ARCHETYPE: Conduction Through Slabs, Rods & Composites

Every insulation decision ever taken is equation (2) with a budget attached — a builder's U-value is Example 1, and the industry chases thin foams rather than thick bricks for the reason doubling the insulation beat doubling the brick: series resistances add, so you attack the biggest term. Double glazing is the same trick in a window. Two $4\ \mathrm{mm}$ panes ($k = 0.8$) sandwiching a $12\ \mathrm{mm}$ sealed air gap ($k = 0.024$) over $2\ \mathrm{m^2}$ have $R = 0.0025 + 0.25 + 0.0025 = 0.255\ \mathrm{K\,W^{-1}}$, of which the air supplies $98\ \%$. Across $20\ \mathrm{K}$ the unit conducts about $78\ \mathrm{W}$ against a single pane's $8000\ \mathrm{W}$ — a factor of roughly $100$, bought with a layer of nothing. Read that factor honestly, though: both figures are pure-conduction numbers, and the single-pane $8000\ \mathrm{W}$ is an upper limit, because in a real window thin stagnant films of air cling to both glass faces and add resistance of their own. Those films matter enormously for the bare pane and hardly at all for the double-glazed unit, so the true saving in a fitted window is a few-fold, not a hundred-fold. The physics of the sandwich is exactly as computed; it is the comparison that needs the caveat. Run the law the other way and it becomes a hunt for high conductivity. A processor dumping $100\ \mathrm{W}$ into a copper spreader through $4\ \mathrm{cm^2}$ of contact is a series stack, and the weakest link decides everything: a $0.05\ \mathrm{mm}$ film of thermal paste ($k \approx 5$) has $R = 0.025\ \mathrm{K\,W^{-1}}$ and costs the chip only $2.5\ \mathrm{K}$, while the same gap filled with air has $R = 5.2\ \mathrm{K\,W^{-1}}$ and would demand $520\ \mathrm{K}$ to shift the same heat. Thermal paste is not a lubricant; it is a bad conductor bought to replace a catastrophic one. The parallel rule bites too: a steel rivet through foam, a concrete balcony slab, an aluminium window frame — each a high-$k$ branch across the same $\Delta T$. Architects call them thermal bridges: Example 2 exactly.

In an archetype the geometry is the physics, and here there are exactly two geometries. Everything else is a costume.

Series. The pieces are stacked one behind another along the direction of flow, like coaches of a train: a composite wall, a lagged pipe, a double-glazed window, two rods soldered end to end. Heat has no choice — it must cross every layer in turn. Hence the most important sentence in this article: in the steady state the same heat current $H$ passes through every layer, because nothing may pile up inside any of them. Whatever enters the brick each second must leave it and enter the insulation, or the brick would still be warming up.

Parallel. The pieces sit side by side, each spanning the same hot face and the same cold face — two rods bridging the same steam and ice baths, a window set in a wall, a rivet piercing a foam panel. Now heat has a choice of routes and takes both: each feels the same $\Delta T$, and the currents add. The test is one question: does the heat have to cross both pieces, or does it choose between them? Both, in turn — series. Either, side by side — parallel. Everything now runs off the two results the previous article established:

where: $R$ — the thermal resistance of one piece, in $\mathrm{K\,W^{-1}}$; $L$ — its thickness along the flow; $k$ — its conductivity; $A$ — the face area perpendicular to the flow; $H$ — the steady current in watts; $\Delta T$ — the difference across that piece. Read $R$ as "the kelvin you must spend to push one watt through": thick, narrow and low-$k$ is expensive; thin, wide and high-$k$ is nearly free. And now the reason $R$, not $k$, is the right currency is visible. Conductivity belongs to the material; resistance belongs to the object. A network is made of objects, so it is made of resistances — which combine, exactly as in a circuit, by rules you already know.

Stare at that graph: it contains the entire archetype. The bar is geometrically symmetric — half copper, half steel, same area, same length — yet the profile is wildly lopsided. The kink is not a drawing error but the physics: at the junction the material changes, so $k$ changes, so the slope must change. The only way to push the same current through a much poorer conductor is to spend a bigger temperature difference on it. That is the sentence to carry into the exam hall: same current, so the drops divide like the resistances. It is Ohm's potential-divider rule, wearing a thermal hat.

Take two pieces joined face to face, outer faces held at $T_{\text{hot}}$ and $T_{\text{cold}}$. In the steady state neither stores energy, so one common $H$ crosses both. Apply equation (1) to each piece using the drop across that piece — $\Delta T_1 = H R_1$, $\Delta T_2 = H R_2$ — and demand that the drops account for the whole difference. Adding them:

where: $R_{\text{series}}$ — the resistance of the whole stack; $L_i, k_i, A_i$ — the thickness, conductivity and area of the $i$-th layer. Obstacles in a queue simply add, so every layer's toll is charged to the journey. This is why one thin bad layer can dominate a thick good one — and why a builder adds insulation rather than removing brick. Now the money question. Divide $\Delta T_1 = HR_1$ by $\Delta T_2 = HR_2$; the common $H$ cancels:

where: each layer's share of the drop is proportional to its own resistance. For copper against steel, $R_{\text{steel}}/R_{\text{Cu}} = 40/5.19 = 7.7$ — exactly $k_{\text{Cu}}/k_{\text{steel}} = 385/50$. The junction temperature follows at once, and mark schemes take either route:

where: $R_1$ touches the hot end, $R_2$ the cold end. Pay the toll of crossing piece $1$ on the way down, or climb back up across piece $2$ from the cold end. The two must agree, and checking that they do is the cheapest sanity check in the topic: $100 - 2.21 \times 5.19 = 88.5\ \mathrm{^\circ C}$, and $0 + 2.21 \times 40 = 88.5\ \mathrm{^\circ C}$. One more series result earns its place on the formula card. If the two pieces share a length and area, the composite behaves like a single uniform rod of some equivalent conductivity $k_{\text{eq}}$. Demand that the replacement — total length $2L$, same $A$ — have the resistance of the real one, $2L/(k_{\text{eq}}A) = L/(k_1A) + L/(k_2A)$; the $L/A$ cancels and

where: $k_{\text{eq}}$ — the conductivity a uniform rod of the same overall dimensions would need. Read the number: bolting copper onto steel does not give a material half as good as copper — it gives something barely better than steel, because a reciprocal combination is always dragged towards the smaller member. It reproduces the master answer, $H = k_{\text{eq}}A\Delta T/(2L) = 88.5 \times 10^{-4} \times 100/0.4 = 2.21\ \mathrm{W}$ ✓, and it sets the trap: the arithmetic mean $217.5$ would give $5.44\ \mathrm{W}$, too big by a factor of $2.46$ — the right answer to the parallel question, offered in a series problem.

Now unstack the rods and lay them side by side, each bridging the same steam and ice baths. The logic inverts term for term: the temperature difference is now shared — both rods run from $100\ \mathrm{^\circ C}$ to $0\ \mathrm{^\circ C}$ — and the current is what splits:

where: $1/R$ is the thermal conductance — watts per kelvin, the opposite of resistance. In parallel the conductances add, never the resistances. A second route can only increase the flow, so $R_{\text{par}}$ must come out smaller than either branch. When the rods share a length $L$, equation (6) collapses to something you can write at sight, since $1/R_i = k_iA_i/L$:

where: $k_1A_1 + k_2A_2$ is the total "conducting power" of the cross-section. Two rods of $A = 1\ \mathrm{cm^2}$, $L = 0.5\ \mathrm{m}$, across $100\ \mathrm{K}$: copper carries $7.7\ \mathrm{W}$ and steel $1.0\ \mathrm{W}$, together $8.7\ \mathrm{W}$ — nothing more mysterious than a sum. And here is the symmetry that makes the archetype click. In series the copper–steel pair gave $88.5\ \%$ of the temperature drop to the steel; in parallel the same pair gives $7.7/8.7 = 88.5\ \%$ of the heat current to the copper. Both are $k_{\text{Cu}}/(k_{\text{Cu}}+k_{\text{steel}}) = 385/435$: series hands the drop to the worst conductor, parallel hands the flow to the best. One fraction, two faces.

1 · Board level — the insulated composite wall (CBSE/ICSE/A-Level pattern, 4 marks)

A wall of area $10\ \mathrm{m^2}$ is built of brick $0.1\ \mathrm{m}$ thick ($k = 0.6$) faced outside with insulation $0.05\ \mathrm{m}$ thick ($k = 0.04$), in $\mathrm{W\,m^{-1}K^{-1}}$. The inner surface is at $20\ \mathrm{^\circ C}$, the outer at $0\ \mathrm{^\circ C}$. Find the heat loss, the interface temperature, the heat lost in one hour, and the U-value.

Step 1. Heat must cross both layers — series. Convert each into a resistance, equation (1). Brick: $R_1 = 0.1/(0.6 \times 10) = 0.0167\ \mathrm{K\,W^{-1}}$. Insulation: $R_2 = 0.05/(0.04 \times 10) = 0.125\ \mathrm{K\,W^{-1}}$ — half as thick, yet $7.5$ times the resistance.

Step 2. Add, then use equation (2): $R_{\text{total}} = 0.0167 + 0.125 = 0.1417\ \mathrm{K\,W^{-1}}$, so

Step 3. The interface: walk in from the hot side and pay the brick's toll, equation (4). $\Delta T_{\text{brick}} = H R_1 = 141 \times 0.0167 = 2.4\ \mathrm{K}$, so

Check from the cold side: $HR_2 = 141 \times 0.125 = 17.6\ \mathrm{K}$ above $0\ \mathrm{^\circ C}$ ✓. The insulation carries $88\ \%$ of the drop, so the brick stays near room temperature — it does not feel cold to lean on, and it does not sweat.

Step 4. Energy in one hour, with $t$ in seconds: $Q = Ht = 141 \times 3600 = \boxed{5.1 \times 10^{5}\ \mathrm{J}}$. And the builder's figure of merit, the U-value $U = H/(A\,\Delta T) = 1/\sum_i (L_i/k_i)$: here $U = 141/200 = \boxed{0.71\ \mathrm{W\,m^{-2}K^{-1}}}$. Bare brick would leak $0.6 \times 10 \times 20/0.1 = 1200\ \mathrm{W}$ at $U = 6.0$ — the foam cuts the loss by a factor of $8.5$.

2 · Board level — two rods side by side (ICSE/JEE-Main pattern)

A copper rod and a steel rod, each $0.5\ \mathrm{m}$ long and $1\ \mathrm{cm^2}$ in cross-section, lie side by side, both left ends in steam at $100\ \mathrm{^\circ C}$ and both right ends in ice at $0\ \mathrm{^\circ C}$. Find the total heat current, the fraction carried by the copper, and the equivalent conductivity of the pair.

Step 1. Must the heat cross both rods, or does it choose? It chooses — independent bridges between the same two baths. Parallel: the same $\Delta T = 100\ \mathrm{K}$ on each, currents add. Convert the area at once, $1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}$.

Step 2. Each rod alone, equation (1): $H_{\text{Cu}} = 385 \times 10^{-4} \times 100/0.5 = 7.7\ \mathrm{W}$ and $H_{\text{steel}} = 50 \times 10^{-4} \times 100/0.5 = 1.0\ \mathrm{W}$.

Step 3. Add: $H = 7.7 + 1.0 = \boxed{8.7\ \mathrm{W}}$, of which copper carries $7.7/8.7 = \boxed{88.5\ \%}$ — removing the steel rod would cost barely a tenth of the heat.

Step 4. Equivalent conductivity, equation (8): the pair is one rod of length $0.5\ \mathrm{m}$ and area $2\ \mathrm{cm^2}$, so $k_{\text{eq}} = (385 + 50)/2 = \boxed{217.5\ \mathrm{W\,m^{-1}K^{-1}}}$, and $217.5 \times 2 \times 10^{-4} \times 100/0.5 = 8.7\ \mathrm{W}$ ✓. Here — and only here, in parallel with equal areas — the conductivities average.

3 · Competitive level — the joined bar, all four parts (JEE-Main/NEET pattern, about 3 minutes)

A copper rod and a steel rod, each $0.2\ \mathrm{m}$ long and $1\ \mathrm{cm^2}$ in cross-section, are soldered end to end and lagged. The free copper end is at $100\ \mathrm{^\circ C}$, the free steel end at $0\ \mathrm{^\circ C}$. Find (a) the heat current, (b) the junction temperature, (c) the equivalent conductivity, (d) the heat conducted in $10$ minutes.

Step 1. (a) This is the master bar of the first derivation card — reuse it rather than rebuild it: $R_{\text{Cu}} = 5.19$, $R_{\text{steel}} = 40$, $R_{\text{total}} = 45.19\ \mathrm{K\,W^{-1}}$, so $H = 100/45.19 = \boxed{2.21\ \mathrm{W}}$.

Step 2. (b) The junction, equation (4): $T_j = 0 + HR_{\text{steel}} = 2.21 \times 40 = \boxed{88.5\ \mathrm{^\circ C}}$, and downward from the hot end $100 - 2.21 \times 5.19 = 88.5\ \mathrm{^\circ C}$ ✓. If your answer came out at $50$, you averaged the ends and walked into the trap.

Step 3. (c) Equal lengths, so equation (5) applies: $k_{\text{eq}} = 2(385)(50)/435 = \boxed{88.5\ \mathrm{W\,m^{-1}K^{-1}}}$ — between $50$ and $385$, much nearer the smaller, as a reciprocal mean must be.

Step 4. (d) $H$ is a rate, so multiply by the time in seconds: $Q = 2.21 \times 600 = \boxed{1.33 \times 10^{3}\ \mathrm{J}}$. Timing note: three minutes for all four parts, none of it spent re-deriving — (a) and (b) are two resistances and two multiplications, (c) is a memorised one-liner, (d) is one product. Both places candidates lose the question are conversions: $1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}$ and $10$ min $= 600\ \mathrm{s}$. And the two $88.5$s are equal by coincidence, not physics — one is a temperature in $\mathrm{^\circ C}$, the other a conductivity in $\mathrm{W\,m^{-1}K^{-1}}$.

4 · Competitive level — the engineer's question (JEE-Advanced pattern, about 4 minutes)

The wall of Example 1 (losing $141\ \mathrm{W}$) is to be improved. You may double the thickness of the brick or of the insulation, at equal cost. (a) Which wins, and by how much? (b) A workman leaves a $1\ \mathrm{mm}$ air gap ($k = 0.024$) between the layers. Show that this "nothing" carries as much thermal resistance as $16\ \mathrm{m}$ of solid copper.

Step 1. Do not rebuild the wall — perturb it. Only one resistance changes in each option, and the total is a sum, so each answer is a single substitution.

Step 2. (a) Double the brick: $R_1$ goes $0.0167 \to 0.0333\ \mathrm{K\,W^{-1}}$, so $R_{\text{total}} = 0.1583$ and $H = 20/0.1583 = 126\ \mathrm{W}$ — ten centimetres of extra masonry for a saving of $15\ \mathrm{W}$.

Step 3. Double the insulation instead: $R_2$ goes $0.125 \to 0.25\ \mathrm{K\,W^{-1}}$, so $R_{\text{total}} = 0.2667$ and $H = 20/0.2667 = \boxed{75\ \mathrm{W}}$, against $\boxed{126\ \mathrm{W}}$ for the brick. Insulate the insulator. Equation (2) is why: the total is a sum, so doubling the big term buys far more than doubling the small one — a principle running from cavity walls to CPU coolers.

Step 4. (b) The gap, per square metre: $R_{\text{gap}} = 0.001/(0.024 \times 1) = 0.0417\ \mathrm{K\,W^{-1}}$. What length of copper matches it? From $R = L/(kA)$, $L_{\text{Cu}} = R\,k_{\text{Cu}}A = 0.0417 \times 385 = \boxed{16\ \mathrm{m}}$. A millimetre of trapped air resists heat as stubbornly as sixteen metres of copper — which is why engineers pay for thermal paste, and why glaziers sell the same gap as a feature. Timing note: four minutes, and the whole of part (a) is one insight — because series resistances add, always attack the largest term. Rebuilding the wall from scratch for each option takes twice as long and slips twice as often. Part (b) needs no conversion: $1\ \mathrm{mm}$ and $1\ \mathrm{m^2}$ are already SI, and the answer is one division and one multiplication.

Adding or averaging conductivities in series. The biggest mark-loser in the topic. $k$ belongs to a material, $R = L/(kA)$ to an object, and only objects can be wired together. In series add the $R$s: for equal-length rods $k_{\text{eq}} = 2k_1k_2/(k_1+k_2) = 88.5$, never $(k_1+k_2)/2 = 217.5$.

Taking the junction temperature as the average of the ends. It is $88.5\ \mathrm{^\circ C}$, not $50$. The drop divides as $R_1 : R_2$, so the junction hugs the good conductor's end. NEET dresses this as an assertion-reason item whose offered "reason" is the symmetry argument — which is false.

Applying Fourier's law to one layer with the total $\Delta T$. Each layer feels only its own drop, $\Delta T_i = HR_i$. Get $H$ from the whole stack first, then walk inward. Unit slips worth a factor of $10^{4}$. $1\ \mathrm{cm^2} = 10^{-4}\ \mathrm{m^2}$, not $10^{-2}$; and $H$ is a rate, so an energy needs the time in seconds ($10$ min $= 600\ \mathrm{s}$).

Ignoring the gap. An unmentioned millimetre of trapped air is worth $16\ \mathrm{m}$ of copper. A "poor joint", an air film or a "loose contact" is never scene-setting — it is the dominant resistance.

The limits worth carrying. If $R_2 \gg R_1$ the bad layer alone sets the flow, $H \to \Delta T/R_2$. If the two resistances are equal, and only then, the junction really is the mean of the ends ($50\ \mathrm{^\circ C}$ for our bar). In parallel with $k_1 \gg k_2$ the good conductor takes essentially all the heat ($88.5\ \%$ here), and adding a branch can only ever lower $R_{\text{par}}$.

Board notes. CBSE and ICSE want the composite wall and junction temperature with working shown; A-Level wants that wall as a U-value; JEE-Main mixes series and parallel in one figure; JEE-Advanced asks for equivalent conductivities and ratios; NEET keeps it conceptual.

Texts. NCERT Physics Class 11, Ch. 11 (Thermal Properties of Matter) — composite rods and walls · HC Verma, Concepts of Physics Vol. 2, "Heat Transfer" — the series/parallel exercises this archetype is built on.

Provenance. All results are standard, and every number on this page is recomputed in `verify_hs2-conduction-slab-problems.py`: the series bar ($R = 5.19 + 40 = 45.19\ \mathrm{K\,W^{-1}}$, $H = 2.21\ \mathrm{W}$, drops $11.5$ and $88.5\ \mathrm{K}$, junction $88.5\ \mathrm{^\circ C}$ both ways, $1.33\times10^{3}\ \mathrm{J}$ in $10$ min, $k_{\text{eq}} = 88.5$ against the wrong $217.5 \to 5.44\ \mathrm{W}$); the parallel pair ($8.7\ \mathrm{W}$, $R_{\text{par}} = 11.49$, copper's share $88.5\ \%$, $k_{\text{eq}} = 217.5$); the composite wall ($H = 141\ \mathrm{W}$, interface $17.6\ \mathrm{^\circ C}$, $U = 0.71$ against bare brick's $1200\ \mathrm{W}$ and $U = 6.0$; the doubling options $126$ and $75\ \mathrm{W}$); the $1\ \mathrm{mm}$ air gap ($0.0417\ \mathrm{K\,W^{-1}} = 16\ \mathrm{m}$ of copper); double glazing ($R = 0.255$, $H = 78\ \mathrm{W}$); the thermal-paste stack ($2.5\ \mathrm{K}$ against air's $520\ \mathrm{K}$).

Prerequisites. Conduction & Thermal Conductivity (Fourier's law, $R = L/(kA)$, the Ohm analogy). Leads to. A terminal archetype — the next network you meet will be electrical, and you already know how to solve it. Go deeper (undergraduate). "Transport Coefficients and Green–Kubo Relations" explains where $385$ and $50$ come from, deriving $k$ from the microscopic motion of the carriers instead of measuring it.