CONDUCTION :

When the atoms or molecules in a hotter part of the material vibrate or move with greater energy than those in a cooler part then by means of collisions, the more energetic molecules pass on some of the their energy to their less energetic neighbors. For example, imagine a gas filling the space between two walls that face each other and are maintained at different temperatures. Molecules strike, the hotter wall, absorb energy from it, and rebound with a greater kinetic energy than when they arrived.

As these more energetic molecules collide with their less energetic neighbors, they transfer some of their energy to them. Eventually, this energy is passed on until it reaches the molecules next to the cooler wall. These molecules, in turn, collide with the wall, giving up some of their energy to it in the process. Through such molecular collisions, heat is conducted from the hotter to the cooler wall.

Steady and variable state :

consider a metal rod \(\mathrm{AB}\), with one end \(\mathrm{A}\) inserted into a chamber containing a heater with other end \(\mathrm{B}\) left free and exposed to the surrounding as shown fig. The rod is thermally insulated sideways with some bad conductor of heat say cotton or felt. Three thermometers are installed in the rod at three distinct sections numbered (1), (2) and (3). Initially, the enitre system is at the room temperature and the three thermometers show the same room temperature. The heater is then switched on. The end A first gets heated up and simultaneously heat is conducted to the adjacent sections towards end B.

Due to heat absorption at each sections, the corresponding temperatures start rising with \(T_{1}>T_{2}>T_{3}\). Such a state, encountered initially, is known as a variable state. It is in this state, the heat coming through end \(\mathrm{A}\), is continuously absorbed at each sections with a temperature rise as time elapses.

After some time when the temperature of end B becomes equal to that of surrounding and thus becomes constant. Consequently, the heat absorption at different sections of the rod ceases. So, there is no more rise in temperatures of any sections. Such a state is known as a steady state. It is in the steady state that temperatures of each section attain steady (constant) value with time.

THERMAL EXPANSION, CALORIMETRY \& HEAT TRANSFER

Consider a portion of the rod of cross sectional area A as shown fig. Let the temperatures of the two sections separated by a length \(L\) be \(T_{1}\) and \(T_{2}\) respectively (with \(T_{1}>T_{2}\) ). Temperature gradient (fall in temperature per unit length) along the length of the rod will be \(\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{~L}}\).

Experiments reveal that, the amount of heat \(\Delta \mathrm{Q}\) flowing in a small time interval \(\Delta \mathrm{t}\) is given by

\( \frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}} \propto \frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{~L}} \)

\(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}} \propto \mathrm{A}\)

Combining eqs. (i) and (ii), \(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}} \propto \frac{\mathrm{A}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{L}}\)

or \(\quad \frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\mathrm{KA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{L}}=-\frac{\mathrm{KA}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)}{\mathrm{L}}\)

where \(\mathrm{K}\) is a positive constant known as the coefficient of thermal conduction or simply thermal conductivity of the material of the rod. Substances having high values of \(\mathrm{K}\) are good conductors of heat, while those having low values are poor conductors.

In the differential form, the above equation can be written as

\(\frac{\mathrm{dQ}}{\mathrm{dt}}=-\mathrm{KA} \frac{\mathrm{dT}}{\mathrm{dx}}\)

where length is measured along \(\mathrm{X}\)-axis. (along the direction of heat flow).

Thermal resistance

From equation \(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\mathrm{KA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{L}}\)

where \(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\) rate of heat flow \(=\) thermal current \(=i\) (say)

\(\frac{\mathrm{L}}{\mathrm{KA}}=\frac{\mathrm{T}_{1}-\mathrm{T}_{2}}{\mathrm{i}}=\frac{\Delta \mathrm{T}}{\mathrm{i}}\)

which is analogous to Ohm's law \(\mathrm{R}=\frac{\Delta \mathrm{V}}{\mathrm{i}}\); where \(\mathrm{R}\) is the electrical resistance, \(\Delta \mathrm{V}\) is potential difference and \(i\) the electric current.

So, \(\frac{\mathrm{L}}{\mathrm{KA}}=\mathrm{R}_{\mathrm{th}}\) is known as the thermal resistance.

The reciprocal of thermal conductivity, i.e., \(\frac{1}{\mathrm{~K}}\) is called thermal resistivity or thermal specific resistance. Note :

In thermodynamics heat current, temperature difference, and thermal resistance resembles, electric current, potential difference and electrical resistance respectively in electrodynamics. Illustration :

Calculate the quantity of heat conducted through \(2 \mathrm{~m}^{2}\) of a brick wall \(12 \mathrm{~cm}\) thick in 1 hour if the temperature on one side is \(8^{\circ} \mathrm{C}\) and on the other side is \(28^{\circ} \mathrm{C}\). (Thermal conductivity of brick \(\left.=0.13 \mathrm{Wm}^{-1} \mathrm{~K}^{-1}\right)\)

Sol. Temperature gradient \(=\frac{28-8}{12 \times 10^{-2}} K^{-1}\) and \(t=3600 \mathrm{~s}\)

\(\therefore \quad  Q=k A t \times \text { temperature gradient } \)

\(=0.13 \times 2 \times 3600 \times \frac{28-8}{12 \times 10^{-2}} \)

\(=  156000 \mathrm{~J}\)

Illustration :

Two parallel plates \(A\) and B are joined together to form a compound plate (figure). The thicknesses of the plates are \(4.0 \mathrm{~cm}\) and \(2.5 \mathrm{~cm}\) respectively and the area of cross-section is \(100 \mathrm{~cm}^{2}\) for each plate. The thermal conductivities are \(K_{A}=200 \mathrm{~W} / \mathrm{m}^{-\circ} \mathrm{C}\) for the plate \(\mathrm{A}\) and \(\mathrm{K}_{B}=400 \mathrm{~W} / \mathrm{m}^{-}\) \({ }^{\circ} \mathrm{C}\) for the plate \(B\). The outer surface of the plate \(A\) is maintained at \(100^{\circ} \mathrm{C}\) and the outer surface of the plate \(B\) is maintained at \(0^{\circ} \mathrm{C}\). Find (a) the rate of heat flow through any cross-section, (b) the temperature at the interface and (c) the equivalent thermal conductivity of the compound plate.

Sol.(a) Let the temperature of the interface be \(\theta\).

The area of cross-section of each pate is \(A=100 \mathrm{~cm}^{2}=0.01 \mathrm{~m}^{2}\). The thicknesses are \(x_{A}=0.04 \mathrm{~m}\) and \(x_{B}=0.025 \mathrm{~m}\)

The thermal resistance of the plate \(A\) is

\(R_{1}=\frac{1}{\mathrm{~K}_{\mathrm{A}}} \frac{\mathrm{x}_{\mathrm{A}}}{\mathrm{A}}\)

and that of the plate \(B\) is

\(R_{2}=\frac{1}{\mathrm{~K}_{\mathrm{B}}} \frac{\mathrm{x}_{\mathrm{B}}}{\mathrm{A}}\)

The equivalent thermal resistance is

\(R=R_{1}+R_{2}=\frac{1}{\mathrm{~A}}\left(\frac{\mathrm{x}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{A}}}+\frac{\mathrm{x}_{\mathrm{B}}}{\mathrm{K}_{\mathrm{B}}}\right) \)

\(\text { Thus, }  \frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\theta_{1}-\theta_{2}}{\mathrm{R}}=\frac{\mathrm{A}\left(\theta_{1}-\theta_{2}\right)}{\mathrm{x}_{\mathrm{A}} / \mathrm{K}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}} / \mathrm{K}_{\mathrm{B}}} \)

 \(=\frac{\left(0.01 \mathrm{~m}^{2}\right)\left(100^{\circ} \mathrm{C}\right)}{(0.04 \mathrm{~m}) /\left(200 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}\right)+(0.025 \mathrm{~m}) /\left(400 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}\right)}=3810 \mathrm{~W} .\)

(b) We have \(\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\mathrm{A}\left(\theta-\theta_{1}\right)}{\mathrm{x}_{\mathrm{B}} / \mathrm{K}_{\mathrm{B}}}\)

or, \(\quad 3810 W=\frac{\left(0.01 \mathrm{~m}^{2}\right)\left(\theta-0^{\circ} \mathrm{C}\right)}{(0.025 \mathrm{~m}) /\left(400 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}\right)}\)

or, \(\quad \theta=25^{\circ} \mathrm{C}\)

(c) If \(K\) is the equivalent thermal conductivity of the compound plate, its thermal resistance is

\(R=\frac{1}{\mathrm{~A}} \frac{\mathrm{x}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}}}{\mathrm{K}}\)

Comparing with (i),

\(\frac{\mathrm{x}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}}}{\mathrm{K}}=\frac{\mathrm{x}_{\mathrm{A}}}{\mathrm{K}_{\mathrm{A}}}+\frac{\mathrm{x}_{\mathrm{B}}}{\mathrm{K}_{\mathrm{B}}}\)

\(o r\)

\(K=\frac{\mathrm{x}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}}}{\mathrm{x}_{\mathrm{A}} / \mathrm{K}_{\mathrm{A}}+\mathrm{x}_{\mathrm{B}} / \mathrm{K}_{\mathrm{B}}}=248 \mathrm{~W} / \mathrm{m}-{ }^{\circ} \mathrm{C}\)

Illustration :

Two thin metallic spherical shells of radii \(r_{1}\) and \(r_{2}\left(r_{1}<r_{2}\right)\) are placed with their centres coinciding. A material of thermal conductivity \(K\) is filled in the space between the shells. The inner shell is maintained at temperature \(\theta_{1}\) and the outer shell at temperature \(\theta_{2}\left(\theta_{1}<\theta_{2}\right)\). Calculate the rate at which heat flows radially through the material.

Sol.

Let us draw two spherical shells of radii \(x\) and \(x+d x\) concentric with the given system. Let then temperatures at these shells be \(\theta\) and \(\theta+d \theta\) respectively. The amount of heat flowing radially inward through the material between \(x\) and \(x+d x\) is

Thus,

or,

\( \frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{\mathrm{K} 4 \pi \mathrm{x}^{2} \mathrm{~d} \theta}{\mathrm{dx}}\)

or,

\(K 4 \pi \int_{\theta_{1}}^{\theta_{2}} \mathrm{~d} \theta=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}} \int_{\mathrm{r}_{1}}^{\mathrm{r}_{2}} \frac{\mathrm{dx}}{\mathrm{x}^{2}}\)

\(K 4 \pi\left(\theta_{2}-\theta_{\nu}\right)=\frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}\left(\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{r}_{2}}\right)\)

\( \frac{\Delta \mathrm{Q}}{\Delta \mathrm{t}}=\frac{4 \pi \mathrm{Kr}_{1} \mathrm{r}_{2}\left(\theta_{2}-\theta_{1}\right)}{\mathrm{r}_{2}-\mathrm{r}_{1}}\)

Temperature distribution along a conductor

In order to study conduction in more detail consider figure (i), which shows a metal bar \(\mathrm{AB}\) whose ends have been soldered into the walls of two metal tanks, \(\mathrm{H}\) and \(\mathrm{C}\). H contains boiling water, and \(\mathrm{C}\) contains ice-water. Heat flows along the bar from A to B, and when conditions are steady, the temperature \(\theta\) of the bar is measured at points along its length.

.

(i) unlagged

(ii) lagged

Figure : Temperature fall along lagged and unlagged bars

The curve in the upper part of the figure shows how the temperature falls along the bar, less and less steeply from the hot end to the cold. So the temperature gradient decreases from the hot end to the cold. The figure (ii) shows how the temperature varies along the bar, if the bar is well lagged with a bad conductor, such as Asbestos wool. It now falls uniformly from the hot to the cold end, so the temperature gradient along the bar is constant.

The difference between the temperature distribution is due to the fact that, when the bar is unlagged, heat escapes from its sides, by convection in the surrounding air, figure (i). So the heat flowing past D per second is less than that entering the bar at Aby the amount which escapes from the surface AD. The arrows in the figure represent the heat escaping per second from the surface of the bar, and the heat flowing per second along its length. The heat flowing per second along the length decreases from the hot end to the cold. But when the bar is lagged, the heat escaping from its sides is negligible, and the flow per second is now constant along the length of the bar, figure (ii).

Growth of Ice on Ponds

When atmospheric temperature falls below \(0^{\circ} \mathrm{C}\) the water in the lake will start frezing. Let at any time \(t\), the thickness of ice in the lake by \(y\) and atmospheric temperature is \(-\theta^{\circ} \mathrm{C}\). The temperature of water in contact with the lower surface of ice will be zero. If \(A\) is the area of the lake, heat escaping through ice in time dt,

\( \mathrm{dQ}_{1}=\mathrm{KA} \frac{[0-(-\theta)]}{\mathrm{y}} \mathrm{dt}\)

Now due to escaping of this heat if dy thickness of water in contact with lower surface of ice freezes,

\( \mathrm{dQ}_{2}=\mathrm{mL}=\rho(\text { dy A) } \mathrm{L}[\text { as } \mathrm{m}=\rho \mathrm{V}=\rho \mathrm{Ady}] \)

But as \(\mathrm{dQ}_{1}=\mathrm{dQ}_{2}\), the rate of growth of ice will be

\( \frac{d y}{d t}=\frac{\mathrm{K \theta}}{\rho \mathrm{L}} \times \frac{1}{\mathrm{y}} \)