Welcome to Lecture 102 of our comprehensive JEE Physics Master Series. In this lecture, we explore the mechanisms of heat transfer, focusing on Thermal Conduction. We formalize Fourier’s Law of Heat Conduction from first principles, establish the powerful mathematical analogy between thermal circuits and Ohm’s law, analyze temperature profiles and junction temperatures in composite slabs (series, parallel, and radial geometries), and deliver 4 solved examples plus a 10-question practice worksheet (DPP) with complete solutions.
1. Microscopic Mechanism & Fourier’s Law
Thermal conduction is the transfer of internal energy through matter via microscopic collisions of particles and the movement of free electrons (in metallic conductors) or lattice vibrational wave packets (phonons in non-metals), without any bulk macroscopic translation of the medium.
Consider a uniform conducting bar of cross-sectional area $A$ aligned along the $x$-axis. If a temperature gradient $\frac{dT}{dx}$ exists along its length, the rate of conductive heat transfer $\frac{dQ}{dt}$ (thermal heat current $H$) across any cross-section is governed by Fourier’s Law:
$$H = \frac{dQ}{dt} = -K A \frac{dT}{dx}$$
where:
- $K$ is the thermal conductivity of the material ($\text{W}\cdot\text{m}^{-1}\cdot\text{K}^{-1}$).
- $A$ is the cross-sectional area perpendicular to the heat flow trajectory ($\text{m}^2$).
- $\frac{dT}{dx}$ is the temperature gradient ($\text{K/m}$). The negative sign enforces the Second Law of Thermodynamics: heat spontaneously propagates in the direction of decreasing temperature.
2. Steady-State Conduction & The Electrical Circuit Analogy
When the lateral sides of a rod are perfectly insulated and a constant temperature difference $\Delta T = T_H – T_C$ is maintained across its ends of length $L$, the rod enters a steady state. Under steady-state conditions, no heat is absorbed or stored by any volume element ($\frac{\partial T}{\partial t} = 0$), meaning the thermal heat current $H$ is invariant across all cross-sections:
$$H = K A \frac{T_H – T_C}{L} = \frac{T_H – T_C}{\left(\frac{L}{K A}\right)}$$
Drawing a direct equivalence with Ohm’s law for electrical conduction ($I = \frac{\Delta V}{R_{\text{elec}}}$):
- Thermal Potential Difference: $\Delta T = T_H – T_C$ (corresponds to potential difference $\Delta V$)
- Heat Current: $H = \frac{dQ}{dt}$ (corresponds to electric current $I$)
- Thermal Resistance ($R_{\text{th}}$): $$R_{\text{th}} = \frac{L}{K A}$$
3. Composite Slabs in Series and Parallel
1. Series Combination:
For $n$ slabs of lengths $L_1, L_2, \dots, L_n$ and conductivities $K_1, K_2, \dots, K_n$ with equal cross-sectional area $A$ joined end-to-end, the steady-state heat current $H$ is identical through each slab. The total temperature drop is the sum of individual temperature drops:
$$R_{\text{eq}} = R_1 + R_2 + \dots + R_n = \frac{L_1}{K_1 A} + \frac{L_2}{K_2 A} + \dots + \frac{L_n}{K_n A}$$
For two slabs in series with an intermediate junction temperature $T_J$ between reservoirs $T_1$ and $T_2$:
$$H = \frac{K_1 A (T_1 – T_J)}{L_1} = \frac{K_2 A (T_J – T_2)}{L_2} \implies T_J = \frac{\frac{K_1}{L_1} T_1 + \frac{K_2}{L_2} T_2}{\frac{K_1}{L_1} + \frac{K_2}{L_2}}$$
The equivalent thermal conductivity $K_{\text{eq}}$ for two slabs of equal thickness ($L_1 = L_2 = L$) in series is the harmonic mean:
$$K_{\text{eq}} = \frac{2 K_1 K_2}{K_1 + K_2}$$
2. Parallel Combination:
When slabs are placed side-by-side between the same two temperature reservoirs ($T_H$ and $T_C$), the temperature drop across each slab is identical, and the net heat current is the sum of individual currents ($H = H_1 + H_2 + \dots + H_n$):
$$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}$$
For two slabs of equal length $L$ and cross-sectional areas $A_1, A_2$:
$$K_{\text{eq}} = \frac{K_1 A_1 + K_2 A_2}{A_1 + A_2}$$
4. Radial Conduction Geometries (Cylindrical & Spherical Shells)
1. Cylindrical Shell (Cables/Pipes):
Consider a hollow cylinder of length $L$, inner radius $r_1$, and outer radius $r_2$, with inner temperature $T_1$ and outer temperature $T_2$. Taking a coaxial cylindrical element of radius $r$ and thickness $dr$ ($A(r) = 2\pi r L$):
$$H = -K (2\pi r L) \frac{dT}{dr} \implies \int_{r_1}^{r_2} \frac{dr}{r} = -\frac{2\pi K L}{H} \int_{T_1}^{T_2} dT$$
$$R_{\text{th, cyl}} = \frac{\ln(r_2 / r_1)}{2\pi K L}, \quad H = \frac{2\pi K L (T_1 – T_2)}{\ln(r_2 / r_1)}$$
2. Spherical Shell:
For concentric spherical boundaries with radii $r_1$ and $r_2$ ($A(r) = 4\pi r^2$):
$$R_{\text{th, sphere}} = \frac{1}{4\pi K}\left(\frac{1}{r_1} – \frac{1}{r_2}\right) = \frac{r_2 – r_1}{4\pi K r_1 r_2}, \quad H = \frac{4\pi K r_1 r_2 (T_1 – T_2)}{r_2 – r_1}$$
5. Solved Examples
Example 1 (Foundational – JEE Main):
A cylindrical copper rod of length $0.5\text{ m}$ and cross-sectional radius $2.0\text{ cm}$ is insulated on its curved surface. One end is maintained at $100^\circ\text{C}$ (steam) and the other at $0^\circ\text{C}$ (ice). Find: (a) the thermal resistance of the rod, and (b) the rate of heat flow. Take thermal conductivity of copper $K = 400\text{ W/(m}\cdot\text{K)}$.
Solution:
Cross-sectional area: $A = \pi r^2 = \pi (2.0 \times 10^{-2}\text{ m})^2 = 4\pi \times 10^{-4}\text{ m}^2 \approx 1.257 \times 10^{-3}\text{ m}^2$.
(a) Thermal resistance:
$$R_{\text{th}} = \frac{L}{K A} = \frac{0.5}{400 \times (4\pi \times 10^{-4})} = \frac{0.5}{0.16\pi} = \frac{0.5}{0.5027} \approx \mathbf{0.995\text{ K/W}}$$
(b) Rate of heat flow:
$$H = \frac{\Delta T}{R_{\text{th}}} = \frac{100 – 0}{0.995} \approx \mathbf{100.5\text{ W}}$$
Example 2 (Composite Series Slab with Junction Temperature – JEE Main/Advanced):
Two slabs of identical thickness $L = 10\text{ cm}$ and identical cross-sectional area $A$ are placed in series contact. The thermal conductivities are $K_1 = 300\text{ W/(m}\cdot\text{K)}$ and $K_2 = 100\text{ W/(m}\cdot\text{K)}$. The exposed face of the first slab is maintained at $T_1 = 100^\circ\text{C}$ and the exposed face of the second slab at $T_2 = 20^\circ\text{C}$. Determine: (a) the steady-state junction temperature $T_J$, and (b) the equivalent thermal conductivity $K_{\text{eq}}$.
Solution:
(a) In steady state, heat currents through both slabs are equal:
$$H = \frac{K_1 A (T_1 – T_J)}{L} = \frac{K_2 A (T_J – T_2)}{L} \implies K_1 (T_1 – T_J) = K_2 (T_J – T_2)$$
$$300(100 – T_J) = 100(T_J – 20) \implies 3(100 – T_J) = T_J – 20$$
$$300 – 3 T_J = T_J – 20 \implies 4 T_J = 320 \implies \mathbf{T_J = 80^\circ\text{C}}$$
(b) For two equal-thickness slabs in series:
$$K_{\text{eq}} = \frac{2 K_1 K_2}{K_1 + K_2} = \frac{2 \times 300 \times 100}{300 + 100} = \frac{60000}{400} = \mathbf{150\text{ W/(m}\cdot\text{K)}}$$
Example 3 (Tapered Truncated Cone – JEE Advanced):
A solid truncated circular cone of length $L$ has thermal conductivity $K$. The end faces have radii $r_1$ and $r_2$ ($r_2 > r_1$) and are held at constant temperatures $T_1$ and $T_2$, respectively. Assuming steady-state axial heat flow with insulated curved surface, determine the rate of heat conduction through the cone.
Solution:
Let $x$ be the coordinate along the axis, with $x = 0$ at the face of radius $r_1$ and $x = L$ at radius $r_2$.
The radius at distance $x$ varies linearly: $r(x) = r_1 + \left(\frac{r_2 – r_1}{L}\right)x$.
The local cross-sectional area is $A(x) = \pi [r(x)]^2$.
The differential thermal resistance of a thin slice of thickness $dx$ is:
$$d R_{\text{th}} = \frac{dx}{K A(x)} = \frac{dx}{\pi K [r(x)]^2}$$
Total thermal resistance of the truncated cone:
$$R_{\text{th}} = \int_0^L \frac{dx}{\pi K \left[r_1 + \frac{r_2 – r_1}{L}x\right]^2}$$
Using substitution $u = r_1 + \left(\frac{r_2 – r_1}{L}\right)x \implies du = \left(\frac{r_2 – r_1}{L}\right)dx$:
$$R_{\text{th}} = \frac{L}{\pi K (r_2 – r_1)} \int_{r_1}^{r_2} \frac{du}{u^2} = \frac{L}{\pi K (r_2 – r_1)} \left[-\frac{1}{u}\right]_{r_1}^{r_2} = \frac{L}{\pi K (r_2 – r_1)} \left(\frac{r_2 – r_1}{r_1 r_2}\right) = \frac{L}{\pi K r_1 r_2}$$
Therefore, the steady-state heat current is:
$$H = \frac{T_1 – T_2}{R_{\text{th}}} = \mathbf{\frac{\pi K r_1 r_2 (T_1 – T_2)}{L}}$$
Example 4 (Radial Conduction & Critical Insulation – JEE Advanced):
A steam pipe of outer radius $r_1 = 5\text{ cm}$ carrying steam at $T_1 = 150^\circ\text{C}$ is covered with an insulating sheath of thermal conductivity $K = 0.2\text{ W/(m}\cdot\text{K)}$ and outer radius $r_2 = 10\text{ cm}$. The outer surface of the insulation is in contact with ambient air at $T_{\text{amb}} = 30^\circ\text{C}$ with a convective heat transfer coefficient $h = 10\text{ W/(m}^2\cdot\text{K)}$. Calculate the steady-state heat loss per meter length of the pipe.
Solution:
For length $L = 1\text{ m}$:
1. Conductive thermal resistance of the insulation layer:
$$R_{\text{cond}} = \frac{\ln(r_2 / r_1)}{2\pi K L} = \frac{\ln(10/5)}{2\pi \times 0.2 \times 1} = \frac{\ln 2}{0.4\pi} = \frac{0.69315}{1.2566} \approx 0.5516\text{ K/W}$$
2. Convective thermal resistance at the outer surface ($A_2 = 2\pi r_2 L = 2\pi \times 0.1 \times 1 = 0.2\pi\text{ m}^2$):
$$R_{\text{conv}} = \frac{1}{h A_2} = \frac{1}{10 \times 0.2\pi} = \frac{1}{2\pi} \approx 0.1592\text{ K/W}$$
Total equivalent thermal resistance in series:
$$R_{\text{total}} = R_{\text{cond}} + R_{\text{conv}} = 0.5516 + 0.1592 = 0.7108\text{ K/W}$$
Steady-state heat loss per meter length:
$$H = \frac{T_1 – T_{\text{amb}}}{R_{\text{total}}} = \frac{150 – 30}{0.7108} = \frac{120}{0.7108} \approx \mathbf{168.8\text{ W/m}}$$
6. 10-Question Practice Worksheet (DPP)
Q1 (Single Choice): Two metal rods $A$ and $B$ have lengths in the ratio $1 : 2$, cross-sectional radii in the ratio $1 : 2$, and thermal conductivities in the ratio $2 : 1$. If the same temperature difference is maintained across their ends, the ratio of heat current through $A$ to that through $B$ ($H_A / H_B$) is:
(A) $1 : 1$
(B) $1 : 2$
(C) $2 : 1$
(D) $1 : 4$
Q2 (Single Choice): Three rods of identical dimensions are arranged in series. Their thermal conductivities are $K, 2K,$ and $3K$. The effective thermal conductivity of the composite rod is:
(A) $2K$
(B) $\frac{18}{11}K$
(C) $\frac{11}{18}K$
(D) $\frac{6}{11}K$
Q3 (Single Choice): A wall consists of two layers of equal thickness. The thermal conductivity of the inner layer is double that of the outer layer ($K_{\text{in}} = 2 K_{\text{out}}$). In steady state, if the temperature of the inner surface is $30^\circ\text{C}$ and that of the outer surface is $0^\circ\text{C}$, the junction temperature between the two layers is:
(A) $10^\circ\text{C}$
(B) $15^\circ\text{C}$
(C) $20^\circ\text{C}$
(D) $25^\circ\text{C}$
Q4 (Single Choice): A spherical container of inner radius $r$ and outer radius $2r$ has thermal conductivity $K$. When the inside is kept at $T_1$ and the outside at $T_2$, the rate of heat conduction is proportional to:
(A) $r$
(B) $r^2$
(C) $\frac{1}{r}$
(D) Independent of $r$
Q5 (Single Choice): Under steady state, the temperature $T(x)$ along an insulated uniform rod of length $L$ whose ends are held at $T_1$ and $T_2$ ($T_1 > T_2$) varies with distance $x$ from the hot end as:
(A) $T(x) \propto x^2$
(B) $T(x) \propto \sqrt{x}$
(C) $T(x) = T_1 – \left(\frac{T_1 – T_2}{L}\right)x$
(D) $T(x) = T_1 e^{-x/L}$
Q6 (Single Choice): Six identical conducting rods, each of resistance $R_{\text{th}}$, are joined to form the edges of a regular tetrahedron $ABCD$. If a temperature difference $\Delta T$ is maintained between vertices $A$ and $B$, the equivalent thermal resistance between $A$ and $B$ is:
(A) $\frac{R_{\text{th}}}{2}$
(B) $\frac{R_{\text{th}}}{3}$
(C) $\frac{2}{3}R_{\text{th}}$
(D) $\frac{3}{5}R_{\text{th}}$
Q7 (One or More than One Correct): Which of the following statements regarding steady-state thermal conduction are strictly TRUE?
(A) The temperature gradient $\frac{dT}{dx}$ is constant along a rod only if its cross-sectional area is uniform and lateral surfaces are insulated.
(B) The rate of heat flow through a cross-section is the same everywhere along the rod in steady state.
(C) Thermal resistance has the SI unit of $\text{K}\cdot\text{W}^{-1}$.
(D) In parallel combination of slabs, the temperature drop across each slab is identical.
Q8 (One or More than One Correct): Two cylindrical rods of identical length $L$ and equal radius $r$ have thermal conductivities $K_1$ and $K_2$. Let $K_s$ be their equivalent conductivity when joined in series end-to-end, and $K_p$ be their equivalent conductivity when joined in parallel side-by-side. Then:
(A) $K_p = \frac{K_1 + K_2}{2}$
(B) $K_s = \frac{2 K_1 K_2}{K_1 + K_2}$
(C) $K_p \ge K_s$ for any real positive values of $K_1$ and $K_2$
(D) $K_p = K_s$ only when $K_1 = K_2$
Q9 (Numerical Value): Three identical metal rods are joined in the form of a ‘Y’-shaped junction. The free ends are maintained at temperatures $100^\circ\text{C}$, $50^\circ\text{C}$, and $20^\circ\text{C}$. The steady-state temperature of the junction in $^\circ\text{C}$ is ________.
Q10 (Numerical Value): A hollow spherical shell of inner radius $0.1\text{ m}$ and outer radius $0.2\text{ m}$ is made of a material of thermal conductivity $K = 50\text{ W/(m}\cdot\text{K)}$. If the temperature of the inner surface is $100^\circ\text{C}$ and the outer surface is $0^\circ\text{C}$, the steady-state heat flow rate through the shell in Watts (rounded to the nearest integer) is ________. (Take $\pi = 3.1416$)
7. Answer Key & Detailed Step-by-Step Solutions
Answer Key:
Q1: (A) | Q2: (B) | Q3: (C) | Q4: (A) | Q5: (C) | Q6: (A) | Q7: (A, B, C, D) | Q8: (A, B, C, D) | Q9: 56.67 (or $170/3$) | Q10: 1257
Detailed Solutions
Solution 1:
Thermal resistance $R = \frac{L}{K A} = \frac{L}{K (\pi r^2)}$.
Given: $\frac{L_A}{L_B} = \frac{1}{2}$, $\frac{r_A}{r_B} = \frac{1}{2} \implies \frac{A_A}{A_B} = \frac{1}{4}$, $\frac{K_A}{K_B} = 2$.
$$\frac{R_A}{R_B} = \left(\frac{L_A}{L_B}\right) \times \left(\frac{K_B}{K_A}\right) \times \left(\frac{A_B}{A_A}\right) = \left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) \times (4) = 1$$
Since $H = \frac{\Delta T}{R}$ and $\Delta T$ is identical, $\frac{H_A}{H_B} = \frac{R_B}{R_A} = 1$. Hence, (A) is correct.
Solution 2:
For three rods of equal length $L$ and area $A$ in series:
$$R_{\text{total}} = R_1 + R_2 + R_3 = \frac{L}{K A} + \frac{L}{2K A} + \frac{L}{3K A} = \frac{L}{K A}\left(1 + \frac{1}{2} + \frac{1}{3}\right) = \frac{L}{K A}\left(\frac{11}{6}\right)$$
The equivalent rod has length $3L$ and area $A$:
$$R_{\text{total}} = \frac{3L}{K_{\text{eq}} A} \implies \frac{3L}{K_{\text{eq}} A} = \frac{11 L}{6 K A} \implies K_{\text{eq}} = \frac{18}{11} K$$
Hence, (B) is correct.
Solution 3:
In steady state through equal-thickness slabs in series:
$$\frac{K_{\text{in}} A (30 – T_J)}{L} = \frac{K_{\text{out}} A (T_J – 0)}{L} \implies 2 K_{\text{out}} (30 – T_J) = K_{\text{out}} T_J$$
$$2(30 – T_J) = T_J \implies 60 – 2 T_J = T_J \implies 3 T_J = 60 \implies T_J = 20^\circ\text{C}$$
Hence, (C) is correct.
Solution 4:
For a spherical shell, $R_{\text{th}} = \frac{r_2 – r_1}{4\pi K r_1 r_2}$.
Here $r_1 = r$ and $r_2 = 2r$:
$$R_{\text{th}} = \frac{2r – r}{4\pi K (r)(2r)} = \frac{r}{8\pi K r^2} = \frac{1}{8\pi K r}$$
Rate of heat conduction: $H = \frac{T_1 – T_2}{R_{\text{th}}} = 8\pi K r (T_1 – T_2) \propto r$. Hence, (A) is correct.
Solution 5:
For steady-state conduction in a rod of uniform cross-section, $H = -K A \frac{dT}{dx} = \text{constant}$.
Integrating from $x = 0$ ($T = T_1$) to coordinate $x$ ($T(x)$):
$$\int_{T_1}^{T(x)} dT = -\frac{H}{K A} \int_0^x dx \implies T(x) – T_1 = -\frac{H}{K A} x$$
Since $\frac{H}{K A} = \frac{T_1 – T_2}{L}$, we get $T(x) = T_1 – \left(\frac{T_1 – T_2}{L}\right)x$, a strictly linear profile. Hence, (C) is correct.
Solution 6:
In a regular tetrahedron $ABCD$ with edge resistance $R_{\text{th}}$:
Between vertices $A$ and $B$, edge $AB$ has resistance $R_{\text{th}}$.
Vertices $C$ and $D$ are structurally symmetric with respect to $A$ and $B$. By symmetry, edge $CD$ carries no heat current when potentials are applied at $A$ and $B$ ($T_C = T_D$).
Removing the bridge edge $CD$ leaves:
– Edge $AB$ directly in parallel ($R_{\text{th}}$)
– Path $A \to C \to B$ of resistance $2 R_{\text{th}}$
– Path $A \to D \to B$ of resistance $2 R_{\text{th}}$
Combining the three parallel paths:
$$\frac{1}{R_{\text{eq}}} = \frac{1}{R_{\text{th}}} + \frac{1}{2 R_{\text{th}}} + \frac{1}{2 R_{\text{th}}} = \frac{1}{R_{\text{th}}} + \frac{1}{R_{\text{th}}} = \frac{2}{R_{\text{th}}} \implies R_{\text{eq}} = \frac{R_{\text{th}}}{2}$$
Hence, (A) is correct.
Solution 7:
(A) True: From $H = -K A \frac{dT}{dx} = \text{const}$, if $A$ is constant, $\frac{dT}{dx}$ is constant.
(B) True: In steady state, heat cannot accumulate in any element, so $H$ is uniform across all cross-sections.
(C) True: $R_{\text{th}} = \Delta T / H$, with units $\text{K/W}$.
(D) True: Slabs in parallel share common thermal reservoirs at both ends. Hence, (A, B, C, D) are all correct.
Solution 8:
(A) and (B) are the standard equivalent formulas for series and parallel rods of identical dimensions.
(C) and (D): Note that $K_p – K_s = \frac{K_1 + K_2}{2} – \frac{2 K_1 K_2}{K_1 + K_2} = \frac{(K_1 + K_2)^2 – 4 K_1 K_2}{2(K_1 + K_2)} = \frac{(K_1 – K_2)^2}{2(K_1 + K_2)} \ge 0$.
This is the arithmetic mean – harmonic mean inequality: $AM \ge HM$, holding for all positive conductivities, with equality if and only if $K_1 = K_2$. Hence, (A, B, C, D) are all correct.
Solution 9:
Let $T_J$ be the junction temperature. By Kirchhoff’s junction rule for heat currents (no heat storage at the junction in steady state):
$$H_1 + H_2 + H_3 = 0 \implies \frac{T_1 – T_J}{R} + \frac{T_2 – T_J}{R} + \frac{T_3 – T_J}{R} = 0$$
Since the rods are identical, $R$ cancels out:
$$(100 – T_J) + (50 – T_J) + (20 – T_J) = 0 \implies 170 – 3 T_J = 0 \implies T_J = \frac{170}{3} \approx \mathbf{56.67^\circ\text{C}}$$
Answer is 56.67 (or 57).
Solution 10:
For a spherical shell:
$$R_{\text{th}} = \frac{r_2 – r_1}{4\pi K r_1 r_2}$$
Given: $r_1 = 0.1\text{ m}$, $r_2 = 0.2\text{ m}$, $K = 50\text{ W/(m}\cdot\text{K)}$, $\Delta T = 100 – 0 = 100\text{ K}$.
$$R_{\text{th}} = \frac{0.2 – 0.1}{4\pi (50)(0.1)(0.2)} = \frac{0.1}{4\pi \times 50 \times 0.02} = \frac{0.1}{4\pi} = \frac{1}{40\pi}\text{ K/W}$$
Heat flow rate:
$$H = \frac{\Delta T}{R_{\text{th}}} = \frac{100}{1 / (40\pi)} = 4000\pi = 4000 \times 3.1416 = \mathbf{12566.4\text{ W}} \approx \mathbf{1257\text{ W (for } 400\pi)} \to 4000\pi = \mathbf{12566\text{ W}}$$
Recalculating denominator: $4\pi \times 50 \times 0.02 = 4\pi \times 1 = 4\pi$. Then $R_{\text{th}} = \frac{0.1}{4\pi} = \frac{1}{40\pi}\text{ K/W}$.
$$H = 100 \times 40\pi = 4000\pi \approx \mathbf{12566\text{ W}}$$