JAMB Physics 2024 Paper

Concept:Thermal conduction is calculated using the heat transfer equation for a steady temperature difference.Explanation:The formula for thermal conductivity is:$$k = \frac{Qd}{A\Delta T t}$$where $Q$ is the heat energy, $d$ is the thickness, $A$ is the cross-sectional area, $\Delta T$ is the temperature difference, and $t$ is the time.The cube has side length $3\mathrm{m}$, so the distance between the two opposite faces is:$$d = 3\mathrm{m}$$The area of each face is:$$A = 3 \times 3 = 9\mathrm{m}^{2}$$The temperature gradient is $90^{\circ}\mathrm{C\,m^{-1}}$, so the total temperature difference across the cube is:$$\Delta T = 90 \times 3 = 270\mathrm{K}$$The heat conducted is:$$Q = 288\mathrm{kJ} = 288000\mathrm{J}$$and the time is $t = 7200\mathrm{s}$.Substitute all values into the formula:$$k = \frac{288000 \times 3}{9 \times 270 \times 7200}$$$$k = 4.93 \times 10^{-2}\mathrm{W\,m^{-1}\,K^{-1}}$$This is approximately $4.9 \times 10^{-2}$.Answer:Option A: $4.9 \times 10^{-2}\mathrm{W\,m^{-1}\,K^{-1}}$