Worked examples — Heat transfer — conduction (Fourier's law k), convection, radiation (Stefan-Boltzmann σT⁴)

This is the practice floor for the parent topic. Before we compute anything, let us map out every kind of question the three heat-transfer laws can throw at you. Then we solve one example per box so you never meet a case you have not already seen.

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The scenario matrix

#	Cell (case class)	Which law	The trap it tests
A	Single slab, plain numbers	Conduction	°C-difference = K-difference
B	Slabs in series	Conduction	resistances add; find interface temperature
C	Slabs in parallel	Conduction	conductances add, not resistances
D	Degenerate: gradient $\to 0$	Conduction	$\Delta T=0\Rightarrow H=0$ (equilibrium)
E	Convection off a surface	Convection	$hA\Delta T$, linear in $\Delta T$
F	Radiation, surroundings negligible	Radiation	when may we drop $T_0^4$?
G	Radiation, surroundings NOT negligible	Radiation	full $T^4-T_0^4$, both in kelvin
H	Sign / net = 0 limiting case	Radiation	body in equilibrium with room, $P_{\text{net}}=0$
I	Real-world word problem	Combined	pick the dominant mode
J	Exam twist: unit sabotage	Radiation	°C used where kelvin required

We will hit A–J below. Prerequisites you may want open: Thermal conductivity k, Newton's law of cooling, Black body radiation, Electrical resistance Ohm's law.

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Example 1 — Cell A: single slab, bare numbers

1. Write the magnitude form of Fourier's law. $H=kA\frac{\Delta T}{\Delta x}$ Why this step? We only need the size of the flow, so we use the magnitude version with $\Delta T=T_{\text{hot}}-T_{\text{cold}}$ — the minus sign of the full law $H=-kAdT/dx$ is already accounted for by taking hot-minus-cold.

2. Compute $\Delta T$ and note the unit trick. $\Delta T=22-2=20{(}^{\circ }\text{C or K, identical})$ Why this step? A difference of 20 °C equals a difference of 20 K exactly — each kelvin equals one Celsius degree in size. No conversion needed for a difference.

3. Plug in. $H=0.6\times 6\times \frac{20}{0.20}=0.6\times 6\times 100=360\text{W}$

Verify: Units: $({\text{W m}}^{-1}{\text{K}}^{-1})({\text{m}}^2)(\text{K})/(\text{m})=\text{W}$ ✓. Magnitude 360 W — hundreds, as forecast. A brick wall lets out far less than the 8000 W glass window in the parent note, because brick is thick and has small $k$. Sensible. ✓

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Example 2 — Cell B: slabs in series + interface temperature

1. Compute each thermal resistance $R=\Delta x/(kA)$. $R_1=\frac{0.20}{0.6\times 6}=0.0556\text{K/W},R_2=\frac{0.05}{0.03\times 6}=0.2778\text{K/W}$ Why this step? In series the same current $H$ passes through both slabs (energy cannot pile up in steady state — see Second law of thermodynamics), so resistances add, exactly like resistors carrying one current in Electrical resistance Ohm's law.

2. Add resistances, find $H$. $R_{\text{tot}}=0.0556+0.2778=0.3333\text{K/W},H=\frac{\Delta T}{R_{\text{tot}}}=\frac{20}{0.3333}=60\text{W}$ Why this step? $H=\Delta T/R_{\text{tot}}$ is Ohm's law for heat. The foam alone cut the flow from 360 W (Example 1) down to 60 W.

3. Find the interface temperature $T_m$. The same $H$ flows through the brick, so the drop across the brick is $H{R}_1$: $\Delta T_{\text{brick}}=H{R}_1=60\times 0.0556=3.33\text{K}$ $T_m=T_{\text{inside}}-\Delta T_{\text{brick}}=22-3.33=18.67^{\circ }\text{C}$ Why this step? Look at the figure: most of the 20 K drop (16.67 K) falls across the foam, the poor conductor. The steep temperature cliff lives in the insulating layer — that is the whole point of insulation.

Verify: Drop across foam $=H{R}_2=60\times 0.2778=16.67$ K, and $3.33+16.67=20$ K = total ✓. Interface at $18.67^{\circ }$C is between 2 and 22 — physically valid ✓.

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Example 3 — Cell C: slabs in parallel

1. Each path sees the full $\Delta T$; heat currents add. $H_{\text{brick}}=\frac{15}{0.5}=30\text{W},H_{\text{glass}}=\frac{15}{0.1}=150\text{W}$ Why this step? Brick and glass sit side by side, both bridging the same two temperatures — that is the definition of parallel. Currents through parallel paths add (they split the load), so we do not add resistances here.

2. Total heat current. $H_{\text{tot}}=30+150=180\text{W}$

3. (Cross-check via combined resistance — conductances add in parallel.) $\frac{1}{R_{\text{tot}}}=\frac{1}{0.5}+\frac{1}{0.1}=2+10=12\text{W/K},R_{\text{tot}}=0.0833\text{K/W},H=\frac{15}{0.0833}=180\text{W}$ Why this step? Confirms the rule: parallel → conductances $1/R$ add. The small window (20% of area) carries 150/180 ≈ 83% of the loss — the low-resistance path dominates.

Verify: Both methods give 180 W ✓. Glass carries the majority of the loss as forecast ✓.

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Example 4 — Cell D: degenerate case, $\Delta T\to 0$

1. Set $\Delta T=0$. $H=kA\frac{\Delta T}{\Delta x}=0.6\times 6\times \frac{0}{0.20}=0\text{W}$ Why this step? Fourier's law is linear in $\Delta T$. No temperature difference means no driving "push," so no net heat flows — the wall is in thermal equilibrium, consistent with the Second law of thermodynamics: heat only flows when a gradient exists.

Verify: $H=0$ when $\Delta T=0$ — the limiting/degenerate case is well-behaved (not undefined, because $\Delta x\ne 0$). ✓

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Example 5 — Cell E: convection off a hot plate

1. Apply Newton's law of cooling. $H=hA(T_s-T_{\text{fluid}})$ Why this step? Convection carries heat away in proportion to how far the surface temperature sits above the fluid — a simple engineering model (see Newton's law of cooling). Again $\Delta T$ in °C = K.

2. Plug in. $H=15\times 0.4\times (80-25)=15\times 0.4\times 55=330\text{W}$

Verify: Units $({\text{W m}}^{-2}{\text{K}}^{-1})({\text{m}}^2)(\text{K})=\text{W}$ ✓. 330 W — hundreds, as forecast ✓.

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Example 6 — Cell F: radiation, surroundings negligible

1. Write the net radiation law. $P_{\text{net}}=e\sigma A\left(T^4-T_0^4\right),\sigma =5.67\times 10^{-8}$ Why this step? A body both emits ($e\sigma A{T}^4$) and absorbs from the room ($e\sigma A{T}_0^4$); the net is the difference (Kirchhoff — same $e$ both ways).

2. Evaluate the fourth powers. $T^4=1200^4=2.0736\times 10^{12},T_0^4=300^4=8.1\times 10^9$ Why this step? $T_0^4$ is $\approx 256\times$ smaller than $T^4$ — about $0.4\%$. Dropping it introduces only a $0.4\%$ error, so it is a legitimate 80/20 shortcut here.

3. Compute. $P_{\text{net}}=0.9\times 5.67\times 10^{-8}\times 4\times 10^{-4}\times (2.0736\times 10^{12}-8.1\times 10^9)$ $=0.9\times 5.67\times 10^{-8}\times 4\times 10^{-4}\times 2.0655\times 10^{12}\approx 42.2\text{W}$

Verify: Dropping $T_0^4$ entirely gives $42.3$ W — differs from the exact $42.2$ W by ~0.2%, confirming the shortcut is safe when $T\gg T_0$ ✓.

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Example 7 — Cell G: radiation with surroundings that matter

1. Both fourth powers, both in kelvin. $T^4=340^4=1.33634\times 10^{10},T_0^4=290^4=7.0729\times 10^9$ Why this step? When surface and surroundings are close, $T_0^4$ is a big fraction of $T^4$ (here ~53%), so we cannot drop it — that would nearly double the answer.

2. Subtract, then multiply. $T^4-T_0^4=1.33634\times 10^{10}-7.0729\times 10^9=6.2905\times 10^9$ $P_{\text{net}}=0.85\times 5.67\times 10^{-8}\times 1.5\times 6.2905\times 10^9\approx 454.9\text{W}$

Verify: Units: $({\text{W m}}^{-2}{\text{K}}^{-4})({\text{m}}^2)({\text{K}}^4)=\text{W}$ ✓. A modest few-hundred watts, as forecast — the small 50 K gap keeps the net well below Example 6's blazing filament even with far larger area ✓.

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Example 8 — Cell H: limiting case, net radiation = 0

1. Apply the net law with $T=T_0$. $P_{\text{net}}=e\sigma A(T^4-T_0^4)=e\sigma A(295^4-295^4)=0\text{W}$ Why this step? The statue emits and absorbs at the same rate when it is at the room's temperature. It still glows with radiation — but the net exchange is zero. This is exactly why bodies stop cooling once they reach ambient temperature (Second law of thermodynamics again: no gradient, no net flow).

Verify: $T^4-T_0^4=0\Rightarrow P_{\text{net}}=0$ regardless of $e$, $A$, $\sigma$ — the equilibrium limit is exact ✓.

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Example 9 — Cell I: real-world word problem (pick the dominant mode)

1. Rule out conduction and convection first. Why this step? A vacuum has no matter, so there is nothing to conduct through and no fluid to convect — both those modes need a medium. Only radiation crosses vacuum. So the entire loss is radiative.

2. Compute the radiative loss with tiny emissivity. $P=e\sigma A(T^4-T_0^4)$ $T^4=350^4=1.500625\times 10^{10},T_0^4=295^4=7.573502\times 10^9$ $T^4-T_0^4=7.432748\times 10^9$ $P=0.05\times 5.67\times 10^{-8}\times 0.06\times 7.432748\times 10^9\approx 1.27\text{W}$ Why this step? The silvering forces $e$ down to $0.05$, slashing radiative loss ~20-fold versus a black surface. Combined with the vacuum killing the other two modes, the coffee loses barely over a watt.

Verify: Just over 1 W — a fraction of what an unsilvered, air-filled container would lose, as forecast. This is the whole engineering point of the thermos ✓.

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Example 10 — Cell J: exam twist, the °C sabotage

1. The wrong calculation (°C, forbidden). $P_{\text{wrong}}=1\times 5.67\times 10^{-8}\times 0.1\times (127{)}^4=5.67\times 10^{-9}\times 2.601\times 10^8\approx 1.475\text{W}$ Why this step? This is the trap: $T^4$ is an absolute quantity, so kelvin is mandatory. Only a difference $\Delta T$ may stay in °C.

2. The correct calculation (kelvin). $T=127+273=400\text{K},P_{\text{correct}}=5.67\times 10^{-9}\times 400^4=5.67\times 10^{-9}\times 2.56\times 10^{10}\approx 145.2\text{W}$ Why this step? Convert first, then take the fourth power. The correct radiated power is nearly 100 times the mistaken one.

3. Ratio. $\frac{P_{\text{correct}}}{P_{\text{wrong}}}=\frac{400^4}{127^4}={\left(\frac{400}{127}\right)}^4\approx (3.15{)}^4\approx 98.4$

Verify: The wrong answer is ~98× too small — a catastrophic error, illustrating why kelvin is non-negotiable in any power of $T$ ✓.

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Recall Self-test before moving on

A slab and its twin are placed (a) in series, (b) in parallel across the same $\Delta T$. Which arrangement passes more total heat? ::: Parallel — resistances halve (conductances add), so current doubles versus a single slab; series doubles resistance, halving current. When is it safe to drop $T_0^4$ in the net radiation formula? ::: When $T\gg T_0$, e.g. $T=1200$ K vs $T_0=300$ K, where $T_0^4$ is under ~1% of $T^4$. A body at room temperature — is it radiating? Is its net radiation zero? ::: Yes it radiates ($e\sigma A{T}^4$); but net is zero because it absorbs an equal amount from the equal-temperature surroundings.