Thermal conductivity of a good conductor: an introductory review for an undergraduate lab

Abstract

In this work, we study the mathematical formulation required in the study of the flow of heat in the metal rod due to conduction process. Furthermore, the complete theory of one of the very popular experiments for the determination of thermal conductivity of a metal rod (Searle’s method) is explained. We also revisit the connection between the Searle’s experiment and nonequilibrium thermodynamics. This experiment is usually performed at the undergraduate level in the thermal physics lab. We believe that this paper may provide useful insight to undergraduate students about the theory and experimental method of determining the coefficient of thermal conductivity of a good conductor.

Keywords:
Thermal conductivity; Searle’s Method; entropy and its production

1. Introduction

We know that heat, a form of energy, may be propagated by three different processes: convection, conduction and radiation. In liquids and gases, convection is a dominating process for the propagation of heat energy. However, the flow of heat is through a conduction process in a solid body (metal bar).

In this work, we have focused on the theory essential for the flow of heat through the conduction process (the metal bar). In 1822, Jean B. J. Fourier laid the foundation for the rate of flow of heat by the conduction method, which can be explained by the physical quantity called “thermal conductivity” of the medium. Conduction is the transfer or propagation of heat through a material from one end to another if there is a temperature difference between them. Basically, it is the transfer of heat energy through matter from a region of higher temperature to another one where it is lower. In this whole process, there is no physical movement of the material itself taking place. However, microscopically, the molecules of the material vibrate about their mean positions as a result of the flow of heat. This vibrational motion of the molecule disturbs the immediate neighboring molecule and hence the heat is transferred from one end to another. The important point is that there is no change in the average position of the molecule during the transfer of heat. One of the factors that controls the flow of heat in a given material is thermal conductivity or the coefficient of thermal conductivity. In this work, we also discuss the experimental method for determining the coefficient of thermal conductivity of good conductors by the Searle method.

The objective of this work is three-fold. First, to introduce the theory and mathematical equations of the heat flow in the conducting rod. Second, apply this theoretical framework to the Searle experiment to determine the thermal conductivity of a good conductor. Third, to revisit the possible extension of this experiment in studying entropy and its production. We believe that this paper will give a coherent review of Searle’s experiment by covering its theory, experimental set-up and its possible extension in calculation of entropy and its production.

The paper is organized as follows. In Section 2, some basic formulae and concepts associated with the conduction of heat in the metal rod are discussed. In this section a detailed theory related to the rectilinear flow of heat along a bar is also described. The Searle method for determining the thermal conductivity of the conducting rod is described in Section 3. Finally, the possible connection of this experiment to statistical mechanics by evaluating the thermodynamical probability is illustrated in Section 4. In this section, we also discuss the concept of entropy production in a heat conducting rod.

2. Theoretical Framework

We first discuss some important facts and definitions related to heat conduction. The necessary mathematical details required for the propagation of heat in the conducting rod are also explained.

2.1 Some basic definitions

(i) Conduction of heat: Consider a heat flow normally across a thin slab of cross- sectional area A, small thickness δx = L, with one face maintained at temperature T and another face at T′ (where δT = T − T′ is a temperature difference between the two faces of the slabs where T > T′). The temperature gradient, the temperature per unit distance, across the slab is $\frac{(T-T^{\prime })}{L}=\frac{\delta T}{\delta x}$. Experimental results have shown that the amount of heat flow in any cross-section of a lagged rod (if the rod is covered with insulating material) in the steady state is proportional to the three factors: time, cross section area and temperature [¹, ²]. Hence the rate of flow of heat normally across the slab in the positive x direction is given by:

(1) $$\begin{array}{cc}rateofflowofheat& =\frac{\delta Q}{\delta t}=-KA\frac{(T-T^{\prime })}{L}\\ & =-KA\frac{\delta T}{\delta x}.\end{array}$$

The negative sign indicates that heat flows in the direction of decreasing temperature. The quantity K is a positive constant of the material called the coefficient of thermal conductivity.

$$K=\frac{\begin{array}{l}\text{rate of flow of heat normal to}\hfill \\ \text{face}\text{of}\text{a slab in steady state}\hfill \end{array}}{\text{area of cross section ×}\text{temperature gradient}}.$$

In the limiting cases where δx and δt are very small, then $\frac{\delta T}{\delta x}$ and $\frac{\delta Q}{\delta t}$ reduce to their differential forms i.e $dT/dx$ and $dQ/dt$ (by using calculus notation). This implies

(2) $$\frac{dQ}{dt}=-KA\frac{dT}{dx}.$$

This is called Fourier’s law of heat conduction, a pure phenomenological law which cannot be derived from the first principles. The other important feature of this law is that it is applied to all forms of matter. If $-dT/dx=1$ degree per unit of length and A = 1 unit of area, then $K=dQ/dt$. Hence, the coefficient of thermal conductivity, K, is defined as: it is the normal rate of flow of heat by a conduction process through a unit area under a unit temperature gradient to the opposite faces of a thin parallel sided slab of the material in steady state condition (see Fig. 1). In S.I. system, the unit of thermal conductivity, K, is $W/m.K$ (in c.g.s. thermal unit system, K is expressed in cal.sec.^-1 cm.^-1 °C^-1).

Figure 1
One-dimensional flow of heat in a thin slab in a unit time.

The other useful quantity is heat flux (J):

(3) $$J=\frac{\text{dQ}}{\text{A}\text{dt}}=-K\frac{dT}{dx}.$$

This is another form of Fourier’s law of heat conduction.

(ii) Thermal conductivity of materials: The thermal conductivity of a material quantifies its ability to conduct heat energy. It is different for the different phases of a material: highest for a material if it is in a solid phase and lowest if it is in a gaseous phase. Heat conduction in solids is primarily due to two modes: lattice vibrational waves (phonon) and motion of free electrons. The transfer of heat energy in metals is mainly due to the movement of free electrons. The contribution of phonons to heat transfer is predominant in the insulators and semiconductors. Further, the thermal conductivity of a material varies with temperature. For solids, it generally decreases with an increase in temperature. This variation is usually very mild or negligible for most of the materials in a certain temperature range (i.e. silver, copper, gold etc.). But for some materials this variation is significant (i.e. diamond, iron, aluminum oxide etc.) [³]. Hence the variation of thermal conductivity is usually approximated by linear function of temperature

$$K\left(T\right)=K_0\left(1+\beta T\right),$$

where β is known as the temperature coefficient of thermal conductivity and K₀ is the thermal conductivity at 0° C. It may be positive or negative depending on the material. For example, β is usually positive for non-metals and negative for metals (there are some exceptions). The thermal conductivity increases for gases as the temperature increases, while it decreases for most liquids with increasing temperature (exceptions i.e. water) [³]. If the variation of thermal conductivity is known in the temperature range for a material, it is often advised to use the average value of thermal conductivity.

Different materials or states of matter exhibit different thermal conductivities that vary over a wide range. The thermal conductivity of pure metals at room temperature usually ranges between 10 and 500 Wm^-1K^-1. For liquids, the typical range of K is 0.1-1.0Wm^-1K^-1 (except that mercury has thermal conductivity, ∼8.3 Wm^-1K^-1). Similarly, the common range of K for gases at room temperature is 0.002-0.05 in SI units (but hydrogen and helium have a high value of thermal conductivity).

(iii) An unlagged bar: If the rod is unlagged (bare), there is a loss of heat from the sides of the rod. In this case, the lines of heat flow are not parallel along the length of rod but are bent towards the sides of rod. In other words, the temperature gradient has a component perpendicular to the axis of the rod. It is also known that the rate of flow of heat decreases through the cross-section of the rod as we move away from the hot end of the given rod. Therefore, the temperature gradient in every section of the unlagged rod decreases from hot to cold end (see Fig 2). In order to explain the temperature variation in the steady state, let Q₁ be the amount of heat entering the rod at point X. Some heat is lost from the side of the rod as it is unlagged and therefore the amount of heat Q₂ that passes through Y is less than Q₁. Similarly, we can explain why Q₃ is less than Q₁. Since the rate of heat flow is proportional to the temperature gradient and the temperature gradient continues to decrease along the length of the unlagged rod from X to Z [², ⁴].

Figure 2
Temperature distribution (top) and flow of heat (middle) in an unlagged rod. The quantity of heat flowing through the cross sections of the rod keep on decreasing along the length of the rod (bottom).

(iv) Lagged bar: Consider the rod that is heated at one end and cooled at the other end. In the lagged bar, (it is surrounded by layers of cotton wool) there is no escape of heat from the sides of it. Hence after some time, the steady state is reached, which means that the temperature of its each section is constant. In other words, the quantity of heat flows per second through each cross section of the rod is the same. The same condition can also be achieved if the metal rod is very thick and unlagged. In the perfect lagged rod, the flow heat lines are parallel to the sides of the rod and the temperature gradient remains the same for all cross sections of the rod along its length (see Fig. 3). Due to the uniformity of the temperature gradient, one can calculate it easily by noting down the steady-state temperature of any two cross sections of the bar and divide it by the distance between them. Experimentally, it is also observed that the temperature of the lagged rod decreases uniformly from the hot end to the cold end [², ⁴].

Figure 3
Temperature distribution (top) and steady flow of heat (middle) in the lagged rod. The quantity of heat flow through all the cross sections of the rod remain same (bottom).

2.2 Flow of heat along a good conductor bar

In this section, we discuss the theory of conduction of heat in a metal rod whose one end is maintained at high temperature. The main objective is to determine the temperature distribution inside the rod that is subjected to different boundary conditions. Assume that the rod is placed along the x axis with the hot end at x = 0. Heat conduction will remain dominant along the x-axis, while it is negligible in other directions. Consider an element of thickness δx along the length of a bar. This cross- sectional area length element A has two planes perpendicular to the axis of the bar at a distance x and x + δx from the hot end as shown in Fig. 4.

Figure 4
Flow of heat in the length element of a rod (along the x-axis).

Let T be the temperature of the plane at x and the temperature of the plane at x + δx is $T+\frac{\partial T}{\partial x}\partial x$. If ${q^{\prime }}_1$ and ${q^{\prime }}_2$, heat per unit time, enter in the plane at x and leave the plane at x + δx respectively, then in the limiting case of δx → 0,

rate of energy conduction into the element

$$={q^{\prime }}_1=q=-KA\frac{\partial T}{\partial x},$$

Similarly, rate of energy conduction out of the element

$$\begin{array}{c}={q^{\prime }}_2=q+\delta q=q+\frac{dq}{dx}\delta x\\ =-KA\left(\frac{\partial T}{\partial x}+\frac{{\partial }^{2}T}{\partial x^2}\delta x\right).\end{array}$$

Hence the extra heat gained per unit time between two planes separated by the thickness δx is

$$\delta q=KA\frac{{\partial }^{2}T}{\partial x^2}\delta x.$$

When thesteady state is not reachedthen this heat will be used not only to increase the temperature of the metal bar, but part of the heat is also lost by radiating it from the surface of the element. If c is the specific heat of the material of the rod, ρ is density (mass per unit volume) and $\partial T/\partial t$ is the rate of increase in temperature, then the heat per unit time used for raising the temperature of the length element is:

$${q^{\prime }}_3=\left(A\delta x\rho c\right)\frac{\partial T}{\partial t}.$$

Furthermore, if ϵ is the emissive power of the rod (rate of loss of heat per unit area per unit excess temperature) and T is the average excess of temperature of the rod element then the rate by which the heat is radiated by the length element δx (By using Newton’s law of cooling)¹:

$${q^{\prime }}_4=ϵ\rho \delta xT.$$

where p is the perimeter of the rod [³, ⁵, ⁶]. Therefore, an energy balance in this thin-length element is:

Heat gained per unit time = Heat used per sec to raise the temperature + Heat lost per sec

(4) $$\begin{array}{c}{q^{\prime }}_1-{q^{\prime }}_2={q^{\prime }}_3+{q^{\prime }}_4,\\ KA\frac{{\partial }^{2}T}{\partial x^2}\delta x=\left(A\delta x\rho c\right)\frac{\partial T}{\partial t}+ϵp\delta xT,\end{array}$$

The general equation of heat flow in the conducting bar can be written as [⁷]:

(5) $$\frac{{\partial }^{2}T}{\partial x^2}=\frac{\left(\rho c\right)}{K}\frac{\partial T}{\partial t}+\frac{ϵp}{KA}T.$$

This equation can be solved exactly for special cases with different initial conditions.

Case I: If the steady state is achieved on an unlagged bar: The steady state implies $\partial T/\partial t=0$ and hence ${q^{\prime }}_3=0$ (There is no change in temperature with time for the length element). Because it is an unlagged bar, the bar is exposed at its sides and there is some loss of heat as a result of radiation. In steady state, T is a function of x only i.e. T = T(x). Then equation (3) reduces to:

$$\frac{d^{2}T}{d{x}^2}=\frac{ϵp}{KA}T={\beta }^{2}T,$$

Where ${\beta }^2=\frac{ϵp}{KA}$. The general solution of this second order differential equation is:

$$T\left(x\right)=C_1{e}^{-\beta x}+C_2{e}^{\beta x},$$

where C₁ and C₂ are constants that can be determined from the boundary conditions [⁸]. After applying the boundary conditions (at x = 0, T = τ₀ and at x = ∞, T = 0), the final solution becomes T(x) = τ₀e^-βx, where τ₀ is the excess temperature at the hot end. If the metal bar is of finite length L, by applying suitable boundary conditions (x = 0, T = τ₀ and $\partial T/\partial x=0$, when x = L), the complete general expression of the steady temperature distribution along the length of the bar can be obtained [¹]. The variation in temperature along the length of the bar is shown in Fig. 2. This is called the steady state in an unlagged bar[⁹].

Case II: If the steady state is achieved in a lagged bar: The lagged bar means that the bar is covered with cotton wool and the radiation loss is negligible. This implies that emissivity is zero. These two conditions (steady state : $\partial T/\partial t=0$ and lagged bar: ϵ = 0) simplify the heat equation and the temperature distribution can be described by Laplace’s equation:

$$\frac{d^{2}T}{d{x}^2}=0.$$

On integrating the above equation w.r.t. x gives $dT/dx=\text{constant}$. It shows that the temperature gradient along the length of the lagged rod is constant. The solution of the above equation is: T(x) = D₁x + D₂, where D₁ and D₂ are arbitrary constants. By using the boundary conditions (when x = 0, T = τ₀ and at x = L, T = τ₁, τ₀ > τ₁), the temperature distribution in the bar is given as

$$T\left(x\right)={\tau }_0-\frac{\left({\tau }_0-{\tau }_1\right)}{L}x,$$

Or

$$\frac{{\tau }_0-T\left(x\right)}{x}=\frac{\left({\tau }_0-{\tau }_1\right)}{L}=\text{constant}\text{.}$$

This shows that the temperature varies linearly along the length of the bar and the temperature gradient is constant in the lagged bar along the length of it. The temperature variation in a lagged bar is shown in Fig. 3. The above relation also indicates the excess temperature at any point along the length of the bar. The rate of flow of heat through the lagged rod is described as:

$$q=-KA\frac{dT}{dx}=\frac{KA}{L}\left({\tau }_0-{\tau }_1\right),$$

Here, $KA/L$ is known as thermal conductance. The reciprocal of thermal conductance is called the thermal resistance [⁸].

Case III: If the steady state is not achieved in the lagged bar: In this case only ${q^{\prime }}_3$ will contribute and ${q^{\prime }}_4=0$ (due to perfect lagging), now the temperature is a function of both time and position coordinates. Then equation (3) becomes:

$$\frac{{\partial }^{2}T\left(x,t\right)}{\partial x^2}=\frac{\left(\rho c\right)}{K}\frac{\partial T\left(x,t\right)}{\partial t},$$

or

$$\frac{\partial T}{\partial t}={\alpha }^2\frac{{\partial }^{2}T}{\partial x^2},$$

Here ${\alpha }^2=\frac{K}{\rho c}$ is known as coefficient of thermal diffusivity which represents rate of change of temperature in a material when it is heated or cooled. It is defined as the ratio of the thermal conductivity to the thermal capacity per unit volume of the material. Materials with a high coefficient of diffusivity (silver) will heat up or cool down quickly as compared to materials with low diffusivity (glass). In order to determine the analytic solution of the above equation (temperature profile in the rod), the initial and boundary conditions must be specified (for different cases, see [³]).

In the next section, we will explain one of the very simple and popular methods used to find the thermal conductivity of metals. This experiment is usually performed by undergraduate students in their thermal physics lab.

3. Searle’s Method for Determination of the Thermal Conductivity of a Copper Rod (Good Conductor)

In general, the determination of thermal conductivity of materials divides itself into three different categories: solids, liquids and gases. The measurement of the thermal conductivity of solids is divided into two subclasses:

• (i)

  Solids which are poor conductors of heat such as glass, cardboard etc. (For more details, see [¹⁰]).

• (ii)

  Solids which are good conductors of heat such as copper, brass etc.

There are two different techniques for measuring the thermal conductivity of solids: the steady-state method and the dynamical method. In the first method, the nonuniform temperature distribution is kept stationary in the sample by the supply of heat (Searle’s apparatus). However, in the dynamical method, the temperature distribution in the sample often varies over time (Angstrom method). For more details on the history of various other methods and modern techniques to measure thermal conductivity, see refs. [¹¹, ¹², ¹³, ¹⁴].

G. F. C. Searle of Cavendish Laboratory designed an apparatus based on the steady state method. It measures the coefficient of thermal conductivity of the metal rod made of copper. The apparatus consists of a thick copper rod of diameter of 4–5 cm (a thick rod is chosen to get a reasonable temperature difference) and has a length of about 20 to 30 cm. Two thermometers (T₁ and T₂) are inserted into the two holes separated by a distance d of approximately 8 to 10 cm into the rod. One end of the rod is connected to the steam box that is maintained at 100°C. The other end is connected to the cooling system in which the water flows from the constant head into the copper coil soldered around the end of the bar. The temperature of the incoming and outgoing water through the coil is measured with the help of thermometers (T₃ and T₄). It enables us to measure the rate of flow of heat in the rod. This method of measuring the rate of flow of heat is called the principle of continuous flow calorimetry [¹⁵, ¹⁶].

The whole apparatus is “lagged” to allow heat flow in one direction (along the x axis) and to prevent the loss of heat by radiation from the surface of the rod. To achieve this, the rod is surrounded by layers of cotton wool (a bad conductor) and placed in a felt-lined wooden box. The apparatus is shown in Fig. 5.

Figure 5
Description of Searle’s Apparatus.

To start the experiment, first allow the steam to pass through the steam box at one end of the rod and steady flow of water at the other end. The temperature of the thermometers begin to rise. After some time, the readings of all thermometers become constant. This indicates that the steady state has been reached. Note down the readings of all four thermometers (${T^{\prime }}_1$, ${T^{\prime }}_2$, ${T^{\prime }}_3$ – ${T^{\prime }}_4$) after the steady state is reached for the given flow of water in the coil C. Now collect the amount of water (say, m gm) in the beaker for a given time (say, t sec).

Since there is no loss of heat from the bar surface (due to lagging) and the steady state is also achieved, hence the rate of flow of heat flowing through any plane in the bar is proportional to the temperature gradient $\frac{\left({T^{\prime }}_1-{T^{\prime }}_2\right)}{d}$ which is a constant quantity (see Section 2, case II). In general, heat transfer from any surface is a combined contribution from convection and radiation. Here in this Searle’s set-up, heat losses due to convection and radiation can be considered negligible relative to axial conduction of heat due to the specific conditions of the set-up (good lagging). Further, radiation loss is negligible in lagged rod and it is completely justified, as the loss of heat due to radiation is significant only if the temperature is very high (usually above 1000 K). In Searle’s method the temperature of the rod is far less than 1000 K and it is also further covered with cotton wool. Hence, radiation loss from the rod surface is negligible. The perfect lagging implies the uniform temperature gradient and therefore as mentioned above, the rate of flow of heat between two planes separated by distance d at P₁ and P₂ is equal to, $heat/time=KA\left({T^{\prime }}_1-{T^{\prime }}_2\right)/d$. Further, if s is the specific heat of the water and ${T^{\prime }}_3$ – ${T^{\prime }}_4$ are the inlet and outlet temperature of the flowing water, then the heat absorbed per unit time by the water is $ms\frac{\left({T^{\prime }}_4-{T^{\prime }}_3\right)}{t}$. Hence, rate of heat flow is

$$q=\frac{Q}{t}=ms\frac{\left({T^{\prime }}_4-{T^{\prime }}_3\right)}{t}=KA\frac{\left({T^{\prime }}_1-{T^{\prime }}_2\right)}{d},$$

or coefficient of thermal conductivity of a metal bar of diameter D can written as:

$$K=\frac{sd}{A}\left[\frac{{T^{\prime }}_4-{T^{\prime }}_3}{{T^{\prime }}_1-{T^{\prime }}_2}\right]\left(\frac{m}{t}\right)=\frac{4sd}{\pi D^2}\left[\frac{{T^{\prime }}_4-{T^{\prime }}_3}{{T^{\prime }}_1-{T^{\prime }}_2}\right]\left(\frac{m}{t}\right).$$

For more details of the method to be followed in the lab, see the appendix. Some more useful points related to Searle’s experiment are as follows:

1. To ensure good thermal contact between the thermometer and the metal rod for effective heat transfer, a thermal paste or thermal grease should be applied. The other standard alternative is to replace the thermometers with thermocouples embedded with epoxy or thermal conducting paste.

2. The holes in the rod should be narrow and not too deep.

3. The zero error in the thermometers can easily be removed from the calculations by interchanging the thermometers. First, interchange the thermometers at T₁ and T₂. Take the average of the difference of temperatures ${T^{\prime }}_1-{T^{\prime }}_2$ (before the interchange of thermometers T₁ and T₂)and ${T^{″}}_1-{T^{″}}_2$ (after the interchange of thermometers T₁ and T₂). Similar methodology is to be used for the thermometers at T₃ and T₄. The idea is that If steady state reading in thermometers T₁ and T₂ can be written as :

$$\begin{array}{c}{T^{\prime }}_1\left(obs\right)={T^{\prime }}_1\left(true\right)+e_1.\\ {T^{\prime }}_2\left(obs\right)={T^{\prime }}_2\left(true\right)+e_2.\end{array}$$

This implies

(6) $${T^{\prime }}_1\left(obs\right)-{T^{\prime }}_2\left(obs\right)={T^{\prime }}_1\left(true\right)-{T^{\prime }}_2\left(true\right)+e_1-e_2,$$

where e₁ and e₂ are zero errors on the thermometers T₁ and T₂ respectively. Now, if we interchange the thermometers for measuring the steady state temperature then:

$$\begin{array}{c}{T^{″}}_1\left(obs\right)={T^{\prime }}_1\left(true\right)+e_2,\\ {T^{″}}_2\left(obs\right)={T^{\prime }}_2\left(true\right)+e_1,\end{array}$$

This implies

(7) $${T^{″}}_1\left(obs\right)-{T^{″}}_2\left(obs\right)={T^{\prime }}_1\left(true\right)-{T^{\prime }}_2\left(true\right)+e_2-e_1,$$

Take the average of equation 5 and equation 6, we get

$$\begin{array}{c}\frac{\left[{T^{\prime }}_1\left(obs\right)-{T^{\prime }}_2\left(obs\right)\right]+\left[{T^{″}}_1\left(obs\right)-{T^{″}}_2\left(obs\right)\right]}{2}\\ ={T^{\prime }}_1\left(true\right)-{T^{\prime }}_2\left(true\right).\end{array}$$

This brief calculation shows how to remove the zero error present in the thermometers.

4. The success of Searle’s method depends on the attainability of a uniform temperature gradient. In order to achieve this, we require very efficient lagging of the rod, which is a very important part of this apparatus. The lagging may be done by several methods. Instead of covering the rod with layers of cotton wool (as explained above), the metal rod can be surrounded by a hollow coaxial cylinder which is made up of the same material as the rod. There should be an air gap between the rod and the coaxial cylinder. As shown in Fig. 6(a) and one end of the coaxial cylinder is in contact with steam along with the conducting rod and the other end of the coaxial cylinder is kept in plain water along with the rod. This arrangement acts as an effective heat insulation if carried out along the entire length of the sample rod. Hence the same temperature gradient exists on the coaxial cylinder as on the rod. This method is called the guard-ring method[¹⁵, ¹⁷].

Figure 6
Guard Ring method: (a) rod is surrounded by hollow coaxial cylinder (b) Thick rod, where outer part of the rod act as a “guard ring”.

5. Another method by which the uniform temperature gradient can be achieved is using the thick metal rod as a specimen (see Fig. 6(b)). In this case, the outer part of the rod acts as a guard ring while in the inner portion of the rod, the heat line flows parallel to the rod. To measure the temperature, the central portion of the bar is to be used. In this region, the flow of heat is normal in the area. By measuring the temperature gradient along the central part of the specimen in the steady state, one can easily calculate the thermal conductivity. This is an example of a guard-ring method [¹].

6. It is essential that water be drawn from the overhead water tank to cool the rod, as there is a strong possibility that the rate of water supply may fluctuate from the direct supply of tap water.

7. To ensure that linear heat flow is attained in the rod, multiple thermometers may be used along the length of the rod. It is essential to have efficient lagging along the length of the rod. Now, by plotting the drop in the temperature of the rod along its length, one obtains the straight-line curve. This shows the quick way to check the linear flow of heat in the rod (see page 221 of [¹⁸]).

4. Discussion: A Study of Entropy and Entropy Production

The study of heat conduction along a one-dimensional conducting rod has been investigated for nonequilibrium thermodynamics which can be further extended to analyze entropy and entropy production [¹⁹, ²⁰, ²¹, ²²]. The main objective of this work is to review the theory associated with the one-dimensional flow of heat in a conducting rod along with the experimental method to determine the thermal conductivity of a good conductor (Searle’s method). Further this experiment can be used as a laboratory exercise for demonstration of important properties related to entropy and entropy production.

(i) Rate of change in entropy: A possible extension of this experiment is to study the change in entropy of the metallic rod during heat conduction. In this system, one end of the rod is connected to steam at temperature T_S and the other end of the rod is cooled with the help of tap water at temperature T_W. During this process, the hot end of the isolated lagged rod loses the heat energy -Q at the temperature T_S and hence the decrease in entropy at the hot end is $-Q/T_S$. The cold ends gain heat +Q at the temperature T_W, which causes the gain in entropy by $+Q/T_W$. The net rate of change in the entropy for an isolated system can be written as [²³]:

(8) $$\begin{array}{c}\frac{\mathrm{\Delta }S}{t}=\frac{\mathrm{\Delta }{S}_{hot}}{t}+\frac{\mathrm{\Delta }{S}_{cold}}{t}>0.\\ \frac{\mathrm{\Delta }S}{t}=\frac{-Q/t}{T_S}+\frac{Q/t}{T_W},\end{array}$$

Here t is the duration of time during which the hot end loses heat energy or the cold end gains heat energy. The rate of heat flow on the metal bar can be written as

$$q=\frac{Q}{t}=KA\frac{\left(T_S-T_W\right)}{L},$$

Here K is the thermal conductivity of the rod obtained in the previous section, A is the cross section area of the rod and L is the length between the hot and the cold end. The rate of change of entropy can be rewritten as:

(9) $$\frac{\mathrm{\Delta }S}{t}=\frac{KA}{L}\frac{{\left(T_S-T_W\right)}^2}{T_S{T}_W}.$$

But this equation is connected to some very important mathematical results:

$$\begin{array}{ll}\left[\frac{{\left(T_S-T_W\right)}^2}{T_S{T}_W}\right]\hfill & =\frac{{\left(T_S+T_W\right)}^2}{T_S{T}_W}-4\hfill \\ \hfill & =4\left[{\left(\frac{T_S+T_W}{2}\right)}^2\frac{1}{T_S{T}_W}-1\right].\hfill \end{array}$$

The rate of change of entropy can be rewritten as:

(10) $$\frac{\mathrm{\Delta }S}{t}=\frac{4KA}{L}\left[{\left(\frac{A.M.}{G.M.}\right)}^2-1\right],$$

where $A.M.=\frac{\left(T_S+T_W\right)}{2}$ is the arithmetic mean and $G.M.=\sqrt{T_S{T}_W}$ is the geometric mean. The above equation can also be rewritten in terms of harmonic mean (H.M.) and quadratic mean (Q.M.) by using the identity A.M. × H.M. = G.M², where $H.M.=\frac{2T_S{T}_W}{\left(T_S+T_W\right)}$ and $Q.M.=\sqrt{\frac{T_S^2+T_W^2}{2}}$ is the quadratic mean.

(11) $$\frac{\mathrm{\Delta }S}{t}=\frac{4KA}{L}\left[\left(\frac{A.M.}{H.M.}\right)-1\right].$$

(12) $$\frac{\mathrm{\Delta }S}{t}=\frac{2KA}{L}\left[{\left(\frac{Q.M.}{G.M.}\right)}^2-1\right].$$

It is very interesting to note that the rate of increase in entropy $\left(\mathrm{\Delta }S/t>0\right)$ in Searle’s experiment is connected to the mathematical inequality: Q.M. > A.M. > G.M. > H.M. This result is discussed partially in the ref. [²⁴].

(ii) Boltzmann definition of entropy: Boltzmann defined the entropy for an isolated macroscopic system at equilibrium by the formula:

$$S=k_{B}ln\Omega ,$$

where k_B is the Boltzmann constant and Ω is the number of states accessible to the system. The change in entropy:

$$\mathrm{\Delta }S=S_f-S_i=k_{B}ln\frac{{\Omega }_f}{{\Omega }_i},$$

where S_f and S_i are the final and initial entropies respectively. Using equation (9)

(13) $$\begin{array}{c}ln\frac{{\Omega }_f}{{\Omega }_i}=\frac{KA}{L}\frac{{\left(T_S-T_W\right)}^2}{T_S{T}_W}\frac{t}{k_B},\\ \frac{{\Omega }_f}{{\Omega }_i}=e^{\frac{KA}{L}\frac{{\left(T_S-T_W\right)}^2}{T_S{T}_W}\frac{t}{k_B}}.\end{array}$$

This expression highlights the ratio of the final and initial microstates that are accessible to a system.

(iii) Verification of the Prigogine Theorem: One of the interesting extensions of the Searle’s set up is verification of the Prigogine Theorem. This theorem states that in the linear regime, the total entropy production attains its minimum if a system has reached the nonequilibrium stationary state. This state can be maintained by a flow of energy and matter [²⁵].

I. Danielewicz-Ferchmin and A. Ryszard Ferchmin have shown that heat conduction in a one-dimensional metal rod can be used to demonstrate the validity of the Prigogine Theorem [²⁶]. Consider the conducting rod of length L whose one end (at x = 0) is in contact with the hot reservoir (steam at temperature T_S) and the other end (at x = L) is in contact with the cold reservoir (tap water at temperature T_W). The rod is perfectly insulated from the surrounding environment. In this externally maintained temperature difference, the main criterion for achieving the steady state is to obtain a minimum entropy production. In this nonequilibrium stationary state the temperature varies linearly with the distance and there is nonzero transfer of energy along with the nonuniform distribution of temperature. Analogously to classical thermodynamics, the equilibrium state is the state at which the entropy production is zero. But in the nonequilibrium steady state, the entropy production is non zero. Hence the entropy production is a measure of a system’s stationary state departure from the thermodynamic equilibrium.

If the heat flow is along the x-direction, the entropy production per unit volume associated with the heat flow is written as product of heat flux density J(x) and ’ thermodynamic force’ [²⁶, ²⁷]:

(14) $$\begin{array}{l}\Lambda \left(x\right)=J\left(x\right)\frac{\partial }{\partial x}\frac{1}{T\left(x\right)}=-\frac{J\left(x\right)}{{\left[T\left(x\right)\right]}^2}\frac{\partial T\left(x\right)}{\partial x}\hfill \\ \begin{array}{cc}& =\left[\frac{K}{T^2\left(x\right)}\right]{\left(\frac{\partial T\left(x\right)}{\partial x}\right)}^2,\end{array}\hfill \end{array}$$

where $J\left(x\right)=-K\frac{\partial T\left(x\right)}{\partial x}$, is the Fourier law of heat conduction. The total rate of entropy production per area in the rod is given by

(15) $$\frac{d{S}_i}{dt}={\displaystyle {\int }_0^L\Lambda \left(x\right)dx=}{\displaystyle {\int }_0^L\frac{K}{{\left[T\left(x\right)\right]}^2}{\left(\frac{\partial T\left(x\right)}{\partial x}\right)}^{2}dx.}$$

In steady state, $\partial T/\partial t=0$, which implies that T(x) is a linear function of x, T(x) = D₁x + D₂ (see Section 2, case-II). The stationary state is a state where the temperature distribution is a linear function of position and the total entropy of the system is constant. This occurs when the entropy flowing out of the system is equal to the entropy produced plus the entropy entered into the system. After substituting the expression of T(x) in the integrand of above equation, we get the entropy production per area in terms of constants D₁ and D₂ is:

(16) $$\frac{d{S}_i}{dt}=\frac{KL{D}_1^{\text{2}}}{D_2\left(D_{1}L+D_2\right)}.$$

After substituting the values of the constants D₁ and D₂ (see case-II), we get

(17) $$\frac{d{S}_i}{dt}=\frac{KL{\left[\frac{T_W}{L}\left(1-\frac{T_S}{T_W}\right)\right]}^2}{T_S\left[\left(\frac{T_W}{L}\right)\left(1-\frac{T_S}{T_W}\right)L+T_S\right]}.$$

Finally, the minimum total entropy production is equal to

(18) $$\frac{\overline{d{S}_i}}{dt}=\frac{KA}{L}\frac{{\left(T_S-T_W\right)}^2}{T_S{T}_W}.$$

This is exactly the same expression as derived above (see equation 9) for the total change in the entropy. In order to demonstrate the minimum entropy production in the non-equilibrium state, the heat conduction along the length of the rod is a simple exercise which can be done in any undergraduate lab.

5. Method

In this problem, first measure the temperature of the rod (placed between two hot and cold reservoirs) along its length, T(x), after equal intervals of time. Here x is the distance from the hot end along the length of the rod. By plotting these data (T(x) versus x for different time intervals, say 2,4,6,8… minutes), we obtain the temperature distribution along the length of the rod for a given time interval. This can be elaborated on as follows.

1. The time begins at the moment the rod is placed between two different reservoirs maintained at temperatures T_S and T_W respectively. After 2 min, note the temperature, T(x), at different points along the length of the rod (at different values of x). We obtain the data of T(x) versus x for time of 2 minutes. This curve is fitted numerically through the general polynomial form (i.e., T(x) = a + bx + cx² + dx³ + ex⁴….).

2. Similarly, the above mentioned step will be followed (observing the temperature along the length of the rod) for the time t = 4, 6, 8, 10… minutes, etc. Plot T(x) versus x for every time t = 4, 6, 8,… minutes and fit the curves with a general polynomial form. One can easily check that as the time interval increases, the coefficients of higher powers of x continue to decrease (see the Appendix).

3. Finally, after a long time period, say after t = 20 minutes, the temperature distribution curve will be fitted by the linear polynomial (T(x) = a + bx, i.e. straight line). All other coefficients with a higher power of x will become zero. Now, the stationary nonequilibrium state is achieved (see the Appendix).

3. Once the functional form of T(x) is obtained for every time t of observation (as explained in the previous step), $\frac{\partial T\left(x\right)}{\partial x}$ can be evaluated and the total entropy production of this system can be calculated by substituting it in equation (15) for every time interval.

4. It can be easily checked that the total entropy production, $\frac{d{S}_i}{dt}$, continues to decrease as the observation of time increases and approaches the minimum value at the nonequilibrium stationary state. The minimum value of the numerically obtained entropy production must match the value obtained using equation (18) (for more details, see [²⁶, ²⁷, ²⁸]).

We believe that this review will provide valuable insight to undergraduate students with a complete theory and an experimental method for determining the thermal conductivity. In addition, we describe the strength of this experiment in establishing the connection between thermodynamics and statistical mechanics, which is an integral part of any undergraduate curriculum.

Acknowledgments

The author expresses his gratitude to the anonymous reviewers for their valuable comments and suggestions, which has significantly improved the paper. The author thanks Darshan Kumar for the assistance provided in drawing the figures and Dr. Anuradha Gupta (S.G.T.B. Khalsa College, Delhi University) for useful discussions.

Appendix A. Practical Lab Guide

As mentioned in the main text, the calculation of thermal conductivity (K) depends crucially on the correct measurement of the temperature and flow rate of the liquid. Therefore, these two should be carefully measured and recorded.

(i) Measurement of temperature: All readings on the thermometer should be steady before the final values are recorded. To ensure steady state is achieved, students are usually advised to record the temperature of all thermometers at intervals of 2 minutes (see Table A.1). Once the difference between the temperatures is constant (the steady state is attained), the final temperature is recorded for the calculation of thermal conductivity. In the experiment of determination of thermal conductivity, it is important that steady state must be achieved otherwise the thermal diffusivity is involved in the calculations, not the coefficient of thermal conductivity, as explained in Section 3 (case III).

Table A.1
Recording of Temperatures along the length of the rod at different time interval.

Time (in mins.)	T1 in °C in °C	T2 in °C in °C	T3 in °C in °C	T4 in °C in °C	Difference (T1−T2)	Difference (T4−T3)
0	…	…	…	…	…	…
2	…	…	…	…	…	…
4	…	…	…	…	…	…
6	…	…	…	…	…	…
…	…	…	…	…	…	…

(ii) Flow rate of water: The rate of flow of water through the coil around the rod should be adjusted in such a way that the temperature difference between incoming and outgoing water is a measurable quantity. Ideally, the flow should be low in the form of a trickle but continuous and should be collected only once the steady state is achieved. The experiment should be repeated by changing the flow rate of the water (see Table A.2).

Table A.2
Time of flow (t), mass of the water (m) collected and rate of flow of water (m/t) in the Searle’s experiment. Here ρ is the density of the water.

No. of obs.	Time t(sec)	Volume of water collected V in cm3	mass of water m = ρV in gm.	m/t in gm/sec	mean m/t in gm/sec
1	…	…	…	…
2	…	…	…	…	…
3	…	…	…	…
4	…	…	…	…

(iii) The error in K can be expressed as [²⁹]:

(A.1) $$\begin{array}{c}{\sigma }_K=K\left[\frac{{\sigma }_m^2}{m^2}+\frac{{\sigma }_{{T^{\prime }}_4-{T^{\prime }}_3}^2}{{\left({T^{\prime }}_4-{T^{\prime }}_3\right)}^2}+\frac{{\sigma }_d^2}{d^2}+\frac{{\sigma }_t^2}{t^2}+\frac{4{\sigma }_D^2}{D^2}\right]\\ {+\frac{{\sigma }_{{T^{\prime }}_1-{T^{\prime }}_2}^2}{{\left({T^{\prime }}_1-{T^{\prime }}_2\right)}^2}]}^{1/2}\end{array}$$

The error in ${\sigma }_{{T^{\prime }}_4-{T^{\prime }}_3}^2$ is equal to ${\sigma }_{{T^{\prime }}_4}^2+{\sigma }_{{T^{\prime }}_3}^2$, which is the sum of the squares of the least count of the thermometer used to measure the temperature of the outgoing and incoming water in the coil C of the setup. Similarly σ_t is the least measured time from the stopwatch and t is the total time taken in seconds to collect the water after the steady state is reached. The error ${\sigma }_{{T^{\prime }}_1-{T^{\prime }}_2}^2$ should be equal to ${\sigma }_{{T^{\prime }}_1}^2+{\sigma }_{{T^{\prime }}_2}^2$, where ${\sigma }_{{T^{\prime }}_1}$ and ${\sigma }_{{T^{\prime }}_2}$ are the least counts of both thermometers. The least count of a vernier caliper and a measuring scale is expressed as σ_D and σ_d respectively. Since the specific heat of the water is a constant quantity, so it will not contribute to the error budget. Here σ_m is the least count of the electronic weighing machine.

(iv) Demonstration of the Prigogine Theorem: It can be verified both graphically and numerically. As shown in Table A.3, the total entropy production can be calculated numerically at different time intervals. After a long interval of time, say after 20 minutes, it attains minimum value. This is known as stationary nonequilibrium state.

Table A.3
Total entropy production at different time intervals. As the time increases, the total entropy production keep on decreases and it becomes minimum after a long time interval.

Time (in min.)	$\left(\frac{d{S}_i}{dt}\right)$
2	max.
4	…
6	…
8	…
…	…
…	…
say at 20	min.

Moreover, the temperature distribution along the length of the rod can be plotted after different intervals of time (see Fig. A.1). Initially, say after 2 minutes, the temperature distribution (curve I, Fig. A.1) is fitted by the higher-order polynomial (i.e., a + bx + cx² + dx³ + ex⁴….). As time increases, the temperature distribution curve gradually becomes straight (curve II, Fig. A.1) and the data are fitted by a polynomial with fewer coefficients (i.e., a + bx + cx²). Finally, after a long period of time, the curve becomes a straight line fitted by a + bx (curve III, Fig. A.1).

Figure A.1
Temperature, T(x), along the length of lagged rod at different intervals of time. Curve I is plotted for a short interval of time (i.e. 2 minutes). Curve II and curve III corresponds to time interval of say 10 minutes and long interval of time (i.e. 20 minutes) respectively.

Since the Prigogine Theorem deals with the total entropy production which can again be shown graphically by plotting the integrand as a function of x, of equation (15). The area under the curve keeps on decreasing as the time increases and finally becomes minimum after a long period of time. This happens when T(x) is represented by a straight line as mentioned above. For more details of this method see Ref. [²⁶].

Data Availability

No data is generated in this manuscript.

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• 1
  L.H.S. of this equation can also be expressed in term of a new variable z = T - T′ and by using boundary conditions it can be rewritten in term of T, where T′ is temperature of surrounding

Edited by

• Editor-in-Chief:
  Marcello Ferreira https://orcid.org/0000-0003-4945-3169

Publication Dates

• Publication in this collection
  23 Feb 2026
• Date of issue
  2026

History

• Received
  01 Aug 2025
• Reviewed
  28 Dec 2025
• Accepted
  13 Jan 2026