1. Conservation and Fourier's law

A hot center and a cool boundary

A solid generates energy throughout its volume while its surface touches an ideal temperature-controlled bath. The bath holds the surface at 300 K. If the interior were also everywhere 300 K, there would be no temperature gradient and no conductive escape route for the generated energy. The interior must warm. The question is how much, and how quickly.

Our model describes a stationary solid, with no transport of matter and no mechanical work. Material properties are constant over the temperature range we investigate. These choices let us isolate conduction and storage.

By the end of this chapter, you can distinguish the quantities in the energy account, derive its local equation, and predict the sign of a heat flux.

Temperature is not the stored quantity

Temperature $T$ measures thermal state. Internal energy is what we account for. For a small temperature change in a fixed solid, write

e-e_{\mathrm{ref}}=c_v(T-T_{\mathrm{ref}}), \qquad c_v=\rho c.

Here $e$ is internal energy per volume, $\rho$ is density, and $c$ is the appropriate specific heat for the solid approximation. We use $c_v$ to mean volumetric heat capacity, not a mass-specific heat symbol. With constant $c_v$ , the energy rate is $\partial_t e=c_v\partial_t T$ . The arbitrary reference energy disappears on differentiation. This is why a storage expression proportional to absolute temperature gives the same balance as one written in temperature rise. For the distinction between energy transfer and temperature, see OpenStax, §1.4.

Symbol	Meaning	SI unit
$T$	Temperature	K
$c_v$	Volumetric heat capacity	J/(m³ K)
$\boldsymbol q$	Conductive energy flux	W/m²
$s$	Volumetric generation	W/m³
$k$	Thermal conductivity	W/(m K)

A flux is a rate per area. Multiplying a uniform normal flux by an area produces power in watts; multiplying power by time produces energy in joules.

Draw the account before writing the PDE

                      outward normal n
                            ↑
                    heat leaves: q · n
                  ┌────────────────────┐
       heat       │ stored energy      │       heat
       enters →   │ generation s       │   →   leaves
                  └────────────────────┘
                       fixed region Ω

Conceptual control-volume diagram. Arrows show energy transfer. The outward normal is defined by the region, regardless of whether energy enters or leaves.

For a fixed region $\Omega$ , the account is

\frac{\mathrm d}{\mathrm dt}\int_\Omega e\,\mathrm dV +\int_{\partial\Omega}\boldsymbol q\cdot\boldsymbol n\,\mathrm dA =\int_\Omega s\,\mathrm dV.

Positive outward flux is an energy loss. With insulated boundaries and positive generation, the boundary integral vanishes and stored energy increases. That simple thought experiment fixes the sign of the equation.

Apply the divergence theorem and require this account on each small region:

c_v\partial_t T+\nabla\cdot\boldsymbol q=s.

This equation still does not tell us how energy moves. Conservation requires a constitutive law, a relationship describing the material’s response.

Fourier’s law supplies the flux

For an isotropic conductor, the constitutive choice is

\boldsymbol q=-k\nabla T,\qquad k>0.

The gradient points toward increasing temperature; the minus sign makes flux point toward decreasing temperature. This is the local version of the slab law in OpenStax, §1.6. Combining the law with the energy account gives

c_v\partial_t T-\nabla\cdot(k\nabla T)=s.

For constant $k$ , define diffusivity $\alpha=k/c_v$ :

\partial_t T-\alpha\nabla^2T=\frac{s}{c_v}, \qquad [\alpha]=\mathrm{m^2/s}.

Conductivity determines flux for a given gradient. Diffusivity compares that transport with the energy needed to change temperature. Doubling $k$ and $c_v$ together preserves source-free diffusion time scales, but doubles the flux carried by any fixed temperature profile.

Three predictions without a solver

At steady state, storage vanishes. If $s=0$ and temperature is uniform, the balance is satisfied. If $s>0$ and every boundary is insulated, no steady state can satisfy the integrated account: zero outward power cannot equal positive generated power. If $s>0$ and the boundary is held cool, the center must become hotter so conduction can remove the power.

For a characteristic length $L$ , balancing $c_v\Delta T/t$ against $k\Delta T/L^2$ suggests a diffusion time $t_d=L^2/\alpha$ . Balancing conduction against generation suggests a temperature rise $\Delta T\sim sL^2/k$ . Geometry and boundary conditions determine the dimensionless factors. The next chapter calculates one exactly.

Exercises

A uniform insulated volume has constant $c_v$ and constant source $s$ . Derive $T(t)$ from initial temperature $T_0$ .
In a one-dimensional body, $T$ increases with $x$ . Give the sign of $q_x$ , and of outward flux at the left face.
Two solids have equal conductivity but one has twice the volumetric heat capacity. Which has the slower source-free diffusion time over equal lengths?
Check the dimensions of every term in the local heat equation.

Solution hints

Uniformity removes the gradient: $T=T_0+st/c_v$ .
$q_x<0$ . The left normal is $-1$ , so $q_n=-q_x>0$ : heat leaves on the left.
The larger-capacity solid has half the diffusivity and twice the time scale.
Storage, divergence of flux, and source all have units W/m³.

References

Samuel J. Ling, Jeff Sanny, and William Moebs, University Physics, Volume 2 (OpenStax, 2016), §1.4, Heat Transfer, Specific Heat, and Calorimetry and §1.6, Mechanisms of Heat Transfer.
Chris Schuh, Materials Processing, MIT 3.044 (2013), Lecture 2: Heat conduction.

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