Derivation

Heat Conduction: Fourier's Law and the Heat Equation

D-434 Home PU-203 Threads energy · matter Depends on The First Law of Thermodynamics and Work in Ideal-Gas Processes

Statement

In a homogeneous, isotropic solid of constant thermal conductivity \(k\), density \(\rho\), and specific heat capacity \(c\), the local heat flux density \(\vec{q}\) (power per unit area, \(\mathrm{W/m^2}\)) is proportional to the negative temperature gradient: \(\vec{q} = -k\,\vec{\nabla}T\). Combining this constitutive law with local energy conservation (first law, no mechanical work) yields the heat equation \(\frac{\partial T}{\partial t} = \alpha\,\nabla^{2}T\), where the thermal diffusivity is \(\alpha = k/(\rho c)\). In one dimension with steady state and no internal sources, the temperature profile is linear, and the heat current through a slab of area \(A\), thickness \(L\), and face temperatures \(T_{\text{hot}}\) and \(T_{\text{cold}}\) is \(H = kA(T_{\text{hot}}-T_{\text{cold}})/L\).

Why it matters

Fourier's law and the heat equation are the foundational transport equations for thermal engineering, materials science, and geophysics. They allow the calculation of heat flow in walls, heat sinks, electronic packages, and the Earth's crust without tracking individual phonons or molecules. The law is the thermal analogue of Ohm's law and Fick's law of diffusion, placing heat conduction within the broader framework of linear irreversible thermodynamics.

The heat equation is a parabolic partial differential equation whose solutions describe how temperature disturbances spread and decay over time. Understanding the characteristic diffusion length \(\sqrt{\alpha t}\) is essential for estimating response times in thermal management, from microelectronics to planetary cooling.

Assumptions

Local thermal equilibrium (LTE).The temperature \(T(\vec{r},t)\) is a well-defined macroscopic field. If the material is not in local equilibrium (e.g., extremely short timescales or nanoscale structures), the concept of a local temperature breaks down, and Fourier's law fails.

Constant material properties.\(k\), \(\rho\), and \(c\) are treated as constants independent of temperature and position. If \(k\) varies strongly with \(T\), the heat equation becomes nonlinear: \(\rho c \frac{\partial T}{\partial t} = \nabla \cdot (k(T) \nabla T)\), which is harder to solve analytically.

No internal heat sources or sinks.The derivation assumes no volumetric heat generation (e.g., from nuclear decay, electrical resistance, or chemical reactions). If a source term \(S(\vec{r},t)\) is present, the equation becomes \(\rho c \frac{\partial T}{\partial t} = \nabla \cdot (k \nabla T) + S\).

Continuum approximation.The material is treated as a continuous medium. This requires the length scale of interest to be much larger than the mean free path of the heat carriers (phonons in insulators, electrons in metals). At nanoscale dimensions, ballistic transport occurs, and Fourier's law is invalid.

Isotropy.The thermal conductivity \(k\) is a scalar, not a tensor. In anisotropic materials (e.g., crystals, composites), \(\vec{q} = -\mathbf{K}\cdot\vec{\nabla}T\), where \(\mathbf{K}\) is a second-rank tensor, and heat flows not necessarily perpendicular to isotherms.

Derivation

\[ \vec{q} = -k\,\vec{\nabla}T \]

Fourier's law: heat flux density is proportional to the negative temperature gradient. The negative sign ensures heat flows from hot to cold. A

\[ \frac{\partial u}{\partial t} = -\vec{\nabla}\cdot\vec{q} \]

Local energy conservation: the rate of change of internal energy density \(u\) equals the negative divergence of the heat flux (net outflow). B

\[ du = \rho c\,dT \;\Longrightarrow\; \frac{\partial u}{\partial t} = \rho c\,\frac{\partial T}{\partial t} \]

For a solid (no expansion work) a temperature rise \(dT\) raises the internal energy per unit volume by \(\rho c\,dT\), with \(\rho\) the density and \(c\) the specific heat capacity; only changes of \(u\) matter, so no absolute zero of energy is needed. A

\[ \rho c \frac{\partial T}{\partial t} = -\vec{\nabla}\cdot\vec{q} \]

Substitute Step 3 into Step 2. This energy balance holds whatever the law relating \(\vec q\) to \(T\). B

\[ \rho c \frac{\partial T}{\partial t} = -\vec{\nabla}\cdot(-k\,\vec{\nabla}T) = \vec{\nabla}\cdot(k\,\vec{\nabla}T) \]

Substitute Fourier's law (Step 1) into Step 4. The double negative cancels. C

\[ \rho c \frac{\partial T}{\partial t} = k\,\nabla^{2}T \]

Assume \(k\) is constant, so it can be pulled out of the divergence operator: \(\vec{\nabla}\cdot(k\,\vec{\nabla}T) = k\,\vec{\nabla}\cdot\vec{\nabla}T = k\,\nabla^{2}T\). B

\[ \frac{\partial T}{\partial t} = \alpha\,\nabla^{2}T \]

Define the thermal diffusivity \(\alpha = k/(\rho c)\). This is the standard heat equation. A

\[ \frac{\partial T}{\partial t} = 0 \quad \text{(steady state)} \]

In steady state, the temperature does not change with time. The heat equation reduces to Laplace's equation. A

\[ \nabla^{2}T = 0 \]

Steady-state heat equation with no sources. This is Laplace's equation for the temperature field. A

\[ \frac{d^{2}T}{dx^{2}} = 0 \quad \text{(1D slab)} \]

In one dimension (slab geometry), the Laplacian reduces to the second derivative with respect to the spatial coordinate \(x\). A

\[ T(x) = Ax + B \]

Integrate twice: the general solution is a linear function of \(x\). A

\[ T(0) = T_{\text{hot}}, \quad T(L) = T_{\text{cold}} \]

Apply boundary conditions: temperature at the hot face (\(x=0\)) and cold face (\(x=L\)). A

\[ T(x) = T_{\text{hot}} - \frac{T_{\text{hot}}-T_{\text{cold}}}{L}x \]

Solve for \(A\) and \(B\): \(B=T_{\text{hot}}\), \(A=-(T_{\text{hot}}-T_{\text{cold}})/L\). B

\[ q_x = -k\frac{dT}{dx} = -k\left(-\frac{T_{\text{hot}}-T_{\text{cold}}}{L}\right) = k\frac{T_{\text{hot}}-T_{\text{cold}}}{L} \]

Compute the heat flux density using Fourier's law and the linear temperature profile. The heat current is \(H = q_x A\). C

Result

\[ \vec{q} = -k\,\vec{\nabla}T, \qquad \frac{\partial T}{\partial t} = \alpha\,\nabla^{2}T, \qquad \alpha = \frac{k}{\rho c} \]

Reading. Fourier's law states that heat flux is proportional to the negative temperature gradient. The heat equation describes how temperature evolves in time due to spatial variations in temperature. In steady state, the temperature profile is linear in a slab, and the heat current is proportional to the temperature difference and inversely proportional to the thickness.

Units check. \([\vec{q}] = \mathrm{W/m^2}\). \([k\,\vec{\nabla}T] = \mathrm{(W/(m\,K)) \cdot (K/m) = W/m^2}\). Consistent. \([\alpha] = \mathrm{(W/(m\,K))/(kg/m^3 \cdot J/(kg\,K)) = (W/(m\,K))/(W/(m^3\,K)) = m^2/s}\). Consistent.

Limiting cases

Steady state, 1D slab: \(T(x)\) is linear, \(q_x = k(T_{\text{hot}}-T_{\text{cold}})/L\), and \(H = q_x A\).
Steady state, cylindrical shell: \(H = 2\pi k L (T_1 - T_2)/\ln(r_2/r_1)\), where \(L\) is the length of the cylinder, and \(r_1, r_2\) are the inner and outer radii.
Diffusion length: The characteristic length over which a temperature disturbance spreads in time \(t\) is \(L \sim \sqrt{\alpha t}\).
Composite wall (series): For slabs in series, the total thermal resistance is \(R_{\text{th}} = \sum R_i = \sum L_i/(k_i A)\), and \(H = \Delta T / R_{\text{th}}\).
Composite wall (parallel): For slabs in parallel, the total thermal conductance is \(G_{\text{th}} = \sum G_i = \sum k_i A_i/L_i\), and \(H = G_{\text{th}} \Delta T\).

Breaks when

Convection or radiation dominate. If heat transfer is primarily by convection (fluid motion) or radiation (electromagnetic waves), Fourier's law does not apply. The boundary conditions must include convective (\(h(T_s - T_\infty)\)) or radiative (\(\epsilon\sigma(T_s^4 - T_{\text{env}}^4)\)) terms.
Very short times or length scales. At nanoscale dimensions or extremely short timescales (femtoseconds), the mean free path of heat carriers is comparable to the system size, and ballistic transport occurs. Fourier's law fails, and the Boltzmann transport equation or phonon hydrodynamics is required.
\(k\) strongly temperature-dependent. If \(k(T)\) varies significantly over the temperature range, the heat equation becomes nonlinear: \(\rho c \frac{\partial T}{\partial t} = \nabla \cdot (k(T) \nabla T)\). Analytical solutions are generally unavailable, and numerical methods are needed.
Anisotropic materials. In crystals or composites, \(k\) is a tensor, and heat flows not necessarily perpendicular to isotherms. The heat equation becomes \(\rho c \frac{\partial T}{\partial t} = \nabla \cdot (\mathbf{K}\cdot\nabla T)\), where \(\mathbf{K}\) is a second-rank tensor.

Failure modes

Forgetting the negative sign in Fourier's law. Writing \(\vec{q} = +k\,\vec{\nabla}T\) implies heat flows from cold to hot, violating the second law of thermodynamics. The negative sign ensures heat flows down the temperature gradient.
Confusing heat flux density \(\vec{q}\) with heat current \(H\). \(\vec{q}\) is power per unit area (\(\mathrm{W/m^2}\)), while \(H\) is total power (\(\mathrm{W}\)). They are related by \(H = \vec{q}\cdot\vec{A}\), where \(\vec{A}\) is the area vector. Forgetting the area factor leads to errors by orders of magnitude.
Assuming the temperature profile is always linear. The linear profile holds only in steady state, 1D, with constant \(k\) and no sources. In transient states, cylindrical geometry, or with sources, the profile is nonlinear (e.g., logarithmic in a cylinder, parabolic with sources).
Mixing up thermal diffusivity \(\alpha\) with thermal conductivity \(k\). \(\alpha = k/(\rho c)\) has units \(\mathrm{m^2/s}\) and governs the rate of temperature change. \(k\) has units \(\mathrm{W/(m\,K)}\) and governs the magnitude of heat flow. They are related but distinct: water conducts heat better than wood (\(k \approx 0.6\) against \(0.15\ \mathrm{W/(m\,K)}\)) yet, because its heat capacity per volume is large, its diffusivity is only \(\alpha \approx 1.4\times10^{-7}\ \mathrm{m^2/s}\), comparable to wood's.

Discussion

Fourier's law is a constitutive relation, not a conservation law. It encodes the empirical observation that heat flows from hot to cold, with the flux proportional to the temperature gradient. The proportionality constant \(k\) is a material property that depends on the microscopic mechanisms of heat transport (phonons, electrons, molecules). The heat equation, on the other hand, is a consequence of the first law of thermodynamics (energy conservation) combined with Fourier's law. This distinction is crucial: the energy balance \(\rho c\,\partial_t T = -\nabla\cdot\vec q\) would survive a different constitutive law (nonlinear or nonlocal), but the heat equation itself would not.

The thermal diffusivity \(\alpha = k/(\rho c)\) measures how quickly a temperature disturbance spreads through a material. A high \(\alpha\) means the material responds quickly to temperature changes (e.g., metals), while a low \(\alpha\) means it responds slowly (e.g., water, concrete). The characteristic diffusion length \(L \sim \sqrt{\alpha t}\) is a powerful scaling law: it tells you how far a heat pulse travels in time \(t\). For example, if \(\alpha = 10^{-5}\ \mathrm{m^2/s}\) (typical for metals), then in \(t = 1\ \mathrm{s}\), \(L \sim \sqrt{10^{-5}} \approx 3\ \mathrm{mm}\).

The steady-state solutions of the heat equation (Laplace's equation) are harmonic functions, which have the mean value property: the temperature at any point is the average of the temperatures on a sphere centered at that point. This property is the basis for many analytical solutions and numerical methods (e.g., finite difference, finite element). In 1D, the linear profile is the simplest harmonic function. In 2D and 3D, the solutions are more complex but still governed by the same principle: no local maxima or minima in the interior (maximum principle).

At a deeper level, the heat equation is a diffusion equation, and its solutions are related to the Green's function of the Laplacian. The fundamental solution in 3D is \(G(\vec{r},t) = \frac{1}{(4\pi\alpha t)^{3/2}} \exp(-r^2/4\alpha t)\), which describes how a point source of heat spreads in time. This solution is a Gaussian whose width grows as \(\sqrt{\alpha t}\), confirming the diffusion length scaling. The heat equation is also the parabolic analogue of the wave equation, and its solutions are smoother and more dissipative than wave solutions. This dissipation is a direct consequence of the second law of thermodynamics: heat flows irreversibly from hot to cold, and the temperature field evolves toward equilibrium.

Common misconceptions. (1) Heat is not a substance that flows; it is energy in transit due to a temperature difference. (2) Temperature is not the same as heat: a small object at high temperature may contain less total heat energy than a large object at low temperature. (3) Thermal conductivity \(k\) is not the same as thermal diffusivity \(\alpha\): \(k\) measures how well a material conducts heat, while \(\alpha\) measures how quickly it responds to temperature changes. (4) The heat equation is not valid in vacuum: it requires a material medium to carry heat. In vacuum, heat transfer is by radiation, not conduction.

Worked examples

Example 1. A copper slab of thickness \(L = 0.02\ \mathrm{m}\), area \(A = 0.5\ \mathrm{m^2}\), has its two faces maintained at \(T_{\text{hot}} = 350\ \mathrm{K}\) and \(T_{\text{cold}} = 300\ \mathrm{K}\). The thermal conductivity of copper is \(k = 400\ \mathrm{W/(m\,K)}\). Find the heat current \(H\) through the slab in steady state.

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

Steady-state 1D heat current through a slab. A

\[ H = 400 \times 0.5 \times \frac{350-300}{0.02} \]

Substitute the given values. A

\[ H = 200 \times \frac{50}{0.02} = 200 \times 2500 = 500{,}000\ \mathrm{W} \]

Arithmetic: \(50/0.02 = 2500\), and \(200 \times 2500 = 500{,}000\). A

\[ H = 5.0 \times 10^{5}\ \mathrm{W} \]

Reading. The heat current is 500 kW, which is large due to the high thermal conductivity of copper and the large area. The temperature gradient is \(2500\ \mathrm{K/m}\), and the flux density is \(q = k\,dT/dx = 400 \times 2500 = 10^{6}\ \mathrm{W/m^2}\).

Units check. \([kA\Delta T/L] = \mathrm{(W/(m\,K)) \cdot m^2 \cdot K / m = W}\). Consistent.

Example 2. A cylindrical pipe of length \(L = 1.0\ \mathrm{m}\), inner radius \(r_1 = 0.05\ \mathrm{m}\), and outer radius \(r_2 = 0.06\ \mathrm{m}\) is made of glass (\(k = 0.8\ \mathrm{W/(m\,K)}\)). The inner surface is at \(T_1 = 370\ \mathrm{K}\), and the outer surface is at \(T_2 = 300\ \mathrm{K}\). Find the heat current \(H\) through the pipe wall in steady state.

\[ H = 2\pi k L \frac{T_1 - T_2}{\ln(r_2/r_1)} \]

Steady-state heat current through a cylindrical shell. B

\[ H = 2\pi (0.8)(1.0) \frac{370-300}{\ln(0.06/0.05)} \]

Substitute the given values. A

\[ \ln(0.06/0.05) = \ln(1.2) \approx 0.1823 \]

Compute the logarithm. A

\[ H = 2\pi (0.8) \frac{70}{0.1823} \approx 5.027 \times \frac{70}{0.1823} \approx 5.027 \times 384.0 \approx 1930\ \mathrm{W} \]

Arithmetic: \(2\pi \times 0.8 \approx 5.027\), \(70/0.1823 \approx 384.0\), and \(5.027 \times 384.0 \approx 1930\). A

\[ H \approx 1.93 \times 10^{3}\ \mathrm{W} \]

Reading. The heat current is 1.93 kW, which is much smaller than in Example 1 due to the low thermal conductivity of glass and the smaller area. The logarithmic dependence on the radius ratio reflects the cylindrical geometry.

Units check. \([kL\Delta T/\ln(r_2/r_1)] = \mathrm{(W/(m\,K)) \cdot m \cdot K / 1 = W}\). Consistent.

Problems

A wall of thickness \(L = 0.10\ \mathrm{m}\) and area \(A = 2.0\ \mathrm{m^2}\) is made of material with \(k = 0.5\ \mathrm{W/(m\,K)}\). The inner surface is at \(20^\circ\mathrm{C}\), and the outer surface is at \(0^\circ\mathrm{C}\). Find the heat current \(H\) in steady state.
Solution
\(H = kA\Delta T/L = 0.5 \times 2.0 \times (20-0)/0.10 = 1.0 \times 200 = 200\ \mathrm{W}\).
A composite wall consists of two slabs in series: slab 1 has \(L_1 = 0.05\ \mathrm{m}\), \(k_1 = 1.0\ \mathrm{W/(m\,K)}\), and slab 2 has \(L_2 = 0.10\ \mathrm{m}\), \(k_2 = 0.5\ \mathrm{W/(m\,K)}\). The area is \(A = 1.0\ \mathrm{m^2}\), and the temperature difference across the composite wall is \(\Delta T = 30\ \mathrm{K}\). Find the heat current \(H\).
Solution
Thermal resistances: \(R_1 = L_1/(k_1 A) = 0.05/(1.0 \times 1.0) = 0.05\ \mathrm{K/W}\), \(R_2 = L_2/(k_2 A) = 0.10/(0.5 \times 1.0) = 0.20\ \mathrm{K/W}\). Total resistance: \(R_{\text{th}} = R_1 + R_2 = 0.25\ \mathrm{K/W}\). Heat current: \(H = \Delta T / R_{\text{th}} = 30/0.25 = 120\ \mathrm{W}\).
A copper rod of length \(L = 0.5\ \mathrm{m}\), cross-sectional area \(A = 1.0 \times 10^{-4}\ \mathrm{m^2}\), has one end at \(350\ \mathrm{K}\) and the other at \(300\ \mathrm{K}\). The thermal conductivity of copper is \(k = 400\ \mathrm{W/(m\,K)}\). Find the heat flux density \(q\) and the heat current \(H\).
Solution
Heat flux density: \(q = k\Delta T/L = 400 \times (350-300)/0.5 = 400 \times 100 = 40{,}000\ \mathrm{W/m^2}\). Heat current: \(H = qA = 40{,}000 \times 1.0 \times 10^{-4} = 4.0\ \mathrm{W}\).
A spherical shell of inner radius \(r_1 = 0.10\ \mathrm{m}\), outer radius \(r_2 = 0.12\ \mathrm{m}\), and thermal conductivity \(k = 0.8\ \mathrm{W/(m\,K)}\) has its inner surface at \(T_1 = 370\ \mathrm{K}\) and outer surface at \(T_2 = 300\ \mathrm{K}\). Find the heat current \(H\) in steady state. (Hint: For a spherical shell, \(H = 4\pi k (T_1 - T_2) / (1/r_1 - 1/r_2)\).)
Solution
\(1/r_1 - 1/r_2 = 1/0.10 - 1/0.12 = 10 - 8.333 = 1.667\ \mathrm{m^{-1}}\). Heat current: \(H = 4\pi (0.8) (70) / 1.667 \approx 10.053 \times 70 / 1.667 \approx 703.7 / 1.667 \approx 422\ \mathrm{W}\).
A material has thermal diffusivity \(\alpha = 1.0 \times 10^{-5}\ \mathrm{m^2/s}\). Estimate the time \(t\) required for a temperature disturbance to spread a distance \(L = 0.01\ \mathrm{m}\) (1 cm). Use the scaling \(L \sim \sqrt{\alpha t}\).
Solution
\(L^2 \sim \alpha t \Rightarrow t \sim L^2/\alpha = (0.01)^2 / (1.0 \times 10^{-5}) = 1.0 \times 10^{-4} / 1.0 \times 10^{-5} = 10\ \mathrm{s}\).