A concrete wall is $15.0 \mathrm{~cm}$ thick and has an area of $10.0 \mathrm{~m}^{2}$. A layer of wood that is $2.50 \mathrm{~cm}$ thick is placed over the wall to reduce the loss of heat by thermal conduction. The thermal conductivity of concrete is 1.70 $\mathrm{W} / \mathrm{m}^{\circ} \mathrm{C}$ and the thermal conductivity of wood is $0.0400 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}$. What is the effective thermal conductivity of the wood-on-concrete system?

Question

Quizplus · Accepted Answer

The effective thermal conductivity $ k_{\text{eff}} $ for materials in series is given by the formula: $\frac{1}{k_{\text{eff}}} = \frac{d_1}{k_1} + \frac{d_2}{k_2}$, where $d_1$ and $d_2$ are the thicknesses, and $k_1$ and $k_2$ are the thermal conductivities of the materials. Substituting the given values, the effective thermal conductivity is approximately $0.245 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}$.

A Concrete Wall Is 15.0 cm15.0 \mathrm{~cm}15.0 cm Thick and Has an Area Of

A Concrete Wall Is $15.0 \mathrm{~cm}$ Thick and Has an Area Of