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Composite slab questions look messy: two or three materials, different thicknesses, an unknown temperature in the middle. They become routine once you see that heat conduction follows the same rules as current in a circuit. This post sets up that analogy, works through a roof, a two-rod junction and a parallel case, and lists the errors examiners see most.
What is the basic law of heat conduction?
NCERT's Thermal Properties of Matter chapter describes a bar of length L and area A with its ends held at temperatures T_C and T_D, sides insulated. In the steady state the rate of heat flow is
H = KA(T_C − T_D)/L
where K is the thermal conductivity in W m⁻¹ K⁻¹. Its Table 10.6 gives values worth knowing: silver 406, copper 385, aluminium 205, steel 50.2, concrete 0.8, glass 0.8, water 0.8, wood 0.12, glass wool 0.04 and air 0.024.
"Heat conduction in one dimension" is in the JEE (Advanced) 2026 syllabus, and the CBSE 2026-27 curriculum lists thermal conductivity. Check the current year's documents.
How is heat flow like electric current?
Rewrite the law as H = ΔT / R_th, with R_th = L/(KA). Compare I = ΔV/R and R = ρL/A.
Heat current H plays the part of electric current.
Temperature difference plays the part of potential difference.
Thermal resistance L/(KA) plays the part of resistance.
So everything you know about resistor combinations carries over. In steady state, no heat piles up anywhere, so the same H flows through every layer in series, just as the same current flows through series resistors.
How much does insulation help a concrete roof?
Example 1. A concrete roof slab is 15 cm thick. Its outer surface is at 45 °C and inner surface at 30 °C. Find the heat flow through each square metre, then repeat with a 5.0 cm layer of glass wool on top, keeping the same outer and inner temperatures. Use K = 0.8 for concrete and 0.04 for glass wool.
Concrete alone: R = 0.15/(0.8 × 1) = 0.1875 K/W. H = 15/0.1875 = 80 W.
Glass wool: R = 0.050/(0.04 × 1) = 1.25 K/W.
In series: R_total = 0.1875 + 1.25 = 1.4375 K/W, so H = 15/1.4375 ≈ 10.4 W.
Interface temperature: the drop across the glass wool is H × R = 10.4 × 1.25 ≈ 13.0 K, so the interface is at about 45 − 13.0 = 32.0 °C. The remaining 2.0 K drop is across the concrete.
A thin layer of a poor conductor dominates the total resistance. Most of the temperature drop happens across it. NCERT makes the same practical point about layers of earth or foam on concrete ceilings.
How do you find the temperature of a junction?
Example 2. A copper rod and a steel rod of the same length and area are joined end to end. The free copper end is at 100 °C and the free steel end at 0 °C. Find the junction temperature T.
Same H through both: K_Cu(100 − T) = K_steel(T − 0), since L and A cancel.
385(100 − T) = 50.2 T, so T = 38500/435.2 ≈ 88.5 °C.
The steel, with much higher thermal resistance, takes most of the drop. NCERT's Example 10.6 solves a version with different lengths and areas; work it after this one.
What changes when slabs are side by side?
When two materials sit next to each other between the same two temperatures, each carries its own heat current, and the totals add: H = H₁ + H₂. Equivalently, 1/R = 1/R₁ + 1/R₂, the parallel rule.
Example 3. A 1.0 m² wall section is half insulating brick (K = 0.15) and half glass (K = 0.8), both 10 cm thick, with a 20 K difference. H = ΔT × [K₁A₁ + K₂A₂]/L = 20 × (0.15 × 0.5 + 0.8 × 0.5)/0.10 = 20 × 0.475/0.10 = 95 W. The glass half carries 80 W of it. This is why a window loses so much more heat than the wall around it.
What mistakes should you avoid?
Using the series rule for parallel paths. Ask: does the heat have to pass through both, or can it choose?
Forgetting the steady-state condition. Equal H in series layers only holds after temperatures stop changing.
Unit slips. Thickness in metres, area in m². A 5 cm layer is 0.05 m.
Mixing conduction with cooling. Rate of cooling of a body in air is a different law. See our post on Newton's law of cooling, and for heat balance problems, calorimetry and heat transfer.
Trusting touch. Metal feels colder than wood at the same room temperature because it conducts heat away from your hand faster. In interviews reported by Lewis and Linn (1994), many students believed metals "conduct" or "hold" cold, and that aluminium foil would keep cold objects colder than wool. Reason from K, not from feel.
Practice routine
Draw the circuit diagram first for every slab problem: a box for each layer, series or parallel.
Solve one three-layer wall, one two-rod junction and one parallel wall from NCERT's table values.
For each, find which layer takes the largest temperature drop before you calculate, then check.
When you get stuck on the set-up rather than the algebra, the Fermi AI tutor can draw the figure and work out what you already know, so the next step is clear.
Redo Example 1 with 2.5 cm of glass wool and predict, before calculating, whether the heat flow roughly doubles.
