Question on: JAMB Physics - 2024
288 kJ is conducted across two opposite faces of a 3 m cube with a temperature gradient of 90 °C/m in 7200 s. Calculate the thermal conductivity.
A
4.9 x 10-2 W/m·K
B
5.0 x 10-2 W/m·K
C
5.2 x 10-2 W/m·K
D
5.1 x 10-2 W/m·K
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Correct Option: A
Step-by-step solution:
- Convert energy to SI units: Q = 288 kJ = 288,000 J
- Cube thickness x = 3 m, time t = 7200 s, temperature gradient dT/dx = 90 °C/m
- Area of cube face: A = 3 x 3 = 9 m²
- Use heat conduction formula: Q = (k x A x dT/dx x t) / x
- Rearrange for k: k = (Q x x) / (A x dT/dx x t)
- Plug in values: k = (288000 x 3) / (9 x 90 x 7200) = 864000 / 5832000 ≈ 0.05 or 5.0 x 10-2 W/m·K
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