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Thermal conductivity is the material response coefficient in Fourier's law. For an isotropic material it is a scalar $k$, with units $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$; larger positive $k$ means a larger conductive heat flux for the same temperature gradient. It can depend on temperature and composition. Direction-dependent response in an anisotropic material is described by a thermal conductivity tensor. The two worked examples use stipulated positive constants, rather than measured values for named materials.

Thermal conductivity describes material response, whereas thermal resistance also includes geometry: for the steady bar, $R_{\mathrm{th}}=L/(kA)$. Doubling $k$ at fixed length, cross-sectional area and end temperatures halves thermal resistance and doubles the rate of heat transfer. Doubling length changes thermal resistance without changing the material's thermal conductivity. This distinction lets a designer improve cooling either by choosing a higher thermal conductivity or by shortening and widening the path. It also explains why the less conductive layer in thermal resistance takes the greater temperature drop.

Thermal diffusivity answers a different question: how rapidly a temperature disturbance spreads. Its usual model $\kappa =k/(\rho c_p)$ also involves mass density $\rho$ and specific heat capacity $c_p$, so equal thermal conductivities need not imply equal rates of warming.

Here thermal diffusion means the spreading of temperature by thermal conduction. In a stationary homogeneous material with constant thermal conductivity $k$, mass density $\rho$ and specific heat capacity $c$, energy balance and Fourier's law give the heat equation ${\partial }_{t}T=\alpha {\nabla }^{2}T$, where $\alpha =k/(\rho c)$ is the thermal diffusivity.

A disturbance spreads over a length of order $\sqrt{\alpha t}$, so equilibration across thickness $L$ takes a time of order $L^2/\alpha$. This is the heat-transport sense of the term; temperature-gradient-driven separation of components in a mixture is a distinct transport effect.

Thermal diffusivity measures how rapidly thermal conduction smooths a temperature difference inside a material. Thermal conductivity measures the heat flux driven by a specified temperature gradient; thermal diffusivity also accounts for the heat capacity that must be warmed. A material can carry substantial heat yet change temperature slowly because it stores substantial internal energy per kelvin.

Write $\alpha =\kappa$ for thermal diffusivity, reserving $k$ for thermal conductivity. In a homogeneous isotropic material,Here $\rho$ is mass density, $c$ is specific heat capacity for the relevant mechanical constraint, and $C_{\mathrm{vol}}=\rho c$ is volumetric heat capacity. The familiar $k/(\rho c_p)$ uses specific heat capacity at constant pressure. For ordinary solids with negligible thermal expansion, the difference between the two constrained specific heat capacities is often neglected. A rigid fixed-volume model instead uses specific heat capacity at constant volume; a compressible moving fluid requires the accompanying work and transport terms. Thus the constraint is part of the approximation, not a choice made solely from the units. ^[ref]

Assume a stationary material in the continuum approximation, local thermal equilibration on times shorter than those resolved, no internal heating, no phase transition, and negligible convection and mechanical work. Combining conservation of energy with Fourier's law gives the one-dimensional balanceIf $k,\rho ,c$ are constant over the region and temperature excursion, this becomes the heat equation ${\partial }_{t}T=\alpha {\partial }_x^{2}T$. If $k$ varies, retain the differentiated product; replacing it with $k{\partial }_x^{2}T$ loses a term. Anisotropic materials need a thermal conductivity tensor. The scalar constant-$\alpha$ examples below deliberately exclude these complications. ^[ref]

For an original comparison, stipulate two model materials with equal thermal conductivity $k=2\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$ and equal mass density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, but specific heat capacities $c_A=1000$ and $c_B=4000\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. ThenAt the same temperature gradient, both have the same instantaneous heat flux by Fourier's law. At the same initial temperature profile and boundary conditions, material B takes four times as long to smooth a given spatial pattern. With fixed end temperatures, identical geometry and no sources, both eventually have the same steady profile and heat flux. The larger heat capacity changes the transient, not this steady state. These are stipulated parameters, not measurements of named substances.

Volumetric heat capacity is heat capacity per unit volume, with units $\mathrm{J}{\mathrm{m}}^{-3}{\mathrm{K}}^{-1}$. For homogeneous material of mass density $\rho =m/V$, it equals $\rho c$, using specific heat capacity $c$ for the same constraint. It measures the heat required per small temperature increase for a chosen occupied volume. In a material model using specific heat capacity at constant pressure, thermal diffusivity is $\alpha =k/(\rho c_p)$, where $k$ is thermal conductivity.

Water calorimetry measures an energy transfer using the thermal response of water. In radiation metrology, it determines absorbed dose to water from a small measured temperature rise. The ideal conversion follows from $Q=m{c}_w\Delta T$ and $D=Q/m$:where $c_w$ is water's specific heat capacity. The sample mass cancels because dose is energy per unit mass.^[ref]

For an original scale estimate, take $c_w=4180\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ as an illustrative near-room-temperature value. An absorbed dose of $1\mathrm{Gy}=1\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$ would then give only $1/4180\mathrm{K}$, about $0.239\mathrm{mK}$. Detecting such a small change requires careful thermometry and separation from background thermal drift.

The physical conversion also has chemical assumptions. Heat defect describes the fraction of absorbed energy that does not appear as heat because of radiation-induced chemical change. Thermal conduction, convection and sensor effects are additional measurement questions, rather than alternate names for that chemical correction. This makes water calorimetry an example of an apparently direct energy measurement whose interpretation still needs a well-defined model.