JAMB - Physics (2024 - No. 23)

288KJ is conducted across two opposite faces of a 3m cube of temperature gradient 90ºCm\(^{-1}\) in 7200s. Calculate the thermal conductivity.

4.9 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)

5.0 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)

5.2 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)

5.1 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)

k = \(\frac{Q.d}{A. ΔT .t}\)

d = thickness = 3m

A = 9m\(^2\)

Q = 288kJ = 288000J, t = 7200s, ΔT = 90ºC/m → 90K/m x 3 = 270K

k = \(\frac{288000 \times 3}{9 \times 270 \times 7200}\)

k = 4.93 x 10\(^{-2}\)Wm\(^{-1}\)k\(^{-1}\)