RADIATION :

Radiation is the process in which energy is transferred by means of electromagnetic waves.

All bodies continuously radiate energy in the form of electromagnetic waves. Even an ice cube radiates energy, although so little of it is in the form of visible light that an ice cube cannot be seen in the dark.

The surface of an object plays a significant role in determining how much radiant energy the object will absorb or emit.

Nature of Radiation :

(i) Radiation is electromagnetic wave ( 0 to \(\infty\) wavelength) and travels with velocity of light

(ii) Radiation travels in straight line

(iii) Radiation can travel through vacuum

(iv) Radiation obeys the laws of reflection, refraction, interference, diffraction, polarization etc (all optical phenomena)

Definitions :

(i) Reflecting power or refracting coefficient (r) :

Ratio of amount of radiation reflected by the surface to the amount of heat incident over it

(ii) Transmitting power or transmitting coefficient (t) : Ratio of amount of radiation transmitted by the surface to the amount of heat incident over it.

(iii) Absorption power or absorption coefficient (a) : Ratio of amount of radiation absorbed by the surface to the amount of heat incident over it

Note :-

(a) A radiation incident over a surface is either refracted, transmitted or absorbed, so

\( \mathrm{a}+\mathrm{t}+\mathrm{r}=1\)

(b) The reflecting, refracting and transmitting powers of a surface depends on nature of surface as well as on the wavelength of radiation falling on them

(c) Emissive power (e) : Emissive power is defined as the amount of heat radiated by unit area of the surface in one second at a particular temperature. It's unit is \(\mathrm{Joule} / \mathrm{m}^{2}\) or watt-sec \(/ \mathrm{m}^{2}\).

Black body

The experiments described before lead us to the idea of a perfectly black body; (one which absorbs all the radiation that falls upon it, and reflects and transmits none). The experiments also lead us to suppose that such a body would be the best possible radiator. Laws of Radiation:

(a) Prevost's theory of heat exchange

(i) Mutual exchange of heat between any object and its' surroundings occurs according to their temperatures.

(ii) This process is continuous

(iii) A body at high temperature emits more radiations and absorbs less. Therefore it experiences cooling. While the body at low temperature absorbs more heat and experiences heating

(iv) At absolute zero temperature the radiated energy is zero.

(v) The rate of radiation of energy from a body depends upon it's area, its' temperature and the nature of surface. This does not depends upon temperature difference between body and its surroundings.

(B) Kirchoff's law

The ratio of emissive power to absorptive power is the same for all bodies at a given temperature and is equal to the emissive power of a blackbody at that temperature. Thus,

\( \frac{\mathrm{E}(\text { body })}{\mathrm{a}(\text { body })}=\mathrm{E}(\text { black body }) \)

Kirchhoff's law tells that if a body has high emissive power, it should also have high absorptive power to have the ratio E/a same. Similarly, a body having low emissive power should have low absorptive power. Kirchhoff's law may be easily proved by a simple argument as described below.

Consider two bodies A and B of identical geometerical shapes placed in an enclosure. Suppose A equilibrium, both the bodies will have the same temperature as the temperature of the enclosure. Suppose an amount \(\Delta \mathrm{U}\) of radiation falls on the body A in a given time \(\Delta \mathrm{t}\). As A and B have the same geometrical shapes, the radiation falling on the blackbody B is also \(\Delta \mathrm{U}\). As the temperature of the blackbody remains constant, it also emits an amount \(\Delta \mathrm{U}\) of radiation in that time. If the emissive power of the blackbody is \(\mathrm{E}_{0}\), we have

\(\Delta \mathrm{U} \propto \mathrm{E}_{0}\) or \(\Delta \mathrm{U}=\mathrm{kE}_{0}\)

where \(\mathrm{k}\) is a constant.

Let the absorptive power of A be a. Thus, it absorbs an amount a \(\Delta \mathrm{U}\) of the radiation falling on it in time \(\Delta \mathrm{t}\). As its temperature remains constant, it must also emit the same amount a \(\Delta \mathrm{U}\) in that time. If the emissive power of the body \(\mathrm{A}\) is \(\mathrm{E}\), we have, \(\mathrm{a} \Delta \mathrm{U}=\mathrm{kE}\)

The same proportionality constant \(\mathrm{k}\) is used in (i) and (ii) because the two bodies have identical geometrical shapes and radiation emitted in the same time \(\Delta \mathrm{t}\) is considered.

From (i) and (ii),

\( a=\frac{E}{E_{0}} \)

 \(\frac{E}{a}=E_{0} \)

 \(\frac{E(\text { body) }}{a(b o d y)}=E(\text { black body })\)

This proves Kirchhoff's law. 

Stefan's radiation law

An idealized body that absorbs all the radiation incident upon it is called a blackbody. A blackbody absorbs not only all visible light, but infrared, ultraviolet, and all other wavelengths of electromagnetic radiation. It turns out that a good absorber is also a good emitter of radiation. A blackbody emits more radiant ower per unit surface area than any real object at the same temperature. The rate at which a blackbody emits radiation per unit surface area is proportional to the fourth power of the absolute temperature.

\( \mathrm{P}=\sigma \mathrm{AT}^{4} \)

In equation, \(A\) is the surface area and \(\mathrm{T}\) is the surface temperature of the blackbody in kelvins. Since Stefan's law involves the absolute temperature and not a temperature difference, \({ }^{\circ} \mathrm{C}\) cannot be substituted. The universal constant \(\sigma\) (Greek letter sigma) is called Stefan's constant :

\( \sigma=5.670 \times 10^{-8} \mathrm{~W} /\left(\mathrm{m}^{2} . \mathrm{K}^{4}\right) \)

The fourth-power temperature dependence implies that the power emitted is extremely sensitive to temperature changes. If the absolute temperature of a body doubles, the energy emitted increases by a factor of \(2^{4}=16\).

Since real bodies are not perfect absorbers and therefore emit less than a blackbody, we define the emissivity (e) as the ratio of the emitted power of the body to that of a blackbody at the same temperature. Then Stefan's law becomes.

Stefan's law of radiation :

\(\mathrm{P}=\mathrm{e} \sigma \mathrm{AT}^{4}\)

The emissivity ranges from 0 to \(1 ; \mathrm{e}=1\) for a perfect radiator and absorber (a blackbody) and \(\mathrm{e}=0\) for a perfect reflector. The emissivity for polished aluminium, an excellent reflector, is about \(0.05\); for soot (carbon black) it is about \(0.95\).

Hot object in enclosure

Consider a black body at a temperature of \(\mathrm{T}_{0}\) where \(\mathrm{T}_{0}\) is the temperature of the room or enclosure containing the body. Since the body is in temperature equilibrium, the energy per second it radiates must equal the energy per second it absorbs. If A is the surface area of the body, then, from Stefan's law, energy per second radiated \(=\sigma \mathrm{AT}_{0}^{4}\)

So the energy per second absorbed from the surroundings or encloser \(=\sigma \mathrm{AT}_{0}^{4}\).

Now suppose the black body \(\mathrm{X}\) is heated electrically by a heater of power \(\mathrm{W}\) watts and finally reaches a constant temperature T. In this case, from Prevost's theory, energy per second from heater, \(\mathrm{W}=\) net energy per second radiated by \(\mathrm{X}\) The net energy per second radiated by \(\mathrm{X}=\sigma \mathrm{AT}^{4}-\sigma \mathrm{AT}_{0}^{4}\). So

\(\mathrm{W}=\sigma \mathrm{AT}^{4}-\sigma \mathrm{AT}_{0}^{4}=\sigma \mathrm{A}\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right)\)

Illustration :

A black body emits 10 watts per \(\mathrm{cm}^{2}\) at \(327^{\circ} \mathrm{C}\). The sun radiates \(10^{5}\) watt per \(\mathrm{cm}^{2}\). Then what is the temperature of the sun?

(1) \(5000 \mathrm{~K}\)

(2) \(6000 \mathrm{~K}\)

(3) \(7000 \mathrm{~K}\)

(4) \(8000 \mathrm{~K}\)

Sol. \(\because \frac{\mathrm{E}_{\text {sun }}}{\mathrm{E}_{\text {body }}}=\left(\frac{\mathrm{T}_{\text {sun }}}{\mathrm{T}_{\text {body }}}\right)^{4}\)

\(\therefore \frac{\mathrm{T}_{\text {sun }}}{\mathrm{T}_{\text {body }}}=\left(\frac{10^{5}}{10}\right)^{1 / 4} \\

 \therefore T_{\text {sun }}=6000 \mathrm{~K} \)

Illustration :

A bulb made of tungsten filament of surface area \(0.5 \mathrm{~cm}^{2}\) is heated to a temperature \(3000 \mathrm{k}\) when operated at \(220 \mathrm{~V}\). The emissivity of the filament is \(e=0.35\) and take \(\sigma=5.7 \times 10^{-8} \mathrm{mks}\) units. Then the wattage of the bulb is .... (calculate)

(1) \(81 \mathrm{~W}\)

(2) \(.81 \mathrm{~W}\)

(3) \(81.2 \mathrm{~W}\)

(4) \(8.12 \mathrm{~W}\)

Sol. The emissive power watt \(/ m^{2}\) is

\(E=e \sigma T^{4}\)

Therefore the power of the bulb is

\(P=e \times \operatorname{area}(\) Watts)

\(\therefore P=e A \sigma T^{4}\)

\(\therefore P=0.35 \times 0.5 \times 10^{-4} \times 5,7 \times 10^{-8} \times(3000)^{4}\)

\(\Rightarrow P=80.8 \mathrm{~W}\)

Newtons' Law of Cooling :

Suppose, a Board of surface area \(A\) at an absolute temperature \(T\) is kept in a surrounding having a lower temperature \(\mathrm{T}_{0}\). The net rate of loss of thermal energy from the body due to radiation is

\( \Delta \mathrm{u}=\mathrm{e} \sigma \mathrm{A}\left(\mathrm{T}^{4}-\mathrm{T}_{0}^{4}\right)\)

If the temperature difference is small, we can write

\( \mathrm{T}=\mathrm{T}_{0}+\Delta \mathrm{T} \)

or,

\(T^{4}-T_{0}^{4}=\left(T_{0}+\Delta T\right)^{4}-T_{0}^{4}\)

\(=\mathrm{T}_{0}^{4}\left(1+\frac{\Delta \mathrm{T}}{\mathrm{T}_{0}}\right)^{4}-\mathrm{T}_{0}^{4} \)

\( =\mathrm{T}_{0}^{4}\left(1+4 \frac{\Delta \mathrm{T}}{\mathrm{T}_{0}}+\text { higherpowersof } \frac{\Delta \mathrm{T}}{\mathrm{T}_{0}}\right)-\mathrm{T}_{0}^{4} \)

 \(\approx 4 \mathrm{~T}_{0}^{3} \Delta \mathrm{T}=4 \mathrm{~T}_{0}^{3}\left(\mathrm{~T}-\mathrm{T}_{0}\right)\)

Thus, \(\Delta u=4 e \sigma A T_{0}^{3}\left(T-T_{0}\right)\)

\(=b\left(T-T_{\theta}\right)\)

\(-\frac{\mathrm{dT}}{\mathrm{dt}}=\frac{\mathrm{b}}{\mathrm{ms}}\left(\mathrm{T}-\mathrm{T}_{0}\right)\)

Illustration :

A metal ball cools from \(62^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in 10 min and to \(42^{\circ} \mathrm{C}\) the next ten minutes. What will be its temperature at the end of next ten minutes?

(1) \(3.67^{\circ} \mathrm{C}\)

(2) \(36.7^{\circ} \mathrm{C}\)

(3) \(.376^{\circ} \mathrm{C}\)

(4) \(367^{\circ} \mathrm{C}\)

Sol. Use \(\frac{\theta_{1}-\theta_{2}}{\mathrm{t}}=\mathrm{K}\left(\frac{\theta_{1}-\theta_{2}}{2} \theta_{0}\right)\)

to get the following equations

\(\frac{62-50}{10}=\mathrm{K}\left(\frac{62+50}{2}-\theta_{0}\right) \text { and } \)

 \(\frac{50-42}{10}=\left(\frac{50+42}{2}-\theta_{0}\right)\)

Divide 2 by 1 and solve to get \(\theta_{0}\)

\(\frac{8}{12}=\frac{46-\theta_{0}}{56-\theta_{0}} \)

\(\Rightarrow \theta_{0}=26\)

Let after the next 10 min the temperature falls to then

\(\frac{42-\theta}{10}=\mathrm{K}\left(\frac{42+\theta}{2}-26\right)\)

From (3) using value of \(\theta_{0}=26\) we get

\(\frac{12}{10}-\mathrm{k}(56-26)\)

Divide (4) by (5) to get

\(\frac{42-\theta}{12}=\frac{42+\theta-52}{2(56-26)} \)

 \(\Rightarrow \theta=36.7^{\circ} \mathrm{C}\)

Illustration :

\(0.15 \mathrm{~kg}\) of water is filled in caloriemeter of copper with water equivalent .01 kg. It's rate of cooling is \(88 \times 10^{-3} \mathrm{kcal} / \mathrm{min}\). If xanthate oil is filled in place of water at the same temperature and the same volume. Now find the rate of cooling and rate of fall of temperature. (density of xanthate oil \(=800 \mathrm{~kg} / \mathrm{m}^{3}, \mathrm{sp}\) heat of xanthate oil \(=.42 \mathrm{kcal} / \mathrm{kg}^{\circ} \mathrm{C}\) )

Sol. Since both liquids are cooled under same conditions, rate of cooling will be same

Now, rate of cooling \(=\frac{\mathrm{dQ}}{\mathrm{dt}}=(\mathrm{ms}+\mathrm{W}) \frac{\mathrm{d} \theta}{\mathrm{dt}}\)

\(=88 \times 10^{-3} \mathrm{kcal} / \mathrm{min}\)

Here, \(m=0.15 \times 10^{-3} \times 8 \times 10^{2}=.12 \mathrm{~kg}\)

\(W=.01\) 

\(\Rightarrow \frac{\mathrm{d} \theta}{\mathrm{dt}}=\frac{1}{\mathrm{~ms}+\mathrm{W}} \frac{\mathrm{dQ}}{\mathrm{dt}}=1.457^{\circ} \mathrm{C} / \mathrm{min}\)

Spectral emissive power

To speak of the intensity of a single wavelength is meaningless. The slit of the spectrometer always gathers a band of wavelengths, the narrower the slit the narrower the band-and we always speak of the intensity of a given band. We express it as follows :

energy radiated \(\mathrm{m}^{-2} \mathrm{~s}^{-1}\), in band \(\lambda\) to \(\lambda+\Delta \lambda=\mathrm{E}_{\lambda} \Delta \lambda\)

The quantity \(\mathrm{E}_{\lambda}\) is called emissive power of a black body for the wavelength \(\lambda\) and at the given temperature ; its definition follows from equation \(\lambda\) to \(\lambda+\Delta \lambda=\mathrm{E}_{\lambda} \Delta \lambda\) :

\(\mathrm{E}_{\lambda}=\frac{\text { energy radiated } \mathrm{m}^{-2} \mathrm{~s}^{-1} \text {, in band } \lambda+\Delta \lambda}{\text { band width }, \Delta \lambda} \)

\( \mathrm{E}_{\lambda}=\frac{\text { power radiated } \mathrm{m}^{-2} \text { in band } \lambda \text { to } \Delta \lambda}{\Delta \lambda}\)

THERMAL EXPANSION, CALORIMETRY & HEAT TRANSFER

In the figure, \(\mathrm{E}_{\lambda}\) is expressed in watts per \(\mathrm{m}^{2}\) per nanometre \(\left(10^{-9} \mathrm{~m}\right)\).

The quantity \(\mathrm{E}_{\lambda} \Delta \lambda\) in equation \(\lambda\) to \(\lambda+\Delta \lambda=\mathrm{E}_{\lambda} \Delta \lambda\) is the area beneath the radiation curve between the wavelength \(\lambda\) and \(\Delta \lambda\) (figure). Thus the energy radiated per meter \({ }^{2}\) per second between those wavelengths is proportional to that area.

Similarly, the total radiation emitted per metre \(^{2}\) per second over all wavelengths is proportional to the area under the whole curve.

Figure : Definition of \(E \cdot \lambda_{m}\) and \(E_{\lambda m}\)

Laws of black body radiation

The curves of figure can be explained only by Planck's quantum theory of radiation, which is outside our scope. Both theory and experiment lead to three generalisations, which together describe well the properties of black body radiation.

(i) If \(\lambda_{\mathrm{m}}\) is the wavelength of the peak of the curve for \(\mathrm{T}\) (in \(\mathrm{K}\) ), then

\(\lambda_{\mathrm{m}} \mathrm{T}=\text { constant }\)

The value of the constant is \(2.9 \times 10^{-3} \mathrm{~m} \mathrm{~K}\). In figure the dotted line is the locus of the peaks of the curves for different temperatures.

The relationship in (2) is sometimes called Wien's displacement law.

(ii) If \(E_{\lambda m}\) is the height of the peak of the curve for the temperature \(T\), then

\(\mathrm{E}_{\lambda \mathrm{m}} \propto \mathrm{T}^{5} \)

Figure : Distribution of intensity in black body radiation 

THERMAL EXPANSION, CALORIMETRY \& HEAT TRANSFER

(iii) If \(E\) is the total energy radiated per metre \({ }^{2}\) per second at a temperature \(T\), which is represented by the total area under the particular \(\mathrm{E}_{\lambda}-\lambda\) curve, then

\(\mathrm{E}=\sigma \mathrm{T}^{4} \)

So in figure, which shows four \(\mathrm{E}_{\lambda}-\lambda\) graphs at different temperatures \(\mathrm{T}\), the total area below the graphs should be proportional to the corresponding value of \(\mathrm{T}^{4}\).

Illustration :

At \(1600 \mathrm{~K}\) maximum radiation is emitted at a wavelength of \(2 \mu_{M}\). Then the corresponding wavelength at \(2000 \mathrm{~K}\) will be :

(1) \(1.6 \mu \mathrm{m}\)

(2) \(16 \mu m\)

(3) \(160 \mu m\)

(4) \(.16 \mu m\)

Sol. Using \(\lambda_{\mathrm{m}_{1}} T_{1}=\lambda_{\mathrm{m}_{2}} T_{2}\)

\(\therefore \quad \lambda_{\mathrm{m}_{2}}=\frac{\lambda_{\mathrm{m}_{1}} \mathrm{~T}_{1}}{\mathrm{~T}_{2}} \)

\( \therefore \quad \lambda_{\mathrm{m}_{2}}=\frac{2 \times 10^{-6} \times 1600}{2000} \)

\( \Rightarrow \quad \lambda_{\mathrm{m}_{2}}=1.6 \mu \mathrm{m}\)