The temperature difference between two sides of metal plate, \(3 \mathrm{~cm}\) thick is \(15^{\circ} \mathrm{C}\). Heat is transmitted through plate at the rate of \(900 \mathrm{kcal}\) per minute per \(\mathrm{m}^2\) at steady state. The thermal conductivity of metal is

\( \begin{aligned} & \frac{\mathrm{Q}}{\mathrm{t}}=\frac{\mathrm{kA} \Delta \theta}{\mathrm{d}} \\ & \therefore \mathrm{k}=\frac{\mathrm{Q}}{\mathrm{tA}} \cdot \frac{\mathrm{d}}{\Delta \theta} \\ & \frac{\mathrm{Q}}{\mathrm{tA}}=900 \mathrm{kcal} \text { per minute per } \mathrm{m}^2=\frac{900}{60}=15 \mathrm{kcal} / \mathrm{s} \cdot \mathrm{m}^2 \\ & \mathrm{~d}=3 \mathrm{~cm}=3 \times 10^{-2} \mathrm{~m}, \Delta \theta=15^{\circ} \mathrm{C} \\ & \therefore \mathrm{k}=\frac{15 \times 3 \times 10^{-2}}{15}=3 \times 10^{-2} \mathrm{kcal} / \mathrm{s} \cdot \mathrm{m} .{ }^{\circ} \mathrm{C} \end{aligned} \)