Radiation And Heat Practice Test

A rod with k = 200 W/mK, cross-sectional area 0.01 m^2, L = 0.02 m, ΔT = 40 K conducts heat. What is Qdot?

400 W 40000 W 400000 W 4000 W

Correct Answer:

4000 W

Explanation:

Heat conduction rate follows Fourier’s law: Q̇ = k A ΔT / L. It says how fast heat flows through a material depends on how well it conducts (k), the area available for transfer (A), the temperature difference driving the flow (ΔT), and the distance the heat must travel (L). Plug in the numbers: k = 200 W/mK, A = 0.01 m^2, ΔT = 40 K, L = 0.02 m. Compute step by step: kA = 200 × 0.01 = 2 W/K. Then kAΔT = 2 × 40 = 80 W. Finally, Q̇ = 80 / 0.02 = 4000 W. So the heat conduction rate is 4000 W (4 kW).

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