• id: JME-10-N11
• title: Thermal Conduction and Conductivity
• tags: conduction, conductivity, fourier

In steady-state conduction through a slab: $dQ/dt = kA(T_1 - T_2)/L$. The thermal conductivity $k$ (W m$^{-1}$ K$^{-1}$) measures how well a material conducts heat. Metals are good conductors (silver 429, copper 401); insulators have low $k$ (wood 0.12, air 0.024). The thermal resistance analogy is powerful: $R = L/(kA)$, then $dQ/dt = \Delta T/R$, exactly like $I = V/R$ in circuits. For composite walls in series: same heat flow, temperatures drop add. For slabs in parallel: same temperature difference, heat flows add.