Table of Content

What is Conduction?

The process of transmission of heat energy in which heat is transferred from one particle of the medium to the other, but each particle of the medium stays at its own position is called conduction, for example if you hold an iron rod with one of its end on a fire for some time, the handle will get hot. The heat is transferred from the fire to the handle by conduction along the length of the iron rod. The vibrational amplitude of atoms and electrons of the iron rod at the hot end takes on relatively higher values due to the higher temperature of their environment. These increased vibrational amplitudes are transferred along the rod, from atom to atom during the collision between adjacent atoms. In this way, a region of rising temperature extends itself along the rod to your hand


Consider a slab of face area A, Lateral thickness L, whose faces have temperatures TH and TC(TH > TC).

\(\frac{Q}{t} = -KA\frac{dT}{dx}\)… (2.1)

Now consider two cross sections in the slab at positions A and B separated by a lateral distance of dx. Let temperature of face A be T and that of face B be T + DT. Then experiments show that Q, the amount of heat crossing the area A of the slab at position x in time t is given by

Here K is a constant depending on the material of the slab and is named thermal conductivity of the material, and the quantity \(\left (\frac{dT}{dx} \right )\) is called temperature gradiant. The (_) sign in equation (2.1) shows heat flows from high to low temperature (DT is a _ve quantity)

Steady State Conduction Heat Transfer

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

Remember steady-state is distinct from thermal equilibrium for which temperature at any position (x) in the slab must be same.

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

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

and \(\frac{Q}{t} = -KA\frac{\Delta T}{L}\)

Þ \(\frac{Q}{t} = KAQ\frac{T_H-T_C}{L}\) …. (3.1)

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

Ex. 1 One face of an aluminum cube of edge 2 meter is maintained at 100ºC and the other end is maintained at 0ºC. All other surfaces are covered by adiabatic walls. Find the amount of heat flowing through the cube in 5 seconds. (thermal conductivity of aluminum is 209 W/m_ºC)

Sol. Heat will flow from the end at 100ºC to the end at 0ºC.

Area of cross-section perpendicular to direction of heat flow,

A = 4m2

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

Q = \(Q = \frac{(209W/m^{o}C)(4m^{2})(100^{o}C – 0^{o}C)(5 sec)}{2m}\) = 209 KJ Ans.

Thermal Resistance to Conduction

If you are interested in insulating your house from cold weather or for that matter 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, Lateral thickness L and thermal conductivity K,

\(R = -\frac{L}{KA}\) … (4.1)

In terms of R, the amount of heat flowing though a slab in steady-state (in time t)

\(\frac{Q}{t}= -\frac{(T_H-T_L)}{R}\)

If we name \(\frac{Q}{t}\) as thermal current iT

then, \(i_{T} = \frac{T_H-T_L}{R}\) (4.2)

This is mathematically equivalent to OHM’s law, with temperature donning the role of electric potential. Hence results derived from OHM’s law are also valid for thermal conduction.

More over, for a slab in steady state we have seen earlier that the thermal current iL remains 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 1m each, having cross-section area of 1cm2 each and made of Aluminium, copper and steel respectively are maintained at temperatures of 12ºC, 4ºC and 50ºC respectively at their separate ends.

Find the temperature of their common junction. C:\Users\hp\Documents\byjus phy\Untitled-1OE18.JPG

[ KCu = 400 W/m-K , KAl = 200 W/m-K , Ksteel = 50 W/m-K ]

Sol: \(R_{A1} =\frac{L}{KA} = \frac{1}{10^{-4}\times 200} = \frac{10^{4}}{200}\)

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

Let temperature of common junction = T C:\Users\hp\Documents\byjus phy\Untitled-1OE24.JPG

then from Kirchoff;s current laws,

iAl + isteel + iCu = 0

Þ \(\frac{T-12}{R_{Ai}}+\frac{T-51}{R_{steel}}+\frac{T-4}{R_{cu}} = 0\)

Þ (T _ 12) 200 + (T _ 50) 50 + (T _ 4) 400

Þ 4(T _ 12) + (T _ 50) + 8 (T _ 4) = 0

Þ 13T = 48 + 50 + 32 = 130

Þ T = 10ºC Ans.

Heat Conduction Through Slab

Slabs in Series (in steady state)

Consider a composite slab consisting of two materials having different thickness 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 states are maintained at TH and TC, and all lateral surfaces are covered by an adiabatic coating.

C:\Users\hp\Documents\byjus phy\Untitled-1OE28.JPG

Let temperature at the junction be T, since steady state has been achieved thermal current through each slab will be equal. Then thermal current through the first slab.

\(i=\frac{Q}{t} = \frac{T_H-T}{R_1}\) or TH _ T = iR1 … (5.1)

and that through the second slab,

\(i=\frac{Q}{t} = \frac{T-T_c}{R_2}\) or T _ TC = iR2 ….(5.2)

adding eqn. 5.1 and eqn 5.2

TH _ TL = (R1 + R2) i

or \(i= \frac{T_H-T_c}{R_1+R_2}\)

Thus these two slabs are equivalent to a single slab of thermal resistance R1 + R2.

If more than two slabs are joined in series and are allowed to attain steady state, then equivalent thermal resistance is given by

R = R1 + R2 + R3 + ……. …(5.3)

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), sandwitching 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 temperature T4 and T3.

C:\Users\hp\Documents\byjus phy\Untitled-1OE34.JPG

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


and that of brick wall

\(R_2 = \frac{L_2}{K_2A}=\frac{5L_1}{5K_1A}=R_1\)

Let thermal resistance of the each sand witch layer = R. Then the above wall can be visualised as a circuit

C:\Users\hp\Documents\byjus phy\Untitled-1OE38.JPG

thermal current through each wall is same.

Hence \(\frac{25-20}{R_1}=\frac{20-T_3}{R}=\frac{T_3-T_4}{R}=\frac{T_4+20}{R_1}\)

Þ 25 _ 20 = T4 + 20

Þ T4 = _15ºC Ans.

also, 20 _ T3 = T3 _ T4

Þ T3 = \(\frac{25+T_4}{2}\) = 2.5ºC Ans.

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

Sol. R1 = R2 = \(R= \frac{10\times10^{-2m}}{0.25W/m-^{o}C(100m^{2})}\) = 32 × 10_4 ºC/w

R = \(R= \frac{10\times10^{-2m}}{0.25W/m-^{o}C(100m^{2})}\) = 40 × 10_4 ºC/w

the equivalent thermal resistance of the entire wall = R1 + R2 + 2R

= 144 × 10_4 ºC/W

\ Net heat current, i.e. amount of heat flowing out of the house per second = \(\frac{T_H-T_c}{R}\)

= \(\frac{25^{o}C-(-20^{o}C)}{144\times10^{-4o}C/w} = \frac{45\times10^{4}}{144}\) watt

= 3.12 Kwatt

Hence the heater must supply 3.12 kW to compensate for the outflow of heat. Ans.

Slabs in Parallel

C:\Users\hp\Documents\byjus phy\Untitled-1OE49.JPG

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

then \(R_1 = \frac{L}{K_1A_1},R_2=\frac{L}{K_2A_2}\) thermal current through slab 1

\(i_1 = \frac{T_H-T_C}{R_1}\)

and that through slab 2

\(i_2 = \frac{T_H-T_C}{R_2}\)

Net heat current from the hot to cold reservoir

\(i = i_1+i_2 = (T_H – T_C)\left ( \frac{1}{R_1}+\frac{1}{R_2} \right )\)

Comparing with i = \(\frac{T_H-T_C}{R_{eq}}\) , we get,

\(\frac{1}{R_{eq}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}\)

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

\(\frac{1}{R_{eq}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+……(5.4)\)


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

C:\Users\hp\Documents\byjus phy\Untitled-1OE61.JPG

If ends A and E are maintained at temperatures 125ºC and 0ºC respectively, calculate the amount of heat flowing per second from the hot to cold function. [ KCu = 400 W/m-K , Ksteel = 50 W/m-K ]

Sol. \(R_{steel} = \frac{L}{Ka} = \frac{10^{-1}m}{50(W/m-^{o}C)\times10^{-4}m^{2}} = \frac{1000}{50}^{o}C/w.\) .

Similarly RCu = \(R_{cu}= \frac{1000}{400}^{o}C/w\)

Junction C and D are identical in every respect and both will have same temperature. Consequently, the rod CD is in thermal equilibrium and no heat will flow through it. Hence it can be neglected in further analysis.

Now rod BC and CE are in series their equivalent resistance is R1 = RS + RCu similarly rods BD and DE are in series with same equivalent resistance R1 = RS + RCu

these two are in parallel giving an equivalent resistance of

\(\frac{R_1}{2} = \frac{R_s+R_{cu}}{2}\)

This resistance is connected in series with rod AB. Hence the net equivalent of the combination is

\(R= R_{steel}+\frac{R_{1}}{2} = \frac{3R_{steel}+R_{cu}}{2}\)

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

Now \(i = \frac{T_H – T_c}{R} = \frac{125^{o}C}{500\left ( \frac{3}{50}+\frac{1}{400} \right )}^{o}C/w\) = 4 watt. Ans.


Ex. 6 Two thin concentric shells made from copper with radius r1 and r2 (r2 > r1) have a material of thermal conductivity K filled between them. The inner and outer spheres are maintained at temperatures TH and TC respectively by keeping a heater of power P at the centre of the two spheres. Find the value of P.

Sol. Heat flowing per second through each cross-section of the sphere = P = i.

Thermal resistance of the spherical shell of radius x and thickness dx,

\(dR = \frac{dx}{K4\pi x^{2}}\)

Þ \(R = \int_{r_1}^{r_2}\frac{dx}{4\pi x^{2}K} = \frac{1}{4\pi K}\left ( \frac{1}{r_1}-\frac{1}{r_2} \right )\)

thermal current

i = \(P = \frac{T_H-T_C}{R} = \frac{4\pi K(T_H-T_c)r_1R_2}{(r_2-r_1)}\) Ans.


Ex. 7 A container of negligible heat capacity contains 1 kg of water. It is connected by a steel rod of length 10 m and area of cross-section 10cm2 to a large steam chamber which is maintaned at 100ºC. If initial temperature of water is 0ºC, find the time after which it becomes 50ºC. (Neglect heat capacity of steel rod and assume no loss of heat to surroundings) (use table 3.1, take specific heat of water = 4180 J/kg ºC)

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

\(i = \left ( \frac{100-T}{R} \right )\)

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

Þ \(i = \frac{dH}{dt} = ms\frac{dT}{dt}\)

Þ \(\frac{100-T}{R} = ms\frac{dT}{dt}\)

Þ \(\int_{0}^{50}\frac{dT}{100-T} = \int_{0}^{1}\frac{dT}{Rms}\)

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

or t = Rms ln2 sec

= \(= \frac{L}{KA1}ms\ln2 sec\) \(=\frac{(10m)(1kg)(4180J/kg-^{o}C)}{46(w/m^{o}C)\times(10\times10^{-4}m^{2})}\ln2\) \(=\frac{418}{46}(0.69)\times10^{5}\)

= 6.27 × 105 sec

= 174.16 hours Ans.

Can you now see how the following facts can be explained by thermal conduction?

(a) In winter, iron chairs appears to be colder than the wooden chairs.

(b) Ice is covered in gunny bags to prevent melting.

(c) Woolen clothes are warmer.

(d) We feel warmer in a fur coat.

(e) Two thin blankets are warmer than a single blanket of double the thickness.

(f) Birds often swell their feathers in winter.

(g) A new quilt is warmer than old one.

(h) Kettles are provided with wooden handles.

(i) Eskimo’s make double walled ice houses.

(j) Thermos flask is made double walled.

What is Convection?

When heat is transferred from one point to the other through 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 greater temperature is less dense, while that at the lower temperature is denser. Hence hot fluids rise up white colder fluid sink down, accounting for convection. In the absence of gravity, convection would not be possible.

Also, the anomalous behavior of water (its density increases with temperature in the range 0-4ºC) give rise to interesting consequences vis-a-vis the process of convection. One of these interesting consequences is the presence of aquatic life in temperate and polar waters. The other is the rain cycle.

Can you now see how the following facts can be explained by thermal convection?

(a) Oceans freeze top-down and not bottom up. (this fact is singularly responsible for the presence of aquatic life is temperate and polar waters.)

(b) The temperature in the bottom of deep oceans is invariably 4ºC, whether it is winter or summer.

(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.

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