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

We can use the formula for steady state heat transfer through a flat wall with constant thermal conductivity:

Q = kA (ΔT / L)

where Q is the rate of heat transfer, k is the thermal conductivity, A is the cross-sectional area, ΔT is the temperature difference, and L is the thickness.

Plugging in the given values:

14.0 W = k (350 mm x 350 mm) / (15 mm) (28 C°)

Simplifying:

k = 0.061 W/(m • C°)

Note that we converted the dimensions from mm to m to match the unit of thermal conductivity.

Therefore, the correct answer is D) 0.061 W/(m • C°).

Under Steady State Conditions,a Piece of Wood 350 Mm by 350