Question

A piece of wood 350 mm × 350 mm and 15 mm thick conducts heat through its thickness under steady state conditions. The rate of heat flow is measured to be 14.0 watts when the temperature difference is 28 C°. Determine the coefficient of thermal conductivity for this wood

Answer 1

The coefficient of thermal conductivity (k) is related to the rate of heat flow (Q), the cross-sectional area (A), the length (L), the temperature difference (ΔT), and the thermal resistance (Rth) by the following equation:

k = Q / (A * ΔT * L) = Rth * (A * ΔT)

Reorganizing this equation gives:

Rth = k / (A * ΔT)

The given information in the problem is:

Rate of heat flow (Q) = 14.0 watts

Thermal resistance (Rth) = (350 mm × 350 mm × 15 mm) / (14.0 watts) = 31.5 mm⁴/C

Temperature difference (ΔT) = 28°C

Substituting these values into the equation, we have:

k = Q / (A * ΔT) = 14.0 W / (0.35 m² * 28°C) = 1.94 W/mK

So the coefficient of thermal conductivity (k) for this wood is approximately 1.94 W/mK.