Modes of Transference of Heat

Heat can be transferred from one place to another by three different methods, namely, conduction, convection and radiation. Conduction usually takes place in solids, convection in liquids and gases and no medium is required for radiation.

Heat Conduction

This transfer takes place due to molecular collisions. The molecules at one end of the rod gain heat from the heat source and their average kinetic energy increases. As these molecules collide with neighboring molecules having less kinetic energy, the energy is shared between these two groups. The kinetic energy of these neighboring molecules increases. As they collide with their neighbors on the colder side, they transfer energy to them. This way, heat is passed along the rod from molecule to molecule. The average position of a molecule does not change and hence, there is no mass movement of matter.
The transfer of energy arising form the temperature difference between adjacent parts of a body is called heat conduction. Consider a slab of material of cross-sectional area A and thickness ΔQ, whose faces are kept at different temperatures. We measure the heat ΔQ that flows perpendicular to the faces during time ΔT. Experiment shows that ΔQ is proportional to ΔT and to the cross-sectional area A for a given temperature difference ΔT, and that ΔQ is proportional to ΔT/ΔX for a given ΔT and A, providing both ΔT and ΔX are small. That is,
ΔQΔtαAΔTΔx
In the limit of a slab of infinitesimal thickness dx, across which there is a temperature difference dT, we obtain the fundamental law of heat conduction

dQdt=−kAdTdx

HeredQdtis the time rate of heat transfer across the areaA,dTdxis called the temperature gradient, and k is a constant proportionality called the thermal conductivity. We choose the direction of heat flow to be the direction in which x increases; since heat flows in the direction of decreasing T, we introduce a minus sign in equation, (i.e., we wish dQdtto the positive when dTdxis negative).
The phenomenon of heat conduction also shows that the concepts of heat and temperature are distinctly different. Different rods, having the same temperature difference between their ends, may transfer entirely different quantities of heat in the same time.

Example 1

A rod of length l with thermally insulated lateral surface consists of a material whose heat conductivity coefficient varies with temperature as K = a/T, where a is a constant. The ends of the rod are kept at temperatures T₁, and T₂ (T₁ > T₂). Find the function T(x) where x is the distance from the end whose temperature is T, and the heat flow density.

Solution:

Use fundamental law of heat conduction.
q=−KAdTdx
Heat flow density, H=qA=−KdTdx
⇒H∫0ldx=−∫T1T2αTdT Hl=αln∣∣T1T2∣∣
Hence H=αlln∣∣T1T2∣∣
Once again,H∫0xdx=−α∫T1TdTT⇒Hx=αln∣∣T1T∣∣⇒αxlln∣∣T1T2∣∣=αln∣∣T1T∣∣
HenceT=T1(T2T1)x/l
Alternative method.
The rate of heat flow through a conductor is given byq=dQdt=KAΔTl
where K is the average conductivity.
kav=∫T1T2KdTT2−T1=αln∣∣T2T1∣∣T2−T1
q=αln∣∣T2T1∣∣(T2−T1)[(T1−T2)Al]
or heat flow density=qA=αlln∣∣T1T2∣∣            (1)
Let T be the temperature at a distance x from the left end as shown in the figure.
Then  k′av=αln∣∣TT1∣∣T−T1 and heat flow density is
q′A=αxln∣∣T1T∣∣
Equating (1) and (2) we get
T = T1(T2T1)x/l

Thermal and Electrical Conductivity

There exists an useful analog between thermal conductivity and electrical conductivity

Using the above analogy, a problem of heat conduction may be transformed into a problem of electrical conduction and can be easily using the formulae of electric circuits. Such as, for a series and parallel combination the equivalent resistance is defined a
R=R1+R2 (series)
1R=1R1+1R2 (parallel)

Example 2

Three rods of same length l and cross-sectional area A are joined in series between two heat reservoirs as shown in the figure. Their conductivity are 2K, K and K/2 respectively. Assuming that the conductors are logged from the surroundings find the temperatures T₁ and T₂ at the junction in the steady state condition.
Solution:

The thermal resistance of the three conductors are
R1=12KA
R2=1KA
R3=2lKALet R₁  = R then R₂ = 2R  and R₃ = 4R
Thus, the electric analogy of the heat conduction system is shown in the figure. The equivalent resistance of the system is
R_eq = R + 2R + 4R = 7R
The heat current is
I=TA−TBReq=100−07R=1007R
The temperature T₁ and T₂ of the junction are
T1=TA−IR=100−(1007R)R=6007oC
T1=TA−I(R+2R)=100−(1007R)(3R)=4007oC

Example 3

A double – pane window consists of two glass sheets each of area 1m²  and thickness 0.01 m separated by a 0.05 m thick stagnant air space. In the steady state,  the room glass interface and the glass outdoor interface are at constant temperatures of 27°C and 0°C , respectively.
(a) Calculate the rate of heat flow through the window pane.
(b) Find the temperatures of other interfaces.
Take, thermal conductivities  as K_glass = 0.8 Wm^-1K^-1 K_air = 0.08 Wm^-1K^-1 

Solution:

(a) Total thermal resistance is
R=2t1K1A1+t2K2A2
Here,  A₁ = A₂ = 1m² , t₁ = 0.01 m,
t₂ = 0.05 m;
K₁ = 0.8 Wm^-1K^-1,
K₂ = 0.08Wm^-1K^-1.
R=2(0.01)(0.8)(1)+0.05(0.08)(1)=0.65W−1K
Heat current = dQdt=ΔTR=27−00.65=41.5W

(b) T1=27−(dQdt)t1K1A1=27−0.52=26.48∘C
T2=0+(dqdt)t1K1A1=0.52∘C

Heat Convection

In convection, heat is transferred from one place to the other by the actual motion of heated material. For example, in a hot air blower, air is heated by a heating element and is blown by a fan. The air carries the heat wherever it goes. When water is kept in a vessel and heated on a stove, the water at the bottom gets heat due to conduction through the vessel’s bottom. Its density decreases and consequently it rises. Thus, the heat is carried from the bottom to the top by the actual movement of the parts of the water. If the heated material is forced to move, say by a blower or a pump, the process of heat transfer is called forced convection. If the material moves due to difference in density, it is called natural or free convection.

Heat Radiation

The process in which heat is transferred from one place to the other without any intervening medium is called radiation. All bodies at all temperatures and at all times radiate energy in the form of electromagnetic waves, called radiant energy. Emission rate is faster at higher temperatures. In this radiant energy, electromagnetic waves of wavelength ranging from 10^-4 mm to 1 mm are called infrared radiation or heat radiation. According to prevost’s theory of exchange besides radiating heat radiation or thermal radiation all bodies also absorb part of the thermal radiation falling on them emitted by  the surrounding bodies. If a body radiates more, what it absorbs, its temperature falls and vice versa.

BLACK BODY RADIATION

Basic Definitions

(i) Perfectly Black Body
A body which absorbs all the radiations incident on it is called a perfectly black body.

(ii) Absorptive Power of a surface (a)
The ratio of the radiant energy absorbed by it in a given time to the total radiant energy incident on it in the same time is called the absorptive power (a) of the surface. Since it is a pure ratio it has no units and dimensions. The absorptive power of a perfectly black body is maximum and its value is unity.
a≤1

(iii) Spectral Absorptive Power (aλ)
Absorptive power refers to all wavelengths (total radiant energy). However, any surface will have different values of absorptive powers for different wavelengths. A surface may be a good absorber for a wavelength λ1 and bad absorber for wavelength λ2, it means its absorptive power for wavelength λ1 is greater than the absorptive power for λ2.
Thus, spectral absorptive power aλ is defined as the ratio of the radiant energy of a given wavelength absorbed by a given surface in a given time to the total radiant energy of that wavelength incident in the same time on the same surface. Or we can say that it is absorptive power for that particular wavelength. It is now obvious that wavelength incident in the same time on the same surface. Or we can say that it is absorptive power for that particular wavelength. It is now obvious that
a=∫0∞aλaλ
aλ≤1
For perfectly black body  aλ=1

Example 4

100 units of energy is incident on a surface. In this 20 units are of wavelength λ1, 30 units are of wavelength λ2 and rest 50 units are other wavelengths. The total 60 units of energy is absorbed by the surface. In this 60 units 5 units are of λ1 and 25 units are of λ2. Find a,  and.

Solution:

Total absorptive power a=60100=0.6
and spectral absorptive power for λ1=aλ1=520=0.25
Spectral absorptive power for λ2=aλ1=2530=0.83
From this example it is clear that total absorption power of the surface is only 0.6 whereas aλ1 = is 0.83 (> 0.6) i.e. the surface is good absorber of wavelength λ2.