Thermal resistance
= Thermal resistance
{title2=$R_{\rm th}=\Delta T/\dot Q$}
<Thermal resistance> measures the <temperature> drop required to drive a given rate of <heat transfer>. For the steady one-dimensional bar model in <thermal conduction>,
$$
R_{\rm th}=\frac{T_h-T_c}{\dot Q}=\frac{L}{kA}
\quad[\mathrm{K\,W^{-1}}],\qquad
r=AR_{\rm th}=\frac{L}{k}
\quad[\mathrm{m^2\,K\,W^{-1}}],\qquad
q=\frac{T_h-T_c}{r}.
$$
Here $r$ relates the <temperature> drop to <heat flux>, while $R_{\rm th}$ relates it to the rate of <heat transfer>. The single bar above has $r=0.050\,\mathrm{m^2\,K\,W^{-1}}$ and $R_{\rm th}=5\,\mathrm{K\,W^{-1}}$, so its $20\,\mathrm K$ drop drives $4\,\mathrm W$.
To see why series <thermal resistances> add, apply <Fourier's law> in each layer to obtain $\Delta T_i=\dot Q R_i$. With no accumulation or side losses the rate of <heat transfer> is the same through both, so adding the drops gives $T_h-T_c=\dot Q(R_1+R_2)$. This lets us size layered walls without solving a new <differential equation> for every layer. The structure matches <Ohm's law>: <temperature> difference plays the role of <electric potential difference>, the rate of <heat transfer> plays the role of <electric current>, and <thermal resistance> plays the role of <electrical resistance>. https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node118.html[E. M. Greitzer, Z. S. Spakovszky and I. A. Waitz, MIT Thermodynamics and Propulsion, section 16.4: series thermal resistances and the electrical analogy]{ref}