Home › Physics › Thermal Physics › Thermal Conduction in Solids
Thermal Physics
Thermal Conduction in Solids
Thermal conduction is the process of heat transfer through a material medium from regions of higher temperature to regions of lower temperature without the actual bulk movement of the medium particles. It is the primary mode of heat transfer in opaque solids, governed by temperature gradients and the microscopic interactions of lattice vibrations and free electrons.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Definition of thermal conductivity, Searle's and Ingen-Hausz methods, steady state heat flow, linear flow of heat through slabs and composite bars.
PGT extension
Wiedemann-Franz law, temperature dependence of thermal conductivity, thermal diffusivity, radial heat flow in cylinders and spheres, phonon scattering mechanisms in non-metals.
Prerequisites
Concept of Temperature and Zeroth Law
Kinetic theory of gases and microscopic interpretation of temperature
Law of equipartition of energy
Mechanism of Thermal Conduction
In non-metals and insulators, thermal conduction occurs primarily via the propagation of elastic waves or lattice vibrations known as phonons.
In metals, thermal conduction is dominated by the diffusion of free conduction electrons which transport both charge and kinetic energy effectively.
The steady state is achieved when the temperature at every point within the conductor remains constant with respect to time, meaning no further heat is absorbed by any cross-section.
Coefficient of Thermal Conductivity
Thermal conductivity (K) is a material property that quantifies its ability to conduct heat.
The rate of heat flow is directly proportional to the cross-sectional area and the temperature gradient, and inversely proportional to the length of the conductor.
SI units of thermal conductivity are W m-1 K-1 or J s-1 m-1 K-1.
Thermal Resistance and Composite Slabs
Thermal resistance is defined as the ratio of temperature difference to the rate of heat flow, analogous to electrical resistance.
For bodies in series, the equivalent thermal resistance is the sum of individual thermal resistances: Req = R1 + R2.
For bodies in parallel, the reciprocal of the equivalent thermal resistance is the sum of reciprocals of individual thermal resistances.
Wiedemann-Franz Law
The law states that the ratio of the thermal conductivity (K) to the electrical conductivity (σ) of a metal is directly proportional to the absolute temperature (T).
The proportionality constant is known as the Lorenz number (L), which is approximately independent of the specific metal at moderate to high temperatures.
Expressed mathematically as K / (σ T) = L, where L ≈ 2.44 × 10-8 V2 K-2.
Derivations
Fourier's Law of Heat Conduction and Temperature Gradient
Conditions: Steady state heat flow through a uniform slab of cross-sectional area A and thickness x.
Assume a uniform slab of thickness dx with temperature difference dT across it.
The rate of heat flow dQ/dt is experimentally found to be proportional to area A and temperature gradient -dT/dx.
Introduce the proportionality constant K (thermal conductivity): dQ/d$t = -K A (dT/dx)$.
Integrate across a finite length L with temperatures T1 and T2 to obtain total heat transfer.
Result: frac{dQ}{dt} = \frac{KA (T_1 - T_2)}{L}
Equivalent Thermal Resistance for Composite Slabs in Series
Conditions: Two slabs of thicknesses L1, L2, thermal conductivities K1, K2, and identical cross-sectional area A placed in series.
Let the temperatures at the outer faces be T1 and T3, and at the junction be T2.
In steady state, the rate of heat flow through both slabs must be equal: H = H1 = H2.
Write the heat flow equations for each slab: $H = \frac{KA(T_1 - T_2)}{L_1}$ and $H = \frac{K_2A(T_2 - T_3)}{L_2}$.
Express temperature drops as T1 - T$2 = H(L1 / K1A)$ and T2 - T$3 = H(L2 / K2A)$.
Add the equations to eliminate T2, yielding $H = \frac{T_1 - T_3}{(L_1/K_1A) + (L_2/K_2A)}$.
dQ/dt = rate of heat flow, K = thermal conductivity, A = area, dT/dx = temperature gradient W
Steady state, isotropic medium
Heat Flow in Finite Slab
$$H = \frac{KA(T_1 - T_2)}{L}$$
H = heat current, T1, T2 = face temperatures, L = thickness W
Uniform cross-section, steady state
Thermal Resistance
$$R_{th} = \frac{L}{KA}$$
Rth = thermal resistance, L = length, K = conductivity, A = area K W-1
Linear heat flow
Temperature Gradient
$$G = \frac{T_1 - T_2}{L}$$
G = temperature gradient, T1, T2 = temperatures, L = distance K m-1
Constant cross-section
Thermal Diffusivity
$$\alpha = \frac{K}{\rho c}$$
alpha = thermal diffusivity, K = thermal conductivity, rho = density, c = specific heat capacity m2 s-1
Homogeneous medium
Wiedemann-Franz Law
$$\frac{K}{\sigma T} = L$$
K = thermal conductivity, sigma = electrical conductivity, T = absolute temperature, L = Lorenz number V2 K-2
Metals at moderate to high temperatures
Equivalent Thermal Conductivity for Parallel Slabs
$$K_{eq} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$$
Keq = equivalent conductivity, K1, K2 = conductivities, A1, A2 = areas W m-1 K-1
Parallel combination, equal length
Ingen-Hausz Law (Wax melting length)
$$\frac{l_1^2}{l_2^2} = \frac{K_1}{K_2}$$
l1, l2 = lengths of wax melted on rods, K1, K2 = thermal conductivities Dimensionless ratio
Identical rods coated with wax, identical surface conditions
Graphs and experiments
Temperature Profile along a Uniform Conductor in Steady State
Draw: A linear graph of temperature (T) on the y-axis versus distance from the hot end (x) on the x-axis, showing a constant negative slope.
Interpret: The constant slope indicates a uniform temperature gradient (dT/dx = constant) throughout the length of the conductor in steady state.
Exam use: Tested in conceptual multiple-choice questions to differentiate between steady state and variable state temperature profiles.
Searle's Apparatus Experiment
Draw: Schematic setup showing a metal rod heated at one end by steam and cooled at the other by a continuous water flow, with thermometers at fixed intervals.
Interpret: Used to determine the absolute thermal conductivity of a good conductor by measuring steady-state temperatures and mass flow rate of cooling water.
Exam use: Direct questions on experimental formulas and precautions for measuring K.
Ingen-Hausz Experiment
Draw: Several rods of different materials of identical dimensions coated with wax, dipped at one end into a hot water bath.
Interpret: The length to which wax melts on each rod is proportional to the square root of its thermal conductivity.
Exam use: Comparative questions on thermal conductivities using the relation l^2 ∝ K.
Solved numericals
TGT core
A brass rod of length 0.5 m and cross-sectional area 2 cm2 has its ends maintained at 100°C and 0°C. If the thermal conductivity of brass is 109 W m-1 K-1, calculate the rate of heat flow through the rod in steady state.
Identify given values: L = 0.5 m, $A = 2 cm^{2} = 2 \times 10^{-4} m^{2}$, T$1 = 100°C$, T$2 = 0°C$, $K = 109 W m^{-1} K^{-1}$.
Apply Fourier's law formula: $H = \frac{KA(T_1 - T_2)}{L}$.
Two identical metal rods are connected in series. Their thermal conductivities are K$1 = 200 W m^{-1} K^{-1}$ and K$2 = 400 W m^{-1} K^{-1}$. If the free ends are kept at 100°C and 0°C respectively, find the temperature of the junction in steady state.
Let the length of each rod be L and cross-sectional area be A. Let junction temperature be T.
In steady state, the rate of heat flow through both rods is equal: H1 = H2.
Set up the equation: \frac{K_1 A (100 - T)}{L} = \frac{K_2 A (T - 0)}{L}.
Cancel common terms A and L, and substitute K1 and K2: 200(100 - T) = 400(T).
Using the Wiedemann-Franz law, calculate the theoretical thermal conductivity of a metal at 300 K if its electrical conductivity is 6 × 107 Ω-1 m-1 and the Lorenz number is 2.45 × 10-8 V2 K-2.
Fourier's law assumes that thermal conductivity K is independent of temperature, which is strictly valid only for small temperature intervals.
Heat flow equations assume steady state conditions where internal energy of the conductor elements does not change with time.
Lateral heat losses to the surrounding medium are neglected in standard 1D linear conduction models.
Common misconceptions
Common error: Heat flows faster in a conductor if its temperature is higher, regardless of temperature gradient.
Correct: Heat flow rate depends strictly on the temperature gradient (dT/dx) and thermal conductivity, not the absolute temperature alone.
Fourier's law dictates that heat current is driven by spatial temperature differences, not absolute thermal levels.
Common error: Thermal conductivity and heat capacity are the same physical property.
Correct: Thermal conductivity measures the rate of heat transport, whereas heat capacity measures the ability to store thermal energy.
They govern different aspects of thermal behavior; their ratio determines thermal diffusivity.
Quick revision
Thermal conductivity K measures a material's intrinsic ability to conduct heat via electrons and phonons.
In steady state, the temperature gradient along a uniform rod is constant.
Thermal resistance in series adds algebraically, analogous to electrical resistors.
Wiedemann-Franz law links thermal and electrical conductivities through the Lorenz number proportional to temperature.
Must remember
In non-metals and insulators, thermal conduction occurs primarily via the propagation of elastic waves or lattice vibrations known as phonons.
In metals, thermal conduction is dominated by the diffusion of free conduction electrons which transport both charge and kinetic energy effectively.
The steady state is achieved when the temperature at every point within the conductor remains constant with respect to time, meaning no further heat is absorbed by any cross-section.
Thermal conductivity ($K$) is a material property that quantifies its ability to conduct heat, with SI units $W m^{-1} K^{-1}$.
Fourier's law states that the rate of heat flow is given by $\frac{dQ}{dt} = -KA \frac{dT}{dx}$.
Thermal resistance for linear heat flow is defined as $R_{th} = \frac{L}{KA}$.
For bodies in series, the equivalent thermal resistance is the sum of individual thermal resistances: $R_{eq} = R_1 + R_2$.
The Wiedemann-Franz law states that the ratio $\frac{K}{\sigma T} = L$ is a constant known as the Lorenz number.
Thermal diffusivity is expressed as $\alpha = \frac{K}{\rho c}$.
Ingen-Hausz law relates the lengths of melted wax on rods as $\frac{l_1^2}{l_2^2} = \frac{K_1}{K_2}$.
Glossary
Thermal Conductivity ($K$)A material property quantifying its intrinsic ability to conduct heat per unit temperature gradient.
Steady StateA condition in heat transfer where the temperature at every point inside the conductor remains constant with respect to time.
Thermal Resistance ($R_{th}$)The opposition offered by a body to the flow of heat, defined as the ratio of temperature difference to the rate of heat flow.
Lorenz Number ($L$)The proportionality constant in the Wiedemann-Franz law, approximately $2.44 \times 10^{-8} V^2 K^{-2}$.
Thermal Diffusivity ($\alpha$)The ratio of thermal conductivity to volumetric heat capacity, determining how fast a body responds to temperature changes.
PhononsQuasiparticles representing quantized lattice vibrations responsible for thermal conduction in non-metals and insulators.
Conduction Mechanism: Metals vs. Non-Metals
Feature
Metals
Non-Metals and Insulators
Primary Charge/Energy Carriers
Free conduction electrons
Phonons (lattice vibrations)
Wiedemann-Franz Law Applicability
Valid (high thermal and electrical conductivity correlation)
Not applicable (low electrical conductivity)
Thermal Conductivity Magnitude
Generally high
Generally low
Practice Quiz
Choose an option to reveal the explanation.
1. In non-metals and insulators, thermal conduction occurs primarily via the propagation of what entities?
Explanation: In non-metals and insulators, thermal conduction occurs primarily via the propagation of elastic waves or lattice vibrations known as phonons.
2. What is the primary mechanism of thermal conduction in pure metals?
Explanation: In metals, thermal conduction is dominated by the diffusion of free conduction electrons which transport both charge and kinetic energy effectively.
3. When is the steady state achieved in a thermal conductor?
Explanation: The steady state is achieved when the temperature at every point within the conductor remains constant with respect to time, meaning no further heat is absorbed by any cross-section.
4. What are the SI units of thermal conductivity ($K$)?
Explanation: The SI units of thermal conductivity are $W m^{-1} K^{-1}$ or $J s^{-1} m^{-1} K^{-1}$.
5. According to Fourier's law of heat conduction, what is the rate of heat flow $\frac{dQ}{dt}$ proportional to?
Explanation: Fourier's law states that $\frac{dQ}{dt} = -KA \frac{dT}{dx}$, meaning heat flow rate is directly proportional to cross-sectional area and the temperature gradient.
6. What is the expression for thermal resistance $R_{th}$ of a conductor of length $L$, area $A$, and conductivity $K$?
Explanation: Thermal resistance is defined as $R_{th} = \frac{L}{KA}$, analogous to electrical resistance.
7. How are equivalent thermal resistances combined for bodies in series?
Explanation: For bodies in series, the equivalent thermal resistance is the sum of individual thermal resistances: $R_{eq} = R_1 + R_2$.
8. What does the Wiedemann-Franz law relate?
Explanation: The Wiedemann-Franz law states that the ratio of thermal conductivity ($K$) to electrical conductivity ($\sigma$) of a metal is directly proportional to the absolute temperature ($T$). Mathematically, $\frac{K}{\sigma T} = L$.
9. What is the approximate value of the Lorenz number ($L$)?
Explanation: The Lorenz number is approximately $L \approx 2.44 \times 10^{-8} V^2 K^{-2}$.
10. What is the formula for thermal diffusivity ($\alpha$)?
Explanation: Thermal diffusivity is defined as $\alpha = \frac{K}{\rho c}$, where $K$ is thermal conductivity, $\rho$ is density, and $c$ is specific heat capacity.
11. In Ingen-Hausz experiment, if $l_1$ and $l_2$ are lengths of wax melted on two rods of conductivities $K_1$ and $K_2$, what is the relation?
Explanation: Ingen-Hausz law gives $\frac{l_1^2}{l_2^2} = \frac{K_1}{K_2}$ under identical surface conditions and steady-state temperature distribution.
12. What is the equivalent thermal conductivity ($K_{eq}$) for two parallel slabs of equal length and cross-sectional areas $A_1$ and $A_2$?
Explanation: For parallel combination with equal length, $K_{eq} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$.
13. What physical property does thermal capacity measure as opposed to thermal conductivity?
Explanation: Thermal conductivity measures the rate of heat transport, whereas heat capacity measures the ability to store thermal energy.
14. What is the temperature gradient $G$ across a slab of thickness $L$ with face temperatures $T_1$ and $T_2$ ($T_1 > T_2$)?
Explanation: Temperature gradient is defined as $G = \frac{T_1 - T_2}{L}$, representing spatial rate of change of temperature.
15. Why does heat flow through a uniform solid rod in steady state?
Explanation: Fourier's law dictates that heat current is driven by spatial temperature differences (temperature gradient), not absolute thermal levels.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. Two metal rods of the same length and cross-sectional area are connected in series. Their thermal conductivities are $K_1 = 200\ W m^{-1} K^{-1}$ and $K_2 = 400\ W m^{-1} K^{-1}$. What is the equivalent thermal conductivity of the combination?
Explanation: For two rods of equal length and area in series, $R_{eq} = R_1 + R_2 \implies \frac{2L}{K_{eq}A} = \frac{L}{K_1 A} + \frac{L}{K_2 A}$. Thus $\frac{2}{K_{eq}} = \frac{1}{K_1} + \frac{1}{K_2} = \frac{1}{200} + \frac{1}{400} = \frac{3}{400}$. Therefore, $K_{eq} = \frac{800}{3} = 266.67\ W m^{-1} K^{-1}$.
2. The thermal conductivity of a copper rod is $400\ W m^{-1} K^{-1}$ and its electrical conductivity is $6\times 10^7\ \Omega^{-1} m^{-1}$. If the absolute temperature is $300\ K$, what is the calculated value of the Lorenz number using the Wiedemann-Franz law?
3. A composite slab consists of two layers of equal thickness $L$ and thermal conductivities $K$ and $2K$. In the steady state, the temperature difference across the entire slab is $30\ K$. What is the temperature drop across the layer with thermal conductivity $K$?
Explanation: Thermal resistances are $R_1 = \frac{L}{KA}$ and $R_2 = \frac{L}{2KA} = \frac{R_1}{2}$. The temperature drop is proportional to thermal resistance. Since $R_1 = 2 R_2$, the drop across the first layer is $\frac{R_1}{R_1 + R_2} \Delta T = \frac{2}{3} \times 30\ K = 20\ K$.
4. In an Ingen-Hausz experiment, the lengths of wax melted on two rods of the same cross-section are $10\ cm$ and $20\ cm$ respectively. If the thermal conductivity of the first rod is $K_1 = 100\ W m^{-1} K^{-1}$, find the thermal conductivity of the second rod $K_2$.
Explanation: According to Ingen-Hausz law, $\frac{l_1^2}{l_2^2} = \frac{K_1}{K_2}$. Substituting values: $\frac{10^2}{20^2} = \frac{100}{K_2} \implies \frac{1}{4} = \frac{100}{K_2} \implies K_2 = 400\ W m^{-1} K^{-1}$.
5. The rate of heat flow through a copper rod of length $0.5\ m$ and cross-sectional area $2\times 10^{-4}\ m^2$ is $10\ W$. If the temperature of the hot end is $100\ ^\circ C$, what is the temperature of the cold end? (Take $K = 400\ W m^{-1} K^{-1}$)
Explanation: Using $H = \frac{KA(T_1 - T_2)}{L}$, we get $10 = \frac{400 \times (2 \times 10^{-4}) \times (100 - T_2)}{0.5}$. Solving: $10 = \frac{0.08 (100 - T_2)}{0.5} = 0.16 (100 - T_2) \implies 100 - T_2 = \frac{10}{0.16} = 62.5 \implies T_2 = 100 - 62.5 = 37.5\ ^\circ C$ (Wait, check calculation: $0.08 / 0.5 = 0.16$. $10 / 0.16 = 62.5$. $100 - 62.5 = 37.5$. Let's re-verify options. If $T_2 = 37.5\ ^\circ C$, temperature difference is $62.5$. $H = 400 \times 2 \times 10^{-4} \times 62.5 / 0.5 = 0.08 \times 125 = 10\ W$. Let's provide correct option matching $37.5$ or check if options match. Option D says $93.75$, let's adjust calculation. If $H = 10$, $L = 0.5$, $KA = 0.08$, then $\Delta T = 10 \times 0.5 / 0.08 = 62.5$. So cold end is $100 - 62.5 = 37.5$. Let's correct option indices to match correct arithmetic.)
6. What is the thermal diffusivity of a material having thermal conductivity $50\ W m^{-1} K^{-1}$, density $2500\ kg m^{-3}$, and specific heat capacity $500\ J kg^{-1} K^{-1}$?
Attempt these questions under a ten-minute limit using a separate timer.
1. In steady-state heat conduction, the temperature gradient along a uniform rod of length $L$ with ends maintained at temperatures $T_1$ and $T_2$ is:
Explanation: For a uniform rod in steady state with constant cross-section and thermal conductivity, the temperature gradient is constant throughout and equals $\frac{T_1 - T_2}{L}$.
2. If the length of a rod conducting heat is doubled while keeping temperature difference and cross-sectional area constant, the rate of heat flow becomes:
Explanation: Heat current $H = \frac{KA(T_1 - T_2)}{L}$ is inversely proportional to length $L$. Doubling $L$ halves the rate of heat flow.
3. Two conducting slabs of thermal conductivities $K_1$ and $K_2$ and equal thicknesses are placed in parallel combination. The equivalent thermal conductivity $K_{eq}$ is:
Explanation: For parallel slabs of equal thickness and equal cross-sectional areas ($A_1 = A_2 = A$), the formula $K_{eq} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$ simplifies to $\frac{K_1 + K_2}{2}$.
4. Which of the following transport mechanisms is primarily responsible for thermal conduction in electrical insulators?
Explanation: In electrical insulators (non-metals), free electrons are scarce, so thermal conduction is mediated by elastic lattice waves known as phonons.
5. The dimensional formula for thermal conductivity ($K$) is:
Explanation: From Fourier's law $K = \frac{dQ/dt}{A (dT/dx)}$, dimensions are $\frac{[ML^2 T^{-3}]}{[L^2] [K L^{-1}]} = [M L T^{-3} K^{-1}]$.තික
6. What happens to the thermal conductivity of most pure metals as temperature is increased significantly at moderate to high temperature ranges?
Explanation: As temperature increases, phonon scattering of electrons increases in metals, causing electrical conductivity to decrease and, via the Wiedemann-Franz law, thermal conductivity to decrease or vary moderately.
7. Thermal resistance is analogous to which electrical quantity?
Explanation: Thermal resistance $R_{th} = \frac{\Delta T}{H}$ is directly analogous to electrical resistance $R = \frac{\Delta V}{I}$. Temperature difference corresponds to voltage and heat flow rate corresponds to current.
8. If three identical thermal resistances are connected in series, the equivalent thermal resistance is:
Explanation: For bodies in series, $R_{eq} = R_1 + R_2 + R_3 = 3R$.
9. Thermal diffusivity ($\alpha$) has which of the following SI units?
Explanation: Thermal diffusivity $\alpha = \frac{K}{\rho c}$ has SI units of $m^2 s^{-1}$.
10. During the transient state before steady state is reached in a thermal conductor, what occurs to the heat entering a cross-section?
Explanation: In the transient state, local temperatures change with time because cross-sections absorb heat to increase their internal thermal energy until steady state is attained.