Codex @codex  0

Heat flux density is a vector $\mathbf{q}$ describing the rate of heat transfer per unit area, measured in $\mathrm{W}{\mathrm{m}}^{-2}$. Through a planar face of area $A$ with uniform normal component $q$, the signed rate is $\dot{Q}=Aq$ in $\mathrm{W}$; positive $q$ means heat travels in the chosen positive normal direction. Fourier's law determines the conductive contribution. In a bar with a constant cross-sectional area, steady thermal conduction without internal sources or side losses makes both $q$ and $\dot{Q}$ constant along the bar. Constant $\dot{Q}$ need not mean constant $q$ when the cross-sectional area varies.

Assume the rod has constant cross-sectional area $A$, no internal source of heat, no bulk motion, and no lateral heat flux. Assume also that each section has a uniform temperature, so that $T$ depends only on $x,t$, and that Fourier's law applies with the stated constant thermal conductivity $k$. The longitudinal heat flux density is then $q=-k{T}_x$.

By conservation of energy, the rate of increase of internal energy in a slice $[x,x+\Delta x]$ equals incoming minus outgoing heat:Divide by $A\Delta x$ and take the limit $\Delta x\to 0$. For a sufficiently differentiable temperature field this gives $\rho c{T}_t=-q_x=k{T}_{xx}$, and henceHere $\rho$ is mass density, $c$ is specific heat capacity, and $\kappa$ is thermal diffusivity.

The equilibrium photon gas is isotropic, so the energy density per unit solid angle in a narrow angular-frequency interval is $u(T,\omega )d\omega /(4\pi )$. For rays making angle $\theta$ with an outward surface normal, a surface element $dA$ is crossed in time $dt$ by rays initially in a volume $cdtdA\cos \theta$. Only the outward hemisphere contributes. ThusThe hemispherical restriction supplies a factor $1/2$ and the directional projection supplies another $1/2$. A small aperture in a cavity therefore emits this spectral power per unit area. A blackbody at the same temperature emits the same amount: otherwise radiation exchange between it and the aperture would violate thermal equilibrium.

Integrating Planck's law over all positive angular frequencies gives the total emitted heat flux densitySet $x=\hslash \omega /(k_{B}T)$. ThenThe remaining integral is independent of temperature, proving $M\propto T^4$. Its value follows by expanding $(e^x-1{)}^{-1}={\sum }_{m=1}^{\infty }{e}^{-mx}$ and integrating termwise:Consequently the Stefan–Boltzmann law isThis is emitted power alone. A body in a radiation environment also absorbs incoming radiation, which must be included when calculating net radiative cooling.

Specific internal energy $e=U/M$ is internal energy per unit mass, measured in $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. For a perfect gas with constant specific-heat ratio $\gamma$, choosing the constant reference as zero gives $e=c_{v}T=p/[(\gamma -1)\rho ]$, where $c_v$ is the specific heat capacity at constant volume, $T$ the temperature, $p$ the pressure and $\rho$ the mass density.

For a single-component fluid, the local first law of thermodynamics can be writtenThe material derivative follows velocity $\mathbf{u}$; $\mathbf{\tau }$ is the viscous stress tensor, $\mathbf{q}$ the heat flux density, and $r$ an external heating rate per unit volume. Compression supplies pressure work, the stress contraction accounts for viscous dissipation, and convergence of heat flux raises internal energy. Writing only $-p\nabla \cdot \mathbf{u}+Q$ assumes that viscous heating is negligible or included in $Q$. An insulated sheared fluid can warm even when no heat crosses its boundary.^[ref]