JEE Mains Physics · Thermal Properties of Matter

Heat Conduction

In the steady state a slab carries a heat current H = KAΔT/L, so it behaves like a resistor of thermal resistance L/(KA): resistances add in series, conductances add in parallel, and the heat flowing into any junction flows out of it.

All Thermal Properties of Matter notes JEE Mains Physics strategy

Why this matters

Thirteen PYQs, five of them asking for a number, and one from 2026. Seven join two rods or slabs end to end and ask for the junction temperature or the equivalent conductivity; six send heat through parallel paths, a junction of three rods, the walls of a box or a spherical shell. Treat each one as a circuit: temperature difference for voltage, heat current for current.

Concept 1 of 2: Slabs and rods in series: thermal resistance and junction temperature

Heat flowing steadily through two slabs in a row has nowhere else to go, so the same heat current passes through both. Each slab resists the flow with R = L/(KA), just as a resistor resists current. The resistances add, and the temperature falls across each slab in proportion to its resistance. A poor conductor, or a thick or thin one, takes the larger share of the drop.

Definition

Heat current: $H = \dfrac{KA\Delta T}{L} = \dfrac{\Delta T}{R}$ , with $R = \dfrac{L}{KA}$ .
For a rod of radius r: $R = \dfrac{L}{K\pi r^{2}}$ , so $R \propto L/(Kr^{2})$ .
Series: $R = R_1 + R_2$ ; the same H flows through each part.
Temperature drop across each part $\propto$ its R. Junction: $\theta = \dfrac{\theta_1R_2 + \theta_2R_1}{R_1 + R_2}$ .
Equivalent conductivity (same area): $K_{eq} = \dfrac{L_1 + L_2}{L_1/K_1 + L_2/K_2}$ ; for equal lengths $K_{eq} = \dfrac{2K_1K_2}{K_1 + K_2}$ .
Unknown conductivity from a measured junction temperature: write $H_1 = H_2$ and solve.

Conduction in series

H = \frac{KA\Delta T}{L} \qquad R = \frac{L}{KA} \qquad \theta = \frac{\theta_1R_2 + \theta_2R_1}{R_1 + R_2} \qquad K_{eq} = \frac{L_1 + L_2}{L_1/K_1 + L_2/K_2}

Worked example

Two slabs of the same area are pressed together. Slab 1 is 2 cm thick with

K_1 = 40\ \text{W m}^{-1}\text{K}^{-1}

; slab 2 is 3 cm thick with

K_2 = 20\ \text{W m}^{-1}\text{K}^{-1}

. The outer faces are at

100^{\circ}C

(slab 1) and

20^{\circ}C

(slab 2). Find the temperature of the interface and the equivalent conductivity.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 7 Apr 2025 · Q98Moderate

Example 1 · Thermal Properties of Matter · Heat Conduction

Two cylindrical rods

A

and

B

made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities of these rods are :

\frac{L_{A}}{L_{B}} = \frac{1}{2},\frac{r_{A}}{r_{B}} = 2

and

\frac{K_{A}}{K_{B}} = \frac{1}{2}

. The free ends of rods A and B are maintained at

400\text{ }K,200\text{ }K

, respectively. The temperature of rods interface is

\_\_\_\_

K, when equilibrium is established.

Adding conductivities in series

In series it is the RESISTANCES L/(KA) that add. Averaging or adding the conductivities gives an equivalent conductivity that is too large.

Forgetting that the area goes as r²

For rods, R = L/(Kπr²). Doubling the radius cuts the resistance to a quarter, not a half. Using a diameter for the radius makes the same slip.

Drill 6 more on slabs and rods in series: thermal resistance and junction temperature

Concept 2 of 2: Parallel paths, junctions, box walls and spherical shells

When heat has two side-by-side paths, each carries its own current and the totals add, so conductances KA/L add. When three or more rods meet at a point, nothing piles up there in the steady state: the heat flowing in equals the heat flowing out. For a box, the walls are all in parallel, so use the total area of all the faces. A spherical shell is a stack of thin shells in series whose area grows with radius.

Definition

Parallel: $\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$ ; conductances $KA/L$ add, and so do the heat currents.
Junction rule: $\displaystyle\sum \frac{\theta_i - \theta}{R_i} = 0$ over all rods meeting at the junction.
A network: reduce series and parallel pieces step by step, as with resistors.
Box: $H = \dfrac{KA_{total}\Delta T}{d}$ with $A_{total} = 2(lb + bh + hl)$ and d the wall thickness; ice melts at $\dfrac{dm}{dt} = \dfrac{H}{L_f}$ .
Spherical shell: $R = \dfrac{r_2 - r_1}{4\pi Kr_1r_2}$ , so $H = \dfrac{4\pi Kr_1r_2\,\Delta\theta}{r_2 - r_1}$ .
Turning a temperature difference into electrical energy needs a material that holds the difference (heat leaks through slowly) while letting charge flow easily.

Parallel paths, junctions and shells

\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \qquad \sum \frac{\theta_i - \theta}{R_i} = 0 \qquad \frac{dm}{dt} = \frac{H}{L_f} \qquad R_{shell} = \frac{r_2 - r_1}{4\pi Kr_1r_2}

Worked example

Two rods, each 0.5 m long with area

2 \times 10^{-4}\ \text{m}^{2}

, connect the same two reservoirs at

120^{\circ}C

and

20^{\circ}C

. Their conductivities are 100 and 300 W m⁻¹ K⁻¹. Find the total heat current and the share carried by each rod.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 26 July 2022 · Q101Moderate

Example 2 · Thermal Properties of Matter · Heat Conduction

An ice cube of dimensions

60\text{ }cm \times 50\text{ }cm \times 20\text{ }cm

is placed in an insulation box of wall thickness

1\text{ }cm

. The box keeping the ice cube at

0^{\circ}C

of temperature is brought to a room of temperature

40^{\circ}C

. The rate of melting of ice is approximately: (Latent heat of fusion of ice is

3.4 \times10^{5}\text{ }J{\text{ }kg}^{- 1}

and thermal conductivity of insulation wall is

0.05\ Wm^{- 1}\ ^{\circ}C^{- 1}

)

Using one face of a box

Heat enters through every wall, so the area is the total of all six faces, 2(lb + bh + hl). Using one face, or forgetting the factor 2, gives a melting rate several times too small.

Treating a spherical shell as a flat slab

The area of a shell grows as r², so L/(KA) with one area is wrong. The resistance is (r₂ − r₁)/(4πK r₁r₂).

Losing the sign at a junction

Write every rod's current as flowing INTO the junction, (θᵢ − θ)/Rᵢ, and set the sum to zero. Mixing in- and out-directions doubles one term or cancels another.

Drill 5 more on parallel paths, junctions, box walls and spherical shells

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Slabs and rods in series: thermal resistance and junction temperature
Conduction in series
$H = \frac{KA\Delta T}{L} \qquad R = \frac{L}{KA} \qquad \theta = \frac{\theta_1R_2 + \theta_2R_1}{R_1 + R_2} \qquad K_{eq} = \frac{L_1 + L_2}{L_1/K_1 + L_2/K_2}$
Parallel paths, junctions, box walls and spherical shells
Parallel paths, junctions and shells
$\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \qquad \sum \frac{\theta_i - \theta}{R_i} = 0 \qquad \frac{dm}{dt} = \frac{H}{L_f} \qquad R_{shell} = \frac{r_2 - r_1}{4\pi Kr_1r_2}$

Watch out for (5)

Adding conductivities in series
→ Slabs and rods in series: thermal resistance and junction temperature
Forgetting that the area goes as r²
→ Slabs and rods in series: thermal resistance and junction temperature
Using one face of a box
→ Parallel paths, junctions, box walls and spherical shells
Treating a spherical shell as a flat slab
→ Parallel paths, junctions, box walls and spherical shells
Losing the sign at a junction
→ Parallel paths, junctions, box walls and spherical shells

Test yourself on Thermal Properties of Matter

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Thermal Properties of Matter chapter test Drill all 13 questions