# Heat Conduction Through Composite Slabs > Thermal resistance L/KA in series and parallel, junction temperatures, cylindrical and spherical shells, and ice growth on lakes — JEE Physics. - Canonical URL: https://prepelephant.com/topics/jee/physics/heat-conduction-composite - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Heat Conduction Through Composite Slabs", PrepElephant, https://prepelephant.com/topics/jee/physics/heat-conduction-composite ## 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.