A copper rod has k = 200 W/m·K, cross-sectional area A = 0.0005 m^2, length L = 0.30 m, and ΔT = 50 K. What is the heat transfer rate Q/t by conduction?

A copper rod has k = 200 W/m·K, cross-sectional area A = 0.0005 m^2, length L = 0.30 m, and ΔT = 50 K. What is the heat transfer rate Q/t by conduction?

• A. 5 W
• B. 10 W
• C. 16.7 W
• D. 20 W

Answer: (C)

The main idea is Fourier’s law for conduction: the heat transfer rate through a solid is proportional to the material’s conductivity, the cross-sectional area, and the temperature difference, and inversely proportional to the length. So Q/t = k A ΔT / L. Plug in the values: kA = 200 × 0.0005 = 0.1 W/K. Multiply by ΔT: 0.1 × 50 = 5 W. Then divide by L: 5 / 0.30 ≈ 16.7 W. So the heat transfer rate is about 16.7 watts from hot to cold. The result makes sense: higher conductivity, larger area, and bigger temperature difference push more heat, while a longer rod reduces the rate. If you don’t divide by the length, you’d get 5 W; if you mis-handle the area or temperature difference, you’d get other incorrect values.

The main idea is Fourier’s law for conduction: the heat transfer rate through a solid is proportional to the material’s conductivity, the cross-sectional area, and the temperature difference, and inversely proportional to the length. So Q/t = k A ΔT / L.

Plug in the values: kA = 200 × 0.0005 = 0.1 W/K. Multiply by ΔT: 0.1 × 50 = 5 W. Then divide by L: 5 / 0.30 ≈ 16.7 W.

So the heat transfer rate is about 16.7 watts from hot to cold. The result makes sense: higher conductivity, larger area, and bigger temperature difference push more heat, while a longer rod reduces the rate. If you don’t divide by the length, you’d get 5 W; if you mis-handle the area or temperature difference, you’d get other incorrect values.