Fourier's Law & Steady-State Conduction

Connect temperature gradients to heat flow, and learn to compute steady 1‑D conduction through a flat wall.

sophomoreheat transferconduction

⏱️ About 18 min

Why does a thin metal sheet feel “colder” than a wooden one at the same room temperature?

💡

The big idea: In steady conduction, heat flows down the temperature gradient; Fourier’s law turns a temperature drop over a distance into a heat rate.

🎯 By the end, you'll be able to

State Fourier’s law for 1-D conduction
Interpret the sign of dT/dx and heat-flow direction
Distinguish heat rate Q from heat flux q''
Compute steady heat transfer through a plane wall

📎 Helpful to know first

Pump Selection Case Study

Fourier’s law (1‑D form)

For one-dimensional conduction through a slab (x-direction), Fourier’s law relates heat transfer to the temperature gradient:

Heat flows from hot to cold, i.e., in the direction of decreasing temperature.

\[ q_x = -kA\frac{dT}{dx} \]

Fourier’s law for 1‑D conduction: heat rate q_x (W) through area A due to gradient dT/dx.

🔑 Heat rate vs heat flux

Heat rate (often written Q or q_x) has units of watts (W) and represents total energy per time.

Heat flux is heat rate per area: q'' = Q/A (W/m²).

Steady 1‑D conduction through a plane wall

For a plane wall of thickness L, constant k, and constant area A under steady conditions, the temperature profile is linear.

If T_hot is on the left face and T_cold on the right face, the magnitude of the heat rate is:

\[ Q = kA\frac{T_{\text{hot}}-T_{\text{cold}}}{L} \]

Steady heat rate through a single homogeneous plane wall (constant k).

✨ Sign convention check

If x increases from the hot side to the cold side, then dT/dx is negative (temperature decreases with x).

The negative sign in Fourier’s law makes Q positive in the +x direction, matching the physical heat-flow direction.

📝 Worked example: A flat glass window (k = 0.8 W/(m·K)) is 8 mm thick and has area A = 1.50 m². The indoor surface is at 22°C and the outdoor surface is at 2°C. Assuming steady 1‑D conduction and constant k, compute the heat rate through the glass.

Given: k = 0.8 W/(m·K), L = 0.008 m, A = 1.50 m², ΔT = 22 − 2 = 20 K.
Use Q = kA(ΔT)/L.
Compute numerator: kAΔT = 0.8 × 1.50 × 20 = 24 W·m.
Divide by thickness: Q = 24 / 0.008 = 3000 W.

✓ 3000 W

✏️ Practice: A brick wall (k = 0.7 W/(m·K)) is 0.10 m thick with area 2.0 m². The two wall surfaces are at 35°C and 15°C. Compute the steady heat rate by conduction.

Solution

ΔT = 35 − 15 = 20 K.
Q = kAΔT/L = (0.7)(2.0)(20)/0.10.
Numerator: 0.7 × 2.0 × 20 = 28.
Q = 28/0.10 = 280 W.

Check your understanding

1. Which statement best describes Fourier’s law in 1‑D steady conduction?

For a plane wall with constant k, Q = kA(T_hot−T_cold)/L: proportional to ΔT and A, inversely to L.

2. If x increases from left to right and temperature decreases in that direction, what is the sign of dT/dx?

A decreasing T with increasing x means dT/dx < 0.

3. Heat flux q'' has units of:

Heat flux is heat rate per area: q'' = Q/A (W/m²).

✅ Key takeaways

Fourier’s law links heat transfer to the temperature gradient and thermal conductivity.
In a plane wall at steady state with constant k, temperature varies linearly in x.
Heat flows from hot to cold; the minus sign ensures the heat-flow direction matches the gradient sign.
Heat rate Q (W) differs from heat flux q'' (W/m²) by a factor of area.

➡️ Next, we rewrite the same physics using thermal resistance, making multi-layer problems as easy as circuit analysis.

Want to test yourself on this? Try the Chemical Aptitude test →