Under steady state conditions, a piece of wood 350 mm by 350 mm and 15 mm thick conducts heat through its thickness and loses no appreciable heat through its well-insulated sides. The rate of heat flow is measured to be 14.0 W when the temperature difference across its thickness is 28 C°. Determine the thermal conductivity of this wood.

Question

Quizplus · Accepted Answer

First, we need to use the formula for steady-state heat transfer through a flat wall of thickness L:

Q = kA(dT/dx)

where Q is the rate of heat flow, k is the thermal conductivity, A is the cross-sectional area of the wall, and dT/dx is the temperature gradient.

We are given Q = 14.0 W, A = (0.350 m)^2 = 0.1225 m^2, L = 0.015 m, and dT/dx = 28 C° / L = 1867 C°/m.

Plugging in these values and solving for k, we get:

k = Q/(A*dT/dx) = 0.061 W/(m∙C°)

Therefore, the correct answer is D.

Under Steady State Conditions, a Piece of Wood 350 Mm