Kirchoff's Law of Heat radiation

 Kirchhoff's Law of Heat radiation

Introduction : Kirchhoff studied the behaviour of heat radiation and based on that result obtained from emissive power and absorptive power he made a understanding that radiation depends on wavelength specific radiative emission and absorption by a body in thermal equilibrium. Note: All body when heated emits the same kind of radiation which they absorbs. Thus Kirchhoff arrived at an conclusion known as kirchhoff's law of heat radiation.

Kirchhoff's law of heat : The coefficient of emission is always equal to coefficient of absorption at any given temperature.

Mathematically,

a = e   -------(1)

but we know that e = E / E_b   -------(2)

From (1) & (2)

a = E / E_b 

E / a = E_b -------(3)

From equation no (3) Kirchhoff's law can be defined as At any temperature, the ratio of  the emissive power(E) to the coefficient of absorption(a) is always equal to emissive power of a perfectly black body (E_b ) at  the same temperature.

---

Theoretical proff of  Kirchhoff's law

Consider a thermally isolated system which can be considered as uniform temperature enclosure. Two body A which is ordinary and body B which is perfectly black body have emissive power E and E_b respectively. Let body A has coefficient of emission (e) and coefficient of absorption (a). Both the body have same area and  let Q be the amount of radiant heat incident on both the body in  a given time and thus heat radiation absorbed by the body A  is given by ,

Q_a  = a Q    ---------(4)           [Q_a  / Q = a]

Q_a  =  Radiant heat absorbed by the body

a = Coefficient of absorption

Q = Total heat incident

Thus we know that,

 Heat absorbed by the body = Heat emitted by the body

a Q  = E -----(5)

For perfectly black body the heat radiation absorbed is given by,

Q_a  = a Q  

 ( for perfectly black body a= 1)

Q_a  =  Q 

Thus we know that ,

Heat absorbed by the perfectly black body  =  Heat emitted by the body

Q  = E_b  -----(6)

from (5) and (6) dividing

a = E / E_b

E / a = E_b 

Hence proved

---

Stefan's Law of radiation

Introduction : The temperature dependence of heat radiation emitted by black body was explained by Josef Stefan in 1879.It was also studied by boltzmann later using thermodynamics. Thus this law is  also known as Stefan -Boltzmann law of heat radiation. They studied the system of heat emitted by a perfectly black body and formulated a empirical relation between the emitted heat and temperature.

Stefan-Boltzmann law : The emissive power of perfectly black body is directly proportional to forth power of it's absolute temperature.

OR

The rate of emission of radiant energy per unit area per unit time of perfectly black body is directly proportional to forth power of it's absolute temperature.

 

Mathematically,

Q / At ∝ T⁴ -----(7)

where,

Q / At = Heat radiated per unit area per unit time

^{We know that,}

(E_b = Q / At) -----(8)

putting  (8) in (7)

E_b   ∝  T⁴ 

E_b   = σ  T⁴ 

σ = stefan's constant

σ = 5.67 x 10^-8  J/m² sK⁴ 

 σ = 5.67 x 10^-8  watt/m²K⁴ 

Dimension [ L⁰ M¹  T^-3 K^-4 ]

For normal body, E = eE_b

    Thus for normal body stefan's law is applicable as  

E = eσT⁴ 

Condition for stefan's law : If  perfectly black body having absolute temperature T is placed in surrounding which are at a lower absolute temperature T₀ then heat radiation emitted per unit time  per unit  area is given by,

E_b   = σ  T⁴ 

^{Energy absorbed by the surrounding per unit area per unit time,}

E_b   = σT₀⁴ 

Thus net loss of energy by perfectly black body per unit area per unit time is ,

E_b   = σ(T⁴ - T₀⁴ )

Thus net loss of energy by ordinary body per unit area per unit time is ,

E_b   = eσ(T⁴ - T₀⁴ )

👇

<<<Wein's Displacement law>>>

If any query or any mistake found in the content please let us no.

---