Heat transfer — conduction (Fourier's law k), convection, radiation (Stefan-Boltzmann σT⁴): Visual walkthrough

We build the rate of heat flow through a solid slab — nothing else. Let's earn every letter.

Step 1 — What are we even measuring? (the heat current $H$ )

WHAT. Picture a solid wall. On the left it is hot, on the right it is cold. Energy leaks steadily from hot to cold. We want a single number for how fast that energy crosses the wall.

WHY. "How hot is it" is not enough — we care about flow, joules crossing per second. Physicists call a "per second" flow of something a current. So we invent the heat current:

$H = \frac{d Q}{d t}$

$Q$ ::: the amount of heat energy (measured in joules, J) that has crossed the wall.
$t$ ::: time in seconds.
$\frac{d Q}{d t}$ ::: "how much $Q$ changes each second" — read it as joules per second, which is a watt (W). The little $d$ 's just mean "a tiny change in".

So $H$ is measured in watts, exactly like a lightbulb's power. That is the whole target: find a formula for $H$ .

PICTURE. Below, the yellow arrows are the heat leaking rightward through the wall. The number we chase is: how many joules cross that dashed line every second?

Step 2 — More wall, more flow ( $H \propto A$ )

WHAT. Keep everything the same but make the wall wider and taller — give it more face area $A$ (measured in square metres, m²).

WHY. Heat crosses through the face of the wall. Double the face and you have simply glued two identical walls side by side; each leaks the same, so together they leak twice as much. There is nothing subtle here — it is pure "two of a thing is twice one thing".

$H \propto A$

$\propto$ ::: "is proportional to" — grows in lock-step with. If $A$ triples, $H$ triples.
$A$ ::: the cross-sectional area the heat flows through, not the thickness.

PICTURE. Two identical wall-tiles stacked make a wall of double area and double the arrow-count.

Step 3 — Steeper temperature drop, more flow ( $H \propto Δ T$ )

WHAT. Now fix the area and change the two face temperatures. Call the hot face $T_{1}$ and the cold face $T_{2}$ , with $T_{1} > T_{2}$ . Their gap is

$Δ T = T_{1} - T_{2} .$

$Δ$ (Greek "delta") ::: "the difference in" — always later minus earlier, or here hot minus cold.
$Δ T$ ::: how many degrees hotter the left face is than the right face.

WHY. Heat flows because of a temperature difference (the second law: energy slides from hot to cold). No difference → no flow. A bigger difference is a steeper "hill" for the energy to roll down, so it flows faster. Experiment confirms this is directly proportional — double the gap, double the flow.

$H \propto Δ T$

PICTURE. A gentle temperature slope drives a trickle; a steep slope drives a torrent. The blue line is temperature plotted across the wall — its steepness is what matters.

Step 4 — Thicker wall, less flow ( $H \propto 1/Δ x$ )

WHAT. Keep area and temperatures fixed, but make the wall thicker. Call the thickness $Δ x$ (metres) — the distance the heat must cross from hot face to cold face.

WHY. Energy hops from atom to atom, like a bucket chain. A thicker wall is a longer bucket chain for the same temperature drop, so at each point the "slope" is gentler and the flow is slower. Double the thickness and (for the same faces) you halve the flow:

$H \propto \frac{1}{Δ x}$

$\frac{1}{Δ x}$ ::: "inversely proportional" — as $Δ x$ grows, $H$ shrinks. This is the opposite of Steps 2 and 3.

PICTURE. Same hot and cold faces, but a fat wall spreads the same $Δ T$ over a longer path — the slope flattens, the arrows thin out.

Step 5 — Bolt the three facts together (birth of $k$ )

WHAT. We now have three separate proportionalities. Multiply them into one statement:

$H \propto A \frac{Δ T}{Δ x} .$

To turn " $\propto$ " into " $=$ " we insert a single constant $k$ that captures which material the wall is made of:

$H = k A \frac{Δ T}{Δ x}$

WHY a constant is needed. The proportionality tells us the shape of the law but not the scale. Copper and wood have the same shape of law but wildly different flow — that difference is baked into $k$ , the thermal conductivity.

$k$ ::: how eagerly this material conducts. Big $k$ (metal) → lots of flow; tiny $k$ (foam, air) → great insulator.
Units of $k$ . Rearrange: $k = \frac{H Δ x}{A Δ T}$ → $\frac{W \cdot m}{m ^{2} \cdot K} = W m^{- 1} K^{- 1}$ . $k$ exists precisely to make the units on both sides match.

PICTURE. The three influences (blue area, pink drop, yellow inverse-thickness) feed a mixing funnel; $k$ is the last ingredient that fixes the final number.

Step 6 — From slab to a point: the derivative and the minus sign

WHAT. So far $Δ T /Δ x$ was the average slope across the whole wall. But temperature might not fall in a straight line. To describe the flow at a single point inside the wall, shrink the slab to zero thickness. When you let $Δ x \to 0$ , the ratio $Δ T /Δ x$ becomes the derivative $\frac{d T}{d x}$ .

WHY the derivative and not just the ratio. A derivative answers exactly the question "how steeply is $T$ changing right here?" — the local slope of the blue temperature curve. Using it lets the law work even when the wall is layered or the slope bends. This is the tool built for "instantaneous steepness", so it is the right one.

WHY the minus sign now appears. Define $x$ as increasing in the direction heat flows. Since it flows toward the cold side, temperature goes down as $x$ goes up, so $\frac{d T}{d x}$ is negative. But a heat current pointing forward should be a positive number. To flip the negative slope into a positive current we stick a minus in front:

PICTURE. The blue curve slopes downhill (negative $d T / d x$ ); the minus flips it so the yellow current arrow comes out pointing forward and positive.

Step 7 — Every case, checked (signs, zero, degenerate walls)

We must never leave a scenario unshown. Walk the corners:

Heat flowing the other way ( $T_{2} > T_{1}$ ): now $d T / d x$ is positive, so $- k A d T / d x$ is negative → $H < 0$ . A negative $H$ simply means "flow is backward", i.e. from right to left. The sign auto-corrects; you never have to guess direction.
No temperature difference ( $T_{1} = T_{2}$ ): $d T / d x = 0 \Rightarrow H = 0$ . Flat temperature, zero flow — matches intuition (nothing driving it).
Perfect insulator ( $k \to 0$ , e.g. vacuum-flask foam): $H \to 0$ no matter how big the gap. This is why insulation works — kill $k$ and you kill the current.
Zero thickness ( $Δ x \to 0$ in the slab form): $H \to \infty$ . A wall with no thickness offers no resistance — infinite flow, which is why the derivative form (a finite local slope) is the honest description of a real wall.
Zero area ( $A = 0$ ): $H = 0$ . No face, no flow. Trivial but real — a wire's end-face conducts through its tiny $A$ , which is why thin wires conduct little total heat.

PICTURE. A small "map" of five corner cases: reversed flow, flat $T$ , dead insulator, zero thickness, zero area — each with its verdict.

The one-picture summary

Everything above in a single frame: the wall, the downhill temperature line, the four influences with their arrows, the constant $k$ , and the final boxed law with each symbol tagged by what it does.

Recall Feynman retelling (say it out loud, no symbols)

Imagine a wall, hot on one side, cold on the other. Heat wants to crawl from hot to cold. How fast it crawls — that's our heat current, joules per second, watts. Three plain facts set the pace: a wider wall lets more through (more doors), a bigger temperature gap shoves it harder (steeper hill), and a thicker wall slows it (longer crawl). Multiply those three together and sprinkle in one material number $k$ — big for metals, tiny for foam — and you have the flow. Finally, if you want the flow at one exact spot, you use the slope of the temperature (the derivative), and you paint a minus sign in front just so the number comes out positive when heat flows the natural way. Kill the gap, or kill $k$ , or shrink the area, and the flow dies — exactly as common sense demands. That whole story is Fourier's law.

Recall Quick self-test

Why does a thicker wall carry less heat for the same face temperatures? ::: Same $Δ T$ spread over larger $Δ x$ → gentler slope $\frac{d T}{d x}$ → smaller $H$ (since $H \propto 1/Δ x$ ). What does a negative value of $H$ tell you? ::: Heat is flowing in the $- x$ direction (backward); the sign encodes direction, not a "negative amount" of heat. Where did the constant $k$ come from and why is it needed? ::: It converts the proportionality $H \propto A Δ T /Δ x$ into an equation and encodes the material; it also fixes the units.