# Kirchoff’s law of Radiation

 Science > You are Here

#### Statement:

• The ratio of the emissive power to the coefficient of absorption is constant for all substances at a given temperature and is equal to the emissive power of a perfectly black body at that temperature.   OR At any given temperature, the emissivity (or coefficient of emission) of a body is equal to the coefficient of absorption
• Explanation: If ‘E’ is the emissive power of a substance and ‘a’ is its coefficient of absorption then by Kirchoff’s law #### Theoretical Proof of Kirchoff’s Law of Radiation: • Let us consider two bodies A and B suspended in a constant temperature enclosure. B is a perfectly black body. After sometime both A and B will attain the same temperature as that of enclosure. By Prevost heat exchange theory In this state also every body will emit and absorb thermal radiations.
• Let E be the emissive power of A and  ‘a’ be its coefficient of absorption. Let Eb be the emissive power of B. Let Q be the radiant heat incident per unit time per unit area of each body.

Heat absorbed per unit time per unit area of   A =   a Q

Heat  emitted per unit time per unit area  of A  =  E.

As the temperature remains constant so the heat emitted will be equal to the heat absorbed

∴   E    =  a Q      …………(1)

Perfectly black body B will absorb all the radiant heat incident on it.

Heat absorbed per unit time per unit area of  B  =   Q.

Heat emitted per unit time per unit area of B = Eb

As the temperature of B remains constant so in case of B also heat emitted is equal to heat absorbed.

Eb  = Q   …………………..(2)

Dividing equation (1) by (2) we get, Thus, the coefficient of emission is equal to coefficient of absorption.  This proves Kirchoff’s law.

#### Example – 3:

• 512 J of radiant heat are incident on a body which absorbs 224 J. What is its coefficient of emission?
• Solution:
• Given: Radiant heat incident = Q = 512 J, radiant heat absorbed = Qa = 224 J
• To Find: Coefficient of emission = e = ?

Coefficient of absorption = a = Qa/Q = 224/512 = 0.4375

By Kirchoff’s law of radiation

Coefficient of emission (e) = Coefficient of absorption (a)

∴  e = 0.4375

Ans: Coefficient of emission = 0.4375

#### Example – 4:

• A body of surface area 15 × 10-3 m² emits 1260 J in 40 s at a certain temperature. What is the emissive power of the surface at that temperature?
• Solution:
• Given: Surface area = A = 15 × 10-3 m², radiant heat emitted = Q = 1260 J, time taken = t = 40 s.
• To Find: Emissive power = E = ?

E = Q/At  = 1260 /(15 × 10-3× 40) = 21oo J/m²s

Ans: Emissive power of surface = 21oo J/m²s

#### Example – 5:

• The emissive power of a sphere of area 0.02 m² is 2100 J/m²s. What is the amount of heat radiated by the spherical surface in 20 seconds?
• Solution:
• Given: Surface area = A = 0.02 m², Emissive power = 2100 J/m²s, time taken = t = 20 s.
• To Find: Heat radiated = Q = ?

E = Q/At

∴  Q = E A t

∴  Q = 2100 × 0.02 × 20

Ans: Heat radiated = 840 J

#### Example – 6:

• The energy of 6000 J is radiated in 5 minutes by a body of surface area 100 cm2. Find the emissive power of the body.
• Solution:
• Given: Radiant heat emitted = Q = 6000 J, Time taken = 5 min = 5 × 60 = 300 s, Surface area = 100 cm² = 100 × 10-4
• To Find: Emissive power = E = ?

E = Q/At  = 6000 / (100 × 10-4× 300)

∴  Q = 2000  J/m²s

Ans: Emissive Power = 2000  J/m²s

#### Apparatus: • The apparatus consists of a U – tube manometer containing some coloured liquid. The two arms of the manometer are connected to two identical cylinders P and Q, having the same axis (co-axially arranged). The same face (either left or right) of each cylinder is coated with lamp black while the other face is kept polished. A third cylinder R can be placed between P  and  Q co-axially.  One face of the cylinder R is coated with lamp black while the other face is kept polished. The cylinder R can be rotated about a vertical axis.

#### Working:

• The cylinder R is kept as shown in the figure such that all black surface point in the same direction. Hot water is poured into the cylinder R due to which its temperature will increase. No change will be observed in the liquid levels in the manometer. This shows that the quantity of heat absorbed by both P and Q from R is the same. Therefore pressure exerted by the air in P and Q on the liquid is the same on both sides.
• Let E and  Eb be the emissive powers of the polished and black surfaces and ‘a’ be the coefficient of absorption of the polished face. Let A be the area of the cross-section of each cylinder.

The amount of heat radiated per unit time by the black face of R = A Eb.

A part of this heat is incident on the polished face of P.

Heat incident per unit time on the polished face of P = k A Eb,

The constant k depends upon the distance between P and R.

Heat absorbed per unit time by P per second =   a k A Eb.

Heat radiated per unit time by the polished face of R = A E

Heat incident per unit time on the black face of Q = k A E

Heat absorbed per unit time by the black face of Q = k A E.

The level of coloured liquid in both the arms of the apparatus is the same.

Hence both P and Q absorbed same quantity of heat per unit time.

∴   a k A Eb = k A E

∴   a Eb =  E

∴    a   = E / Eb = e

Thus, the coefficient of absorption  = coefficient of emission.

This is Kirchoff’s Law. Thus the Kirchoff’s Law is experimentally verified.

 Science > You are Here