Conduction is the process of heat transfer between objects that are in direct physical contact with each other.

Conduction is the process by which energy is transmitted from one particle of a medium to another without the particles moving from their original positions. In Physics and Chemistry, conduction usually refers to the transfer of heat energy or electric charge through a material, and can occur in solids, liquids, and gases.

When conduction of heat occurs, the heat energy is usually transferred from molecule to another molecule as they are in direct contact with each other; however, there is no change in the position of the molecules; they simply vibrate amongst each other.

During the conduction of electricity, there is a movement of electrically charged particles in the medium. As such, the electric current is typically carried and moved by ions or electrons.

Conduction Examples

The heat is being transferred from the tip of the iron rod, which is kept in the fire, to the handle by conduction along the length of the rod. This is due to the increased vibrational amplitude of the atoms and electrons in the rod at the hot end, which is caused by the higher temperature of the environment.

The vibrational amplitudes increase and are transferred along the rod from atom to atom during collisions between adjacent atoms. This causes a region of rising temperature to extend along the rod to your hand.

A slab with face area A, lateral thickness L, and faces with temperatures T_H and T_C (T_H > T_C) should be considered.

The amount of heat, Q, crossing the area A of the slab at position x in time t, which has two cross-sections at positions A and B separated by a lateral distance of dx, with the temperature of face A being T and that of face B being T + DT, is given by experiments.

$\begin{array}{l}-K\frac{A}{t}\frac{dT}{dx}=Q\end{array}$

The (_) sign in equation (1) shows heat flows from high to low temperature.

Types of Conduction

There are two main types of conduction:

1. Thermal conduction
2. Electrical conduction

Steady-State Conduction

If the temperature of a cross-section at any position x in the above slab remains constant over time (remember, it does vary with position x), the slab is said to be in a steady-state.

Remember that steady-state is different from thermal equilibrium, where the temperature at any position (x) in the slab must be the same.

The left and right faces of a conductor in steady-state are maintained at constant temperatures T_H and T_C respectively, and all other faces must be covered with adiabatic walls so that no heat escapes through them. As a result, there is no absorption or emission of heat at any cross-section and the same amount of heat (Q₁ = Q = Q₂) flows through each cross-section in a given Interval of time. Therefore, the temperature gradient is constant throughout the slab.

Hence, $$\frac{dT}{dx}=\frac{\mathrm{\Delta }T}{L}=\frac{T_1-T_2}{L}=\frac{T_c-T_H}{L}$$

$\frac{Q}{t}=-KA\frac{\mathrm{\Delta }T}{L}$

$\begin{array}{l}\frac{Q}{t}=KA\frac{T_H-T_C}{L}\end{array}$

Here, Q is the amount of heat flowing through a cross-section of a slab at any position within a time interval of t.

Ex. 1 Find the amount of heat flowing through an aluminium cube of edge 2 meters, with one face maintained at 100ºC and the other end maintained at 0ºC, with all other surfaces covered by adiabatic walls, in 5 seconds. (thermal conductivity of aluminium is 209 W/mºC).

Sol. The heat will flow from the end with a temperature of 100ºC to the end with a temperature of 0ºC.

Area of Cross-Section Perpendicular to Direction of Heat Flow

A = 4m²

Then $$\frac{Q}{t}=KA\frac{(T_H-T_C)}{L}$$

$\begin{array}{l}Q=\frac{(209W/m^{o}C)(4m^2)({100}^{o}C–0^{o}C)(5sec)}{2m}\end{array}$

= 209 KJ

Transient Conduction or Non-Steady-State Conduction

During transient conduction, temperatures can vary at any point within an object at a given time. This mode of conduction is also known as “non-steady-state” conduction. The key point to consider here is the object’s time-dependence of temperature.

Non-steady-state conduction usually occurs when a temperature change is introduced within the outer areas or inside an object, brought about by the sudden entry of a new source of heat.

An example of transient conduction is the starting of an engine in a vehicle. Initially, a new source of heat is added when the engine is turned on, but this only lasts for a brief moment. As the engine reaches its operating temperature, the steady-state phase is established.

Fourier’s Law

Fourier’s law, also known as the law of heat conduction, states that the rate of heat transfer through a material is proportional to the negative gradient in the temperature and to the area, at right angles to that gradient, through which the heat flows.

The law can further be expressed in two forms: the integral form, which considers the amount of energy that is moving in or out of a body in total, and the differential form, which considers the rate of flow or the energy fluxes locally.

Meanwhile, Fourier’s law also has various representations, such as Newton’s law of cooling which is its discrete analogue and Ohm’s law which is its electrical analogue.

Thermal Resistance to Conduction

If you are interested in insulating your house from cold weather or keeping the meal hot in your tiffin-box, you are more interested in poor heat conductors, rather than good conductors. For this reason, the concept of thermal resistance R has been introduced.

For a slab of cross-section A, with lateral thickness L and thermal conductivity K,

$\begin{array}{l}R=\frac{L}{KA}\end{array}$

In terms of R, what is the amount of heat flowing through a slab in steady-state (at time t)?

$\begin{array}{l}Q=-\frac{(T_H-T_L)}{R}\cdot t\end{array}$

If we name Q/t as thermal current i_T, then

$$i_T=\frac{T_H-T_L}{R}$$

Ohm’s law is mathematically equivalent when temperature is used in place of electric potential. Therefore, results obtained from Ohm’s law are applicable to thermal conduction.

Moreover, for a slab in steady-state, we have seen earlier that the thermal current i_L remains the same at each cross-section. This is analogous to Kirchoff’s current law in electricity, which can now be very conveniently applied to thermal conduction.

Ex. 2 Three identical rods of length 1 m each, having cross-section area of 1 cm² each and made of Aluminium, Copper, and Steel respectively are maintained at temperatures of 12ºC, 4ºC, and 50ºC respectively at their separate ends.

Determine the temperature at the junction of the two components.

K_Cu = 400 W/m-K, K_Al = 200 W/m-K, K_steel = 50 W/m-K

Sol: R_A1 = 10⁴/200

Similarly $$R_{steel}=\frac{{10}^4}{50};and;R_{copper}=\frac{{10}^4}{400}$$

Let the temperature of the common junction be equal to T.

Then from Kirchoff’s Current Laws.

i_Al + i_Steel + i_Cu = 0

$$\begin{array}{l}\frac{T-12}{R_{Ai}}+\frac{T-51}{R_{steel}}+\frac{T-4}{R_{cu}}=0\end{array}$$

(T_12) 200 + (T_50) 50 + (T_4) 400

4T - 12 + T - 50 + 8T - 4 = 0

13T = 48 + 50 + 32 = 130

$ÞT=10ºC$

Heat Conduction Through a Slab

Slabs in Series (in Steady-State)

Consider a composite slab composed of two materials with different thicknesses (L₁ and L₂), different cross-sectional areas (A₁ and A₂), and different thermal conductivities (K₁ and K₂). The temperature at the outer surface of the slab is maintained at T_H and T_C, and all lateral surfaces are covered by an adiabatic coating.

The thermal current through the first slab will be equal to the temperature at the junction, T, since steady-state has been achieved.

$\begin{array}{l}i=\frac{Q}{t}=\frac{T_H-T}{R_1}\end{array}$

or TH _ T = iR1 ——(1)

And, through the second slab,

$\begin{array}{l}{T}_c=\frac{Q}{t}{R}_2-T\end{array}$

or T _ TC = iR2 ——-(2)

adding eqn.1 and eqn 2

T_H - T_C = (R₁ + R₂) * i

or

$\begin{array}{l}i=\frac{T_H-T_c}{R_1+R_2}\end{array}$

Thus, the total thermal resistance of these two slabs can be represented as R₁ + R₂.

If more than two slabs are joined in series and are allowed to attain steady-state, then the equivalent thermal resistance is calculated using the sum of the individual thermal resistances.

R = R₁ + R₂ + R₃ + … + R_n

Ex. 3 The figure shows the cross-section of the outer wall of a house built in a hill-resort to keep the house insulated from the freezing temperature of outside. The wall consists of teak wood of thickness L₁ and brick of thickness (L₂ = 5L₁), sandwiching two layers of an unknown material with identical thermal conductivities and thickness. The thermal conductivity of teak wood is K₁ and that of brick is (K₂ = 5K). Heat conduction through the wall has reached a steady-state with the temperature of three surfaces being known (T₁ = 25ºC, T₂ = 20ºC and T₅ = - 20ºC). Find the interface temperatures T₄ and T₃.

Sol. Let the interface area be A, then the thermal resistance of wood,

$$\begin{array}{l}{R}_1=\frac{L_1}{K_{1}A}\end{array}$$

And that of the brick wall,

$$\begin{array}{l}{R}_2=\frac{5L_1}{5K_{1}A}=\frac{L_2}{K_{2}A}=R_1\end{array}$$

The above wall can be visualised as a circuit, with the thermal resistance of each layer equal to R.

The thermal current through each wall is the same.

Hence $\begin{array}{l}\frac{25-20}{R_1}=\frac{20-T_3}{R}=\frac{T_3-T_4}{R}=\frac{T_4+20}{R_1}\end{array}$

25 - 20 = T₄ + 20

T4 = 15°C

20 _T3 = T3_T4

$T_3=\frac{25+T_4}{2}={2.5}^{\circ }C$

Ex. 4 In example 3, K1 = 0.125 W/mºC, K2 = 5K1 = 0.625 W/mºC and thermal conductivity of the unknown material is K = 0.25 W/mºC. L₁ = 4cm, L₂ = 5L₁ = 20cm and L = 10cm. If the total wall area of the single room in the house is 100 m₂, then find the power of the electric heater being used in the room.

Sol.

i = Heat Transfer Rate
T_H = Hot Temperature
T_C = Cold Temperature
R_i = Thermal Resistance

i = (T_H - T_C)/Ri

T_H = 250ºC

T_C = 200º

$R1=\frac{L1}{K1A}$

= 4 x 10^-2/ (0.125 x 100) = 4 x 10^-2/125 = 32 x 10^-4 ºC/W

Therefore,

$i=\frac{25-20}{32\times {10}^{-4}}$

5/32 * 10^-4

1562.5 Watt

1.56 KWatt

Slabs in Parallel

Consider two slabs held between the same heat reservoirs, with thermal conductivities K₁ and K₂, and cross-sectional areas A₁ and A₂.

Then $R_1=\frac{L}{K_1{A}_1},R_2=\frac{L}{K_2{A}_2}$

$\begin{array}{l}{i}_1=\frac{T_H-T_C}{R_1}\end{array}$

And that, through Slab 2.

$\begin{array}{l}{i}_2=\frac{T_H-T_C}{R_2}\end{array}$

Net heat current from the cold to hot reservoir

$$\begin{array}{l}i=i_1+i_2=(T_H-T_C)(\frac{1}{R_1}+\frac{1}{R_2})\end{array}$$

$$\begin{array}{l}\text{We can calculate }i=\frac{T_H-T_C}{R_{eq}}\text{ to compare.}\end{array}$$

$$\begin{array}{l}{R}_{eq}=\frac{R_1{R}_2}{R_1+R_2}\end{array}$$

If more than two rods are joined in parallel, the equivalent thermal resistance is given by:

$$\begin{array}{l}\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\cdots (5.4)\end{array}$$

Ex. 5 Three copper rods and three steel rods, each of length l = 10 cm and area of cross-section 1 cm², are connected as shown:

The amount of heat flowing per second from the hot to cold function can be calculated using the following formula:

$$Q=KCu\cdot A\cdot (T_H-T_C)$$

where

• Q is the amount of heat flowing per second (in W)
• KCu is the thermal conductivity of copper (in W/m-K)
• A is the area of the two ends (in m^2)
• T_H is the temperature of end A (in ºC)
• T_C is the temperature of end E (in ºC)

Given that $KCu=400\text{ W/m-K}$, $Ksteel=50\text{ W/m-K}$, $T_H=125\text{ ºC}$, and $T_C=0\text{ ºC}$, the amount of heat flowing per second is:

$$Q=400\text{ W/m-K}\cdot A\cdot (125\text{ ºC}-0\text{ ºC})$$

Sol. $$\begin{array}{l}{R}_{steel}=\frac{1000}{50\times (W/m{-}^{o}C)}=20W/m{-}^{o}C\end{array}$$

$$\begin{array}{l}{R}_{cu}={\frac{1000}{400}}^{o}C/w\end{array}$$

Since Junction C and D are identical and in thermal equilibrium, the rod CD can be neglected in further analysis.

Now, rods BC and CE are in series, and their equivalent resistance is R₁ = R_S + R_Cu. Similarly, rods BD and DE are in series with the same equivalent resistance R₁ = R_S + R_Cu.

These two are in parallel, giving an equivalent resistance of

$$\begin{array}{l}\frac{R_1}{2}=\frac{R_s+R_{cu}}{2}\end{array}$$

The net equivalent resistance of the combination of this resistance and rod AB connected in series is

$$\begin{array}{l}R=\frac{3R_{steel}+R_{cu}}{2}\end{array}$$

$500{(\frac{3}{50}+\frac{1}{400})}^{o}C/w$

Now $i=\frac{T_H-T_c}{R}={\frac{{125}^{\circ }C}{500(\frac{3}{50}+\frac{1}{400})}}^{\circ }C/W=4W$

Ex. 6 Find the value of P given two thin concentric shells made from copper with radius r1 and r2 (r2 > r1) filled with material of thermal conductivity K, and the inner and outer spheres maintained at temperatures TH and TC, respectively.

Sol. The rate of heat flow per second through each cross-section of the sphere is equal to P = i.

The thermal resistance of a spherical shell with a radius of x and a thickness of dx

$$\begin{array}{l}dR=\frac{K4\pi x^2}{dx}\end{array}$$

$$\begin{array}{l}R=\frac{1}{4\pi K}(\frac{1}{r_2}-\frac{1}{r_1})={\int }_{r_1}^{r_2}\frac{dx}{4\pi x^{2}K}\end{array}$$

Thermal Current

$$\begin{array}{l}i=P\cdot \frac{R}{T_H-T_C}=\frac{4\pi K{r}_1{R}_2}{(r_2-r_1)(T_H-T_c)}\end{array}$$

Ex. 7 If the initial temperature of the 1 kg of water in a container of negligible heat capacity is 0ºC, and the container is connected by a steel rod of length 10 m and area of cross-section 10cm² to a large steam chamber which is maintained at 100ºC, then the time after which the water becomes 50ºC can be calculated using Table 3.1 and taking the specific heat of water as 4180 J/kg ºC. (Neglect heat capacity of steel rod and assume no loss of heat to surroundings).

Sol. Let the temperature of water at time t be T, then the thermal current at time t is,

$$\begin{array}{l}i=\left(\frac{T-100}{R}\right)\end{array}$$

This increases the temperature of water from T to T + dT

$$\begin{array}{l}i=\frac{dH}{dt}=m\frac{dT}{dt}\end{array}$$

$$\begin{array}{l}ms\frac{dT}{dt}=\frac{100-T}{R}\end{array}$$

$$\begin{array}{l}{\int }_0^{50}\frac{dT}{100-T}={\int }_0^1\frac{dT}{Rms}\end{array}$$

$\ln \left(\frac{1}{2}\right)=\frac{t}{Rms}$

$t=\mathrm{Rms}\ln (2\sec )$

$\frac{L}{KA1}\cdot \mathrm{ms}\cdot \ln (2)\sec$

$\frac{(10m)(1kg)(4180J/kg{-}^{o}C)}{46(w/m^{o}C)\times (10\times {10}^{-4}{m}^2)}\ln 2$

$\frac{418}{46}(0.69)\times {10}^5$

6.27 x 10⁵ sec

174.16 hours

How does thermal conduction explain the following facts?

• (a) Iron chairs seem to be colder than wooden chairs during the winter.
• (b) Ice is covered in gunny bags to prevent its melting.
• (c) Clothes made from wool are warmer.
• (d) We feel warmer in a fur coat.
• (e) Two thin blankets are warmer than one thick blanket of double the * thickness.
• (f) Birds often fluff their feathers in winter.
• (g) The old quilt is not as warm as the new one.
• (h) Kettles are provided with wooden handles.
• (i) Eskimos make double-walled ice houses.
• (j) A thermos flask is made with double walls.

What is Convection?

Convection is the transfer of heat through the movement of fluids or particles. It is the most efficient way to transfer heat from one place to another.

When heat is transferred from one point to another through the actual movement of heated particles, the process of heat transfer is called Convection. In liquids and gases, some heat may be transported through Conduction. But most of the transfer of heat in them occurs through the process of convection. Convection occurs through the aid of Earth’s gravity. Normally, the portion of fluid at a greater temperature is less dense, while that at a lower temperature is denser. Hence, hot fluids rise up while colder fluids sink down, accounting for convection. In the absence of gravity, convection would not be possible.

The anomalous behaviour of water (its density increases with temperature in the range 0-4ºC) has interesting consequences for convection. One of these is the presence of aquatic life in temperate and polar waters, and the other is the rain cycle.

How can thermal convection explain the following facts?

• (a) The fact that oceans freeze top-down and not bottom-up is singularly * responsible for the presence of aquatic life in temperate and polar waters.
• (b) The temperature at the bottom of deep oceans is always 4ºC, * regardless of the season.
• (c) You cannot illuminate the interior of a lift in free fall or * an artificial satellite of earth with a candle.
• (d) You can Illuminate your room with a Candle.

Frequently Asked Questions on Conduction

An example of conduction in everyday life is when you hold a metal spoon that has been sitting in hot water; the heat from the water is transferred to your hand through the spoon.

Heating a pan placed on a stove.

What are the three types of conduction?

• The three types of conduction are:
  • 1. Thermal conduction
  • 2. Electrical conduction
  • 3. Conduction of sound

What are the three types of conduction?

• Heat Conduction
• Electric Conduction
• Photoconductivity

What is Convection?

Convection is the transfer of heat by the movement of a fluid, such as air or water. It is one of the three main ways that heat is transferred (the other two being conduction and radiation). Convection occurs when particles with a lot of heat energy move and take the place of particles with less heat energy. This movement of particles creates circulation in the fluid, which helps to spread the heat.

The process of heat transfer occurs through the movement of heated fluids, such as water or air.

The three modes of heat transfer are conduction, convection, and radiation.

• Conduction
• Convection
• Radiation