partial S/partial u )du. ...(5)

Comparing equations (4) and (5), we have

➣(∂S/∂V )= 4/3 u T and

➣{∂S/∂u} = V/T

Again

➣(∂S/∂u )= 4/3 u/T

➣And (∂S/∂u )=V/T

➣(∂²S/∂u∂V ) =∂²S/∂V∂u

➣∂/∂u (∂S/∂V)=∂/∂V)(∂S/∂u )

➣∂/∂u(4/3.u/T² ) =∂/∂V(V/T )

➣1/3.1/T=4/3.u/T².(∂T/∂u )

➣(∂u/∂)=4(∂T/T )

➣log u = 4 log T+a

where a is the integration constant.

We know that the energy E radiated per second per unit area from a perfectly black body at absolute temperature T. and the energy of radiation inside an enclosure at the same temperature, are related by

➣E= ¼uc

where c is velocity of light.

Substituting the value of u in this expression

➣E=¼acT⁴

➣E=σT⁴

Where σ=1/4ac and is called Stefan's constant. This is Stefan's law.

Ex. 1. A black body at 500"C has a surface area of 0.5m" and radiates heat at the rate of 1.02 x 10 J/s. Calculate Stefan's constant.

According to Stefan's law, the total heat radiated per second by a black body of surface area A and at temperature TK is given by

U=σAT⁴

Substituting the given values, we get

(1.02x10 joule/sec)/ 0.5 m²x(773 K⁴)

=5.7x10 joule (m²-sec.K⁴)

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