A concrete wall of a cold storage room measures 3.0 m high,5.0 m wide,and 20 cm thick.The room temperature is maintained at -10°C and the outside temperature is 20°C The inside wall is to be covered by a layer of wood in order to reduce the rate of heat loss through the wall BY 90 percent.The thermal conductivities of concrete and wood are 0.80 W/m • K and 0.040 W/m • K,respectively.Under steady state conditions,the thickness of the layer of wood required is closest to

Question

Quizplus · Accepted Answer

First, calculate the rate of heat loss through the concrete wall without the layer of wood:
Q = (k x A x ΔT)/t
where
k = thermal conductivity
A = surface area
ΔT = temperature difference
t = thickness

Q(concrete) = (0.80 W/m • K x 3.0 m x 5.0 m x (20°C - (-10°C)))/0.20 m = 3600 W

To reduce the rate of heat loss by 90%, we need to find the new rate of heat loss with the layer of wood and solve for the thickness of the wood required:
Q(wood) = 0.1 x Q(concrete) = 360 W

Q(wood) = (0.040 W/m • K x 3.0 m x 5.0 m x (20°C - (-10°C)))/t
t = (0.040 W/m • K x 3.0 m x 5.0 m x (20°C - (-10°C)))/360 W = 0.90 m

Convert to millimeters:
t = 900 mm

Therefore, the thickness of the layer of wood required is closest to 90 mm (choice D).

A Concrete Wall of a Cold Storage Room Measures 3