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Chapter Notes: Heat Transfer Physics Class11

Notes for Heat Transfer chapter of class 11 physics. Dronstudy provides free comprehensive chapterwise class 11 physics notes with proper images & diagram.

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 $\Delta Q$, whose faces are kept at different temperatures. We measure the heat $\Delta Q$ that flows perpendicular to the faces during time $\Delta T$. Experiment shows that $\Delta Q$ is proportional to $\Delta T$ and to the cross-sectional area A for a given temperature difference $\Delta T$, and that $\Delta Q$ is proportional to $\Delta T$/$\Delta X$ for a given $\Delta T$ and A, providing both $\Delta T$ and $\Delta X$ are small. That is,
${\textstyle{{\Delta Q} \over {\Delta t}}}\alpha A{\textstyle{{\Delta T} \over {\Delta x}}}$
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

${{dQ}\over {dt}} =- kA{{dT} \over {dx}}$

Here${{dQ} \over {dt}}$is the time rate of heat transfer across the area$A,{{dT} \over {dx}}$is 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 ${{dQ} \over {dt}}$to the positive when ${{dT} \over {dx}}$is 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.
$q =- KA{{dT} \over {dx}}$
Heat flow density, $H={q \over A} =- K{{dT} \over {dx}}$
$\Rightarrow$$H\int\limits_0^l {dx}=- \int\limits_{{T_1}}^{{T_2}} {{\alpha \over T}dT}$ $Hl = \alpha\ln \left| {{{{T_1}} \over {{T_2}}}} \right|$
Hence $H = {\alpha\over l}\ln \left| {{{{T_1}} \over {{T_2}}}} \right|$
Once again,$H\int\limits_0^x {dx}=-\alpha \int\limits_{{T_1}}^T {{{dT} \over T}}$$\Rightarrow$$Hx=\alpha\ln \left| {{{{T_1}} \over T}} \right|$$\Rightarrow$${{\alpha x} \over l}\ln \left| {{{{T_1}} \over {{T_2}}}} \right|= \alpha \ln \left| {{{{T_1}} \over T}} \right|$
Hence$T={T_1}{\left( {{{{T_2}} \over {{T_1}}}} \right)^{x/l}}$
Alternative method.
The rate of heat flow through a conductor is given by$q = {{dQ} \over {dt}} = {{KA\Delta T} \over l}$
where K is the average conductivity.
${k_{av}} = \int\limits_{{T_1}}^{{T_2}} {{{KdT} \over {{T_2} - {T_1}}}} = {{\alpha \ln \left| {{{{T_2}} \over {{T_1}}}} \right|} \over {{T_2} - {T_1}}}$
$q = {{\alpha \ln \left| {{{{T_2}} \over {{T_1}}}} \right|} \over {\left( {{T_2} - {T_1}} \right)}}\left[ {{{\left( {{T_1} - {T_2}} \right)A} \over l}} \right]$
or heat flow density$= {q \over A} = {\alpha \over l}\ln \left| {{{{T_1}} \over {{T_2}}}} \right|$            (1)
Let T be the temperature at a distance x from the left end as shown in the figure. Then  $k{'_{av}} = {{\alpha \ln \left| {{T \over {{T_1}}}} \right|} \over {T - {T_1}}}$ and heat flow density is
${{q'} \over A} = {\alpha \over x}\ln \left| {{{{T_1}} \over T}} \right|$
Equating (1) and (2) we get
T = ${T_1}{\left( {{{{T_2}} \over {{T_1}}}} \right)^{x/l}}$

Thermal and Electrical Conductivity

There exists an useful analog between thermal conductivity and electrical conductivity

 Thermal Conduction ${{dQ} \over {dt}} = kA\left( {{{{T_1} - {T_2}} \over l}} \right)$  Heat flows from higher temperature to lower temperature The rate of heat flow is called the heat current.         I = ${{dQ} \over {dt}}$  Thermal resistance is defined as        RT  =${l \over {kA}}$ Ohm’s law for heat conduction may be stated as           I =${{{T_1} - {T_2}} \over {{R_T}}}$ Electrical Conduction ${{dq} \over {dt}} = {A \over \rho }\left( {{{{V_1} - {V_2}} \over l}} \right)$  Charges flow from higher potential to lower potential The rate of charge flow is called the electric current          I =${{dq} \over {dt}}$  Electrical resistance is defined as       RE =${{\rho \,l} \over A}$  Ohm’s law for electric conduction may be stated as             I =${{{V_1} - {V_2}} \over {{R_E}}}$ 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
$R = {R_1} + R{}_2$ (series)
${1 \over R} = {1 \over {{R_1}}} + {1 \over {{R_2}}}$ (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
${R_1} = {1 \over {2KA}}$
$R{_2} = {1 \over {KA}}$
$R{}_3 = {{2l} \over {KA}}$ Let 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

Req = R + 2R + 4R = 7R
The heat current is

$I = {{{T_A} - {T_B}} \over {{R_{eq}}}} = {{100 - 0} \over {7R}} = {{100} \over {7R}}$
The temperature T1 and T2 of the junction are
${T_1} = {\rm{ }}{T_A}-{\rm{ }}IR = {\rm{ }}100{\rm{ }} - \left( {{{100} \over {7R}}} \right)R = {{{600} \over 7}^o}C$
${T_1} = {\rm{ }}{T_A}-{\rm{ }}I(R + 2R) = {\rm{ }}100{\rm{ }} - \left( {{{100} \over {7R}}} \right)\left( {3R} \right) = {{{400} \over 7}^o}C$

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 Kglass = 0.8 Wm-1K-1 Kair = 0.08 Wm-1K
-1

Solution:

(a) Total thermal resistance is $R = {{2{t_1}} \over {{K_1}{A_1}}} + {{{t_2}} \over {{K_2}{A_2}}}$
Here,  A1 = A2 = 1m2 , t1 = 0.01 m,

t2 = 0.05 m;
K1 = 0.8 Wm-1K-1,
K2 = 0.08Wm-1K-1.

$R = {{2\left( {0.01} \right)} \over {\left( {0.8} \right)\left( 1 \right)}} + {{0.05} \over {\left( {0.08} \right)\left( 1 \right)}} = 0.65{W^{ - 1}}K$
Heat current = ${{dQ} \over {dt}} = {{\Delta T} \over R} = {{27 - 0} \over {0.65}} = 41.5W$

(b) ${T_1} = {\rm{ }}27{\rm{ }} - \left( {{{dQ} \over {dt}}} \right){{{t_1}} \over {{K_1}{A_1}}} = 27 - 0.52 = {26.48^ \circ }C$
${T_2} = {\rm{ }}0{\rm{ }} + \left( {{{dq} \over {dt}}} \right){{{t_1}} \over {{K_1}{A_1}}} = {0.52^ \circ }C$

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.

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.
$a \le 1$

(iii) Spectral Absorptive Power (${a_\lambda }$)
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 ${\lambda _1}$ and bad absorber for wavelength ${\lambda _2}$, it means its absorptive power for wavelength ${\lambda _1}$ is greater than the absorptive power for ${\lambda _2}$.
Thus, spectral absorptive power ${a_\lambda }$ 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
$a = \int\limits_0^\infty{{a_\lambda }a\lambda }$
${a_\lambda } \le1$
For perfectly black body  ${a_\lambda } = 1$

Example 4

100 units of energy is incident on a surface. In this 20 units are of wavelength ${\lambda _1}$, 30 units are of wavelength ${\lambda _2}$ 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 ${\lambda _1}$ and 25 units are of ${\lambda _2}$. Find a,  and.

Solution:

Total absorptive power $a = {{60} \over {100}} = 0.6$
and spectral absorptive power for ${\lambda _1} = {a_{{\lambda _1}}}={{5}\over {20}} = 0.25$
Spectral absorptive power for ${\lambda _2} = {a_{{\lambda _1}}} = {{25} \over {30}} = 0.83$
From this example it is clear that total absorption power of the surface is only 0.6 whereas ${a_{{\lambda _1}}}$ = is 0.83 (> 0.6) i.e. the surface is good absorber of wavelength ${\lambda _2}$.