= Thermal boundary resistance
{title2=$r_B=\Delta T/q$}
{wiki=Interfacial_thermal_resistance}
<Thermal boundary resistance> describes a <temperature> drop between the two sides of a material interface carrying <heat flux>. Efficient <thermal conduction> within each material does not guarantee efficient <heat transfer> between them. The solid–<helium> case is <Kapitza resistance>; the broader concept also includes solid–solid interfaces. <Thermal boundary resistance by E. T. Swartz and Robert Otto Pohl>[Swartz and Pohl (1989), abstract: helium–solid and solid–solid interfaces]{ref}
Let $T_s$ and $T_l$ be the <temperatures> extrapolated to the interface from its solid and liquid sides, each treated as internally in <thermal equilibrium>. For uniform normal <heat flux> $q>0$ through <area> $A$, define the finite-drop, area-specific <thermal resistance> and the resistance of the whole interface by
$$
\Delta T=T_s-T_l,\qquad \dot Q=Aq,\qquad
r_B=\frac{\Delta T}{q}\quad[\mathrm{K\,m^2\,W^{-1}}],\qquad
\mathcal R_B=\frac{\Delta T}{\dot Q}=\frac{r_B}{A}\quad[\mathrm{K\,W^{-1}}].
$$
The macroscopic jump represents an unresolved interfacial region; it is not a claim that one point has two equilibrium <temperatures>. At nonzero <heat flux>, the two sides are outside mutual <thermal equilibrium>. A measured difference between distant <thermometers> also contains bulk <temperature gradients>, which must be measured or bounded before assigning all of it to the interface.
An <energy> balance explains why a <temperature> jump can coexist with continuous <heat flux>. For a stationary interfacial region with no local <work> or <heat> source,
$$
\frac{dU_{\rm int}}{dt}=Aq_{\rm in}-Aq_{\rm out}.
$$
The <first law of thermodynamics> therefore requires $q_{\rm in}=q_{\rm out}=q$ in a <steady state>. It does not require $T_s=T_l$. For a heated sample with electrical <power> $P=VI$, an independently determined outward parasitic loss $P_{\rm leak}$ and no other inputs,
$$
\frac{dU_s}{dt}=P-P_{\rm leak}-Aq,\qquad
q=\frac{P-P_{\rm leak}}{A}\quad\text{in a steady state}.
$$
Here $V$ is the heater's <electric potential difference> and $I$ its <electric current>. Subtracting a lead loss means subtracting <power> before dividing by <area>; $q$ is not the total heating <power>. Unknown losses and stored <internal energy> prevent that simple steady inference.
For a one-dimensional comparison, put an interface between two slabs with the same <area> $A$, positive constant <thermal conductivities> $k_s,k_l$ and <lengths> $L_s,L_l$. Neglect side losses. Integrating <Fourier's law> across each slab gives
$$
T_H-T_s=q\frac{L_s}{k_s},\qquad
T_l-T_C=q\frac{L_l}{k_l},\qquad
T_H-T_C=q\left(\frac{L_s}{k_s}+\frac{L_l}{k_l}\right)+(T_s-T_l).
$$
Thus, for a linear interface with constant $r_B$, the series result is
$$
q=\frac{T_H-T_C}{L_s/k_s+r_B+L_l/k_l},\qquad
\mathcal R_{\rm total}=\frac1A\left(\frac{L_s}{k_s}+r_B+\frac{L_l}{k_l}\right).
$$
These slab formulas are a chosen <thermal conduction> model, not a bulk transport law for moving liquid <helium>. If the interface law is nonlinear, substitute its own relation between $q$ and $T_s-T_l$ into the preceding balance; a single constant resistance cannot describe every load. Increasing bulk <thermal conductivity> has diminishing benefit when the interfacial drop dominates. Increasing usable contact <area> reduces the whole-interface <thermal resistance> if the local interface law and the other conditions remain fixed.
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