A composite wall of two materials — find heat current

Question

A composite wall consists of two slabs in series. Slab 1 has thickness $L_1 = 0.1$ m and thermal conductivity $k_1 = 2$ W/m·K. Slab 2 has thickness $L_2 = 0.2$ m and thermal conductivity $k_2 = 0.5$ W/m·K. The outer surface of slab 1 is at $T_1 = 100°C$ and the outer surface of slab 2 is at $T_2 = 20°C$ . Both slabs have cross-sectional area $A = 1$ m². Find the heat current (rate of heat flow) through the wall, and the temperature at the interface.

Solution — Step by Step

Heat flow through a conducting slab follows Fourier’s law of heat conduction:

H = \frac{dQ}{dt} = \frac{kA\Delta T}{L}

We can define thermal resistance $R_{th}$ exactly like electrical resistance:

R_{th} = \frac{L}{kA}

For slabs in series, total thermal resistance adds up: $R_{total} = R_1 + R_2$

This analogy with electrical circuits (Ohm’s law: $I = V/R$ ) is the most powerful tool for composite wall problems.

R_1 = \frac{L_1}{k_1 A} = \frac{0.1}{2 \times 1} = 0.05 \text{ K/W}

R_2 = \frac{L_2}{k_2 A} = \frac{0.2}{0.5 \times 1} = 0.4 \text{ K/W}

R_{total} = R_1 + R_2 = 0.05 + 0.4 = 0.45 \text{ K/W}

Notice that slab 2 dominates the total resistance even though it’s only twice as thick — its thermal conductivity is 4 times lower, making it the thermal “bottleneck.”

The temperature difference across the entire wall drives the heat current:

\Delta T_{total} = T_1 - T_2 = 100 - 20 = 80 \text{ °C} = 80 \text{ K}

H = \frac{\Delta T_{total}}{R_{total}} = \frac{80}{0.45} \approx \mathbf{177.8 \text{ W}}

In steady state, the same heat current flows through both slabs (no heat accumulates at the interface). This is exactly like current being the same through resistors in series.

Let $T_{int}$ be the temperature at the junction between slab 1 and slab 2.

Using slab 1: $H = \frac{T_1 - T_{int}}{R_1}$

T_1 - T_{int} = H \times R_1 = 177.8 \times 0.05 = 8.89 \text{ °C}

T_{int} = 100 - 8.89 \approx \mathbf{91.1°C}

Check with slab 2: $\Delta T_2 = H \times R_2 = 177.8 \times 0.4 = 71.1°C$ , so $T_{int} = 20 + 71.1 = 91.1°C$ ✓

The small temperature drop across slab 1 (only 8.9°C) vs. the large drop across slab 2 (71.1°C) makes physical sense — slab 1 has very low resistance, so little temperature is “lost” crossing it.

Heat current: $H \approx 177.8$ W (or $\frac{1600}{9}$ W exactly)
Interface temperature: $T_{int} \approx 91.1°C$

Why This Works

The thermal resistance analogy works because both Fourier’s law and Ohm’s law describe a linear relationship between a “driving potential” (temperature difference / voltage) and a “flow” (heat current / electric current) opposed by a “resistance.”

In series, conservation of energy demands that the same heat flow rate pass through every cross-section — just as charge conservation demands the same current through series resistors. This principle directly gives us the interface temperature without needing to solve differential equations.

The key insight: a material with low k and large L has high thermal resistance and will take the largest temperature drop. Here, slab 2 accounts for $0.4/0.45 \approx 89\%$ of total resistance and takes $71.1/80 \approx 89\%$ of the temperature drop. The numbers align perfectly.

Alternative Method — Direct Fourier’s Law Application

Since heat current is the same through both slabs in steady state:

\frac{k_1 A (100 - T_{int})}{L_1} = \frac{k_2 A (T_{int} - 20)}{L_2}

\frac{2 \times 1 \times (100 - T_{int})}{0.1} = \frac{0.5 \times 1 \times (T_{int} - 20)}{0.2}

20(100 - T_{int}) = 2.5(T_{int} - 20)

2000 - 20T_{int} = 2.5T_{int} - 50

2050 = 22.5 T_{int}

T_{int} = 91.1°C

Then $H = \frac{k_1 A (100 - 91.1)}{L_1} = \frac{2 \times 8.9}{0.1} = 178$ W ✓

Common Mistake

The most frequent error is assuming the interface temperature is the average of $T_1$ and $T_2$ , i.e., 60°C. That would only be true if both slabs had equal thermal resistance. Always compute thermal resistances first and use them to find the actual interface temperature.

Another common slip: using Celsius temperature differences — this is fine since we are computing $\Delta T$ , and the difference in Celsius equals the difference in Kelvin. But if you ever need absolute temperature in a radiation problem, always convert to Kelvin.

In JEE Main, composite wall problems appear 1–2 times per year. They often add a twist: three slabs in series, or parallel slabs, or one slab with convection on a surface. The thermal resistance method handles all of these — series resistances add, parallel resistances follow $1/R_{total} = 1/R_1 + 1/R_2$ . Practise setting up the circuit diagram before computing.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Fourier’s Law Application

Common Mistake

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