Conduction of heat: 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. A) 9.2 × 10^-4 W/(m ∙ °C) B) 270 W/(m ∙ °C) C) 16 W/(m ∙ °C) D) 0.061 W/(m ∙ °C) E) 33 W/(m ∙ °C)

Question

Quizplus · Accepted Answer

We can use the formula for heat flow through a flat wall with steady state conditions:

Q = kA (T2-T1)/d

Where:

Q = rate of heat flow (given as 14.0 W) 
k = thermal conductivity (unknown, to be solved for) 
A = area of wood (350 mm x 350 mm = 0.1225 m^2) 
T2-T1 = temperature difference across thickness (given as 28°C) 
d = thickness of wood (15 mm = 0.015 m)

Plugging in the given values and solving for k, we get:

k = Qd/(AT2-T1) 
k = (14.0 W)(0.015 m)/[(0.1225 m^2)(28°C)] 
k = 0.061 W/(m ∙ °C)

Therefore, the thermal conductivity of the wood is 0.061 W/(m ∙ °C), which corresponds to choice D.

Conduction of Heat: Under Steady State Conditions, a Piece of Wood