• Radiation is a process of transfer of heat in the form of electromagnetic waves for which material medium is not necessary.

• In the process of radiation thermal energy or heat energy is transferred from one point to other in the form of electromagnetic waves.
• As radiation is due to electromagnetic waves and electromagnetic waves are capable of passing through a vacuum, there is no necessity of material medium for radiation.
• Due to electromagnetic nature of radiation has the same properties as that of light, such as rectilinear propagation, reflection, refraction, interference etc.
• The velocity of radiant energy in air or vacuum is the same as that of light in vacuum i.e  3 × 108 m/s. Due to this high-speed radiation is the most rapid process of heat transfer.
• When radiant heat is incident on a matter, it is partly absorbed and converted into heat.
• Radiations have the wavelength greater than that of red colour and thus radiation form infrared region of the electromagnetic spectrum.

#### Coefficient of Absorption (a):

• The coefficient of absorption of a body is defined as the ratio of the quantity of radiant heat absorbed by it to the total quantity of radiant heat incident upon it.
• Let Q  be the radiant heat incident on the body and Qa be the radiant heat absorbed. Thus,  a = Q/Q

#### Coefficient of Reflection (r):

• The coefficient of reflection of a body is defined as the ratio of the quantity of radiant heat reflected by it to the total quantity of radiant heat incident upon it.
• Let Q  be the radiant heat incident on the body and Qr be the radiant heat reflected. Thus, r = Q/Q

#### Coefficient of Transmission (t):

• The coefficient of transmission of a body is defined as the ratio of the quantity of radiant heat transmitted by it to the total quantity of radiant heat incident upon it.
• Let Q  be the radiant heat incident on the body and Qt be the radiant heat transmitted. Thus, t = Q/Q

#### Rrelation Between Coefficient of Absorption, Reflection,  and Transmission:

• Let Q be the total radiant heat incident on the surface of a body. Let Qa, Qr, Qt be the radiant heat absorbed, reflected and transmitted respectively by the body.

Then , Qa +   Qr +   Qt  =    Q

Dividing both sides of equation by Q

∴ a    +   r   +  t   = 1

Thus the sum of the coefficient of absorption, the coefficient of reflection and the coefficient of transmission is unity.

#### Diathermanous Substance:

• The substances which can transmit the radiant heat incident upon their surfaces are called diathermanous substances.
• e.g. glass, quartz, gases

#### Characteristics of Diathermanous Substance:

• The substances which can transmit the radiant heat incident upon their surfaces are called diathermanous substances.
• For such substance, the value of the coefficient of transmission (t) is not zero.

• The substances which cannot transmit the radiant heat incident upon their surfaces are called adiathermanous (athermanous) substances.
• e.g. wood, iron copper etc.

• The substances which cannot transmit the radiant heat incident upon their surfaces are called adiathermanous (athermanous) substances.
• For such substance, the value of the coefficient of transmission (t) is zero.

#### Example – 1:

• The coefficients of reflection and transmission of the surface of a thin plate are 0.22 and 0.04 respectively. If 250 J of radiant heat is incident on the surface of the plate, how much heat is absorbed by the surface.
• Solution:
• Given: Coefficient of reflection = r = 0.22, coefficient of transmission = 0.04, radiant heat incident = Q = 250 J
• To Find: Radient heat absorbed = Qa = ?

We have, a + r + t = 1

∴   a + 0.22 + 0.4 = 1

∴   a  = 1 – 0.26 = 0.74

∴   coefficient of absorption = a = 0.74

Now, Qa = aQ = 0.74 × 250 = 185 J

Ans: Radiant heat absorbed = 185 J

#### Example – 2:

• Coefficient of absorption of a body is 0.54 and coefficient of reflection is 0.16. The amount of radiant heat incident on a body is 4000 J. Calculate the amount of heat transmitted through the body.
• Solution:
• Given: Coefficient of absorption = r = 0.54, coefficient of reflection = 0.16, radiant heat incident = Q = 4000 J
• To Find: Radient heattransmitted = Qt = ?

We have, a + r + t = 1

∴   0.54 + 0.16 + t = 1

∴   t  = 1 – 0.70 = 0.30

∴   coefficient of transmission = t = 0.30

Now, Qt = tQ = 0.30 × 4000 = 1200 J

Ans: Radiant heat transmitted = 1200 J

#### Emissive Power:

• Emissive power of a surface is defined as the amount of heat radiated per unit time per unit area of that surface at a given temperature.

• It is found that emissive power of a perfectly black body is greater than the emissive power of any other surface at the same temperature.
• It is denoted by E and its S.I. unit is J/m²s  or  W/m².
• Emissive power of a body depends on
• the temperature of a body
• the nature of the body
• the surface area of the body
• the nature of the surroundings.

#### Coefficient of Emission (Emissivity):

• The coefficient of emission of a surface is defined as the ratio of heat radiated per unit time per unit area of the surface to the heat radiated per unit time per unit area of a perfectly black surface at the same temperature. OR Coefficient of emission of a surface is the ratio of the emissive power of the surface to the emissive power of a perfectly black surface at the same temperature.

• The coefficient of emission is also known as emissivity. It is denoted by letter ‘e’ and it is unitless quantity.  For perfectly black body e = 1, for perfect reflector e = 0 and for all other bodies 0 < e < 1.
• Let E be the emissive power of the body and Eb be the emissive power of the perfectly black body at the same temperature,  then

#### Prevost’s Theory of Heat Exchanges:

• The exchange of heat between a body and its surroundings is explained on the basis of Prevost’s theory.
• The main points of Prevost’s heat exchange theory are
• Every body emits and absorbs thermal radiations at all temperatures except at absolute zero.
• The rate at which thermal radiations are emitted depends upon the absolute temperature of the body.
• The rate of radiation is independent of the temperature of the surrounding.
• Case -1: When a body is at a higher temperature than that of the surroundings then it radiates heat at a faster rate than it absorbs. Therefore it loses heat and its temperature falls.
• Case – 2: When a body is at a lower temperature than that of surroundings then it absorbs heat at a faster rate than the rate at which it radiates heat. Therefore, it gains heat and its temperature rises.
• Case – 3: When the temperature of the body is same as that of the surroundings then the body radiates heat at the same rate at which it absorbs. Therefore, the body neither loses nor gains heat and hence its temperature remains constant. So, even in this state the processes of radiation and absorption continues.

• 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 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

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.