Heat Conduction Through Composite Slabs
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Direct answer
Steady conduction through slabs obeys an exact electrical analogy: each slab is a thermal resistance R = L/(KA), the heat current is H = ΔT/R (the Fourier law H = KA ΔT/L rewritten), series slabs add their resistances while parallel slabs add their conductances. In a series composite, the same H flows through every layer and the interface temperatures partition the total ΔT in proportion to the resistances — the question JEE asks most. Radial geometry replaces the flat formula with H = 2πKL(T1 − T2)/ln(r2/r1) for a cylindrical shell and H = 4πK r1r2(T1 −T2)/(r2 − r1) for a sphere, the results behind lagged-pipe and heat-loss problems. The transient cousin of these steady results is the ice-on-lake law: thickness grows as the square root of time.
What you must remember
- Basic law: H = KA ΔT/L in watts; thermal resistance R_th = L/(KA) in kelvin per watt, and H = ΔT/R_th exactly mirrors Ohm's law.
- Series slabs: R = R1 + R2 + ...; the same H crosses all layers; the interface temperature sits at T_hot − H × R1, that is, temperature drops partition in proportion to resistances.
- Parallel slabs (side by side, same faces): H = H1 + H2; effective conductivity K_eq = (K1A1 + K2A2)/(A1 + A2); for equal areas, the arithmetic mean.
- Series equal-thickness slabs: K_eq = 2K1K2/(K1 + K2), the harmonic mean — always below the arithmetic mean, so a layered wall insulates worse than the average of its materials.
- Cylindrical shell: H = 2πKL(T1 − T2)/ln(r2/r1) — the area grows with radius, so the inner surface controls the resistance; adding insulation to a thin wire can even increase its heat loss (the critical-radius effect).
- Spherical shell: H = 4πK r1r2(T1 − T2)/(r2 − r1), the geometry of lagged storage spheres and planetary heat-flow estimates.
- Ice-on-lake growth: with water below at 0 and air above at −θ, thickness grows as y = √(2Kθt/(ρL_f)) — doubling the ice takes four times the time, a classic JEE Advanced integration.
Two slabs and where the temperature lands
A wall of two equal-thickness slabs, K1 = 2K (conductivity 2K) and K2 = K, holds 100 degrees on the hot side and 0 on the cold. The resistances stand in ratio 1:2 (R = L/(KA) halves when K doubles), so of the 100-degree drop, one-third falls across the good conductor and two-thirds across the poor one: the interface temperature is 100 − 100 × (1/3) = 66.7 degrees Celsius. No heat current was computed — the resistance-partition instinct alone answered the question, and it generalises to any number of layers. The equivalent conductivity of the pair, K_eq = 2 × 2K × K/(3K) = 4K/3, sits below the arithmetic 3K/2, quantifying the folk wisdom that a chain of layers insulates as its worst member.
The radial versions reward the same resistance thinking. Steam at 120 degrees flows in a pipe of inner radius 5 cm wrapped in lagging of K = 0.1 W/mK out to 10 cm, with the outer surface at 30 degrees: per metre of length, H = 2π × 0.1 × 90/ln 2 ≈ 81.6 W per metre. Doubling the lagging thickness would not halve the loss — the logarithm compresses returns, which is why industrial lagging tables quote thickness versus percentage saving rather than naive proportionality.
Where students slip
The series/parallel identification is the first filter: slabs stacked along the heat path are series (same H), slabs side by side sharing the two temperatures are parallel (same ΔT) — students who match them to the visual "one after another = series" usually get it right, but stacked-but-mixed-areas problems need the area carried explicitly. The second error is computing the junction temperature by averaging temperatures rather than resistance-weighting: the interface drifts toward the poorer conductor's side, exactly as voltage divides toward the bigger resistor. In radial problems, forgetting that A varies with r (the integral gives the logarithm, not a simple KA/L) is endemic; and the critical-radius twist — that a thin wire's bare radius below r_c = K/h means added insulation initially raises heat loss — has appeared as an Advanced assertion–reason item. For the ice-growth problem, remember the growing thickness means the same air-to-ice drop drives an ever-thicker slab, so dy/dt ∝ 1/y and y ∝ √t; students who write constant dy/dt get linear growth and every subsequent number wrong.
Frequently asked questions
What is the thermal resistance of a slab?
R_th = L/(KA) in kelvin per watt, so the heat current is H = ΔT/R_th — the Fourier law cast in Ohm's-law form for series-parallel bookkeeping.
How is the junction temperature of two series slabs found?
The same H crosses both, so ΔT divides in the ratio of resistances: T_interface = T_hot − H R1, equivalently the drop fractions R1/(R1 + R2) and R2/(R1 + R2).
What is the equivalent conductivity of slabs in parallel?
K_eq = (K1A1 + K2A2)/(A1 + A2) — an area-weighted arithmetic mean; in series it is the harmonic mean, always smaller.
How does conduction through a cylindrical shell differ from a slab?
The area varies with radius, so H = 2πKL(T1 − T2)/ln(r2/r1); resistance grows only logarithmically with outer radius, and a critical radius exists below which added insulation increases loss.
Why does ice on a lake thicken as the square root of time?
The growing ice layer itself is the insulation, so dy/dt = Kθ/(ρL_f y), inversely proportional to current thickness — integrating gives y ∝ √t, four times the time to double the ice.