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Section 23.1 Thermal Conduction

The heat transfer mechanism that requires a physical contact between the bodies is called conduction. Materials differ in their abilities to conduct heat. Heat flows more readily through a good conductor of heat such as aluminum than through a poorer conductor such as wood. The larger conductivity of aluminum over wood is easily demonstrated by touching an aluminum sheet and a wooden piece, which had been in the same oven for some time so that they have the same temperature.

The aluminum sheet feels hotter because it easily transfers heat from other points of the aluminum sheet to the point of touch. As a result, a large amount of energy is given to your skin in a short amount of time as compared to what happens when you touch wood. This quick transfer of heat raises the temperature of your hand giving the sensation that aluminum sheet is hotter than the wood, even though they are at the same temperature.

Just because a particular object is sensed hotter than another does mean the hotter object is at a higher temperature. We sense the rate at which heat energy is deposited at the site of touch to ascertain the sense of hot and cold. We do not necessarily sense the temperature of a substance. Of course, if you touch two identical aluminum sheets at different temperatures, you will be able to correctly tell which sheet is at a higher temperature, since the two sheets have the same conductivity.

To study the conductivity quantitatively, consider a bar of length \(L\) and uniform cross-section area \(A\) whose ends are kept in contact with two large objects, also called heat baths, of constant temperatures \(T_1\) and \(T_2\) as shown in Figure 23.1.1.

The “heat baths” may not even be baths at all; they may be an oven or some other device so that the temperature can be kept steady at \(T_1\) or \(T_2\) on the two sides of the bar under study.

We will assume that the heat baths are so large that their temperatures do not change during the experiment. We also assume that heat passing from one bath to the other through the bar does not leak out from the side surface of the bar. This can be accomplished by insulating the bar on the sides.

Figure 23.1.1. Study of conductivity of heat for the material connecting the two blocks at temperature \(T_1\) and \(T_2\) . As \(T_1>T_2\) heat flows in the direction from \(T_1\) towards \(T_2\text{.}\)

Let \(\Delta Q\) amount of heat flows from one bath to the other in time duration \(\Delta t\text{.}\) The power flux \(\Phi\) of heat transfer from one bath to the other is defined as the rate of heat transfer per unit cross-sectional area of the bar. Note the use of letter \(t\) here: the letter \(t\) in this section stands for the time and not the temperature.

\begin{equation} \Phi = \dfrac{\Delta Q}{A\Delta t}.\label{eq-thermal-flux-def-1}\tag{23.1.1} \end{equation}

Experiments show that the power flux is directly proportional to the difference in the temperature across the ends and inversely proportional to the length of the rod. That is,

\begin{equation} \dfrac{\Delta Q}{A\Delta t} = k \left( T_1-T_2\right) \dfrac{1}{L}, \ \ \ \ (T_1>T_2).\label{eq-thermal-flux-def-2}\tag{23.1.2} \end{equation}

The proportionality constant \(k\) is called the thermal conductivity which has units of J/s.m.K. The constant \(k\) in distinguishes different materials by their conduction rates. Thermal conductivity of some common materials are given in Table 23.1.2.

A material with larger thermal conductivity is a better conductor of heat. From the table you can see that thermal conductivities of metals are \(10^1\) to \(10^4\) times the conductivities of other materials. Therefore, metals are considered good conductors of heat. Materials whose thermal conductivities are small compared to good conductors are called insulators. Wood and air are good example of insulators. Water is not as good a conductor as metals but a much better conductor than air. That is why you feel much colder in the lake water than air of the same temperature.

This equation has a direct analogy with the electric current \(I\) (in place of \(\Delta Q/\Delta t\)) through a wire of resistance \((R)\) when a voltage difference \(\Delta V\) (in place of \(T_1-T_2\)) is applied across the wire. In electric circuits, the relation among \(I\text{,}\) \(R\) and \(V\) is given by Ohm's law.

\begin{equation*} I = \dfrac{1}{R}V \end{equation*}

Therefore, it is a common practice to define a thermal resistance \(R_\text{th}\) of a slab of thickness \(\Delta x\) and area of cross-section \(A\text{,}\) made up of a material of conductivity \(k\text{,}\) by the following

\begin{equation} R_\text{th} = \dfrac{\Delta x}{k A}\tag{23.1.3} \end{equation}

The insulating properties of building materials, such a fiberglass, are indicated by the thermal resistance of unit area slab, called the \(R\)-factor.

\begin{equation} R \text{-factor} = \dfrac{\Delta x}{k}\tag{23.1.4} \end{equation}

A commonly used fiberglass in the US is R-19, which corresponds to \(\dfrac{1}{19}\) BTU per hour per square foot area of the insulation. In calculating the R-value of a multi-layered installation, the R-values of the individual layers are added. Table 23.1.3 shows the \(R\)-factors of several common insulations used in the construction industry.

Subsection 23.1.1 (Calculus) Thermal Conduction

From the definition of heat flux in Eq. (23.1.1) and the experimental results of conduction in Eq. (23.1.2), we have the following relation for the rate of flow of heat through a bar of length \(L\) and cross-section area \(A\text{.}\)

\begin{equation} \dfrac{1}{A}\dfrac{\Delta Q}{\Delta t} = k \left( T_1-T_2\right) \dfrac{1}{L} \ \ \ \ (T_1>T_2) \label{eq-heat-flux-1}\tag{23.1.5} \end{equation}

The relation actually holds for each thin section of the connecting rod. Therefore, it should be more appropriately written in terms of time-derivative of \(Q\text{.}\)

Let the heat flow be in the \(x\)-direction as shown in the figure. In Eq. (23.1.5), we set \(L = \Delta x\text{,}\) \(\Delta T = T_2-T_1\) and take the limit \(\Delta t \rightarrow 0\) on the left and \(\Delta x \rightarrow 0\) on the right side.

This results in the following differential equation for the dynamical flow of heat.

\begin{equation} \dfrac{dQ}{dt} = - k A\dfrac{dT}{dx}. \label{eq-cunductivity-law}\tag{23.1.6} \end{equation}

Here the minus sign ensures that \(dQ\) flows from a high \(T\) to a low \(T\) region.

Table 23.1.2. Thermal conductivities at \(25^{\circ}\)C (Handbook of chemistry and physics, 82nd edition, 2001-2002)
Substance Conductivity Substance Conductivity
Metals Fluids
Silver 429 Mercury 8.25
Copper 401 Water 0.6
Gold 317 Ethanol 0.169
Aluminum 237 Water vapor 0.027
Iron 80 Air 0.023
Common Solids
Asbestos 0.7 Porcelain 1
Asphalt 0.06 Pyrex glass 1
Brass 120 Rock 1
Brick, Dry 0.04 Rubber 0.05
Cork 0.04 Sand 0.33
Glass wool 0.04 Steel 52
Mica 0.2-0.7 Styrofoam 0.03
Polyurethane 0.06 Wood 0.04-0.35
Table 23.1.3. \(R\)-factor of Common Materials
Material R-factor
Roof(no insulation) 3.3
3" fiberglass 10
6" fiberglass 20
6" cellulose 23
8" thick concrete block 2.0
4" thick brick 4
Wooden frame 5
Wooden wall with 3 1/2" fiberglass 12
Concrete slab with 1" foam permimeter 45
Floor with 6" insulation 25
Single pane window 1
Double pane window 2
1/2" air space between panes 2

A \(0.5\text{-m}\) copper rod of cross-sectional area \(3.14\times 10^{-6}\text{ m}^2\) is insulated so that heat cannot escape from the sides. One end of the rod is free of insulation and placed in a steam container maintained at \(150^{\circ}\text{C}\text{.}\) The other end is also free of insulation and placed in ice water maintained at \(0^{\circ}\text{C}\text{.}\) Find the amount of heat transferred from the steam container to the ice water in 30 minutes.

Data: \(k_\text{Cu} = 401\text{ J/s.m.K}\text{.}\)

Hint

Use \(\dfrac{\Delta Q}{A\Delta t} = k\, \dfrac{T_1-T_2}{L}\ \ \ (T_1>T_2). \text{.}\)

Answer

\(680\ \text{J}\text{.}\)

Solution

This problem is a direct application of the relation between heat flux and the difference in temperature at two points in a conductor.

\begin{equation*} \dfrac{\Delta Q}{A\Delta t} = k\, \dfrac{T_1-T_2}{L}\ \ \ (T_1>T_2). \end{equation*}

Here, the following quantities are given:

\begin{align*} \amp A = 3.14\times 10^{-6}\ \text{m}^2 \\ \amp \Delta t = 30\ \text{min} = 1800\ \text{sec}\\ \amp k = 401\ \text{J/s.m.K}\\ \amp T_1-T_2 = 150\text{K}\\ \amp L = 0.5\ \text{m} \end{align*}

Therefore,

\begin{equation*} \Delta Q = k A \Delta t \dfrac{(T_1-T_2)}{L} = 680\ \text{J}. \end{equation*}

An aluminum and a steel cylindrical rod, each of diameter 1 cm and length 25 cm, are welded together. The other end of the aluminum rod is placed in a large tank of ice water at \(0^{\circ}\text{C}\text{,}\) while the other end of the steel rod is in a boiling water at \(100^{\circ}\text{C}\) as shown in Figure 23.1.6. The rods are insulated so that no heat escapes from the surface. What is the temperature at the joint when a steady state has reached so that the rate of flow of heat is same throughout?

Data: \(k_\text{Al} = 237\text{ J/s.m.K}\text{,}\) \(k_\text{steel} = 80\text{ J/s.m.K}\text{.}\)

Figure 23.1.6. Figure for Checkpoint 23.1.5.
Hint

Balance the flow of heat into the middle with the flow out of the middle.

Answer

\(298.4\ \text{K} \text{ or } 25.2^{\circ} \text{C}\text{.}\)

Solution

Let \(T\) be the temperature at the joint. The steady state condition requires that rate at which heat is transferred to the joint from the hotter end of the steel must be same as the heat transferred by aluminum from the joint to the colder end.

\begin{align*} \amp \text{Hot end to middle: } \dfrac{\Delta Q}{\Delta t} = A_{\text{steel}}k_{\text{steel}} \dfrac{(T_{\text{hot}}-T)}{L_{\text{steel}}}, \\ \amp \text{middle to cold end: } \dfrac{\Delta Q}{\Delta t} = A_{\text{Al}}k_{\text{Al}} \dfrac{(T-T_{\text{cold}})}{L_{\text{Al}}}. \end{align*}

Equating the two rates gives

\begin{equation*} A_{ \text{steel}}k_{ \text{steel}} \dfrac{(T_{ \text{hot}}-T)}{L_{ \text{steel}}} = A_{ \text{Al}}k_{ \text{Al}} \dfrac{(T-T_{ \text{cold}})}{L_{ \text{Al}}}. \end{equation*}

The areas of cross-section and lengths of the rods cancel out from the two sides, and we find the following for the temperature at the joint.

\begin{align*} T \amp = \dfrac{k_{ \text{steel}}T_{ \text{hot}} + k_{ \text{Al}} T_{ \text{cold}}}{k_{ \text{steel}} + k_{ \text{Al}}},\\ \amp = \dfrac{80\times 373.15+237\times 273.15}{40+237}\ \text{K} = 298.4\ \text{K} = 25.2^{\circ} \text{C}. \end{align*}

Note the temperature in the middle of the two metals is not the average of the temperatures at the two ends since the metals have different conductivities.

A 5 m long copper pipe of diameter 2.0 cm carries steam at \(150^{\circ} \text{C}\text{.}\) The pipe is insulated by Styrofoam of thickness 1 cm, and the outside space is at room temperature of \(20^{\circ}\)C (see Figure 23.1.8). Find the rate of heat loss from the pipe.

Data: \(k_\text{Styrofoam} = 0.03\text{ J/s.m.K}\)

Figure 23.1.8. Figure for Checkpoint 23.1.7.
Hint

Use Eq. (23.1.6) and set up the problem in cylindrical coordinates.

Answer

\(177\ \text{W} \text{.}\)

Solution

Here, heat is not transferred in one direction but radially outward in all directions from an axis. Therefore, we use the cylindrical coordinate system.

At the radial distance \(r\text{,}\) the area \(A \) through which heat passes is the area of a cylindrical shell of length \(L \) and width \(2\pi r\text{.}\) Therefore, Eq. (23.1.6) for rate of flow through the shell at radial distance \(r \) would become

\begin{equation*} \dfrac{dQ}{dt} = - k(2\pi r L) \dfrac{dT}{dr} \end{equation*}

Since, there is no source or sink of heat in the insulation, the rate of flow of heat must be independent of the distance from the center. Rearranging the equation, and integrating from \(T = T_1\) when \(r = R_1\) (the inside radius of the insulation) to \(T= T_2\) when \(r =R_2\) (the outer radius of the insulation) will give the following.

\begin{equation*} \dfrac{dQ}{dt} \int_{R_1}^{R_2}\dfrac{dr}{r}= - 2\pi k L \int_{T_1}^{T_2} dT. \end{equation*}

This gives the following relation for the rate of flow of heat from the inner surface at radius \(R_1\) to the outer surface at radius \(R_2\text{.}\)

\begin{equation*} \dfrac{dQ}{dt} = \dfrac{2\pi k L(T_1-T_2)}{\ln{(R_2/R_1)}}. \end{equation*}

Note that we do not need to convert the temperature into Kelvin since the change in Kelvin is the same in degree Celsius. Putting the numerical values for our case, we get

\begin{equation*} \dfrac{dQ}{dt} = \dfrac{2\pi \times 0.03( \text{W/m.K})\times 5 \text{m}\times 130\ \text{K}}{\ln{(2\ \text{cm}/1\ \text{cm})}} = 177\ \text{W}. \end{equation*}