One wall of a house consists of 0.019m thick plywood backed by 0.076 m thick ins
ID: 1701825 • Letter: O
Question
One wall of a house consists of 0.019m thick plywood backed by 0.076 m thick insulation. The temperature at the inside surface (near the insulation) is 25.0oC, while the temperature on the outside surface (near the plywood) is 4.0oC, both being constant. The thermal conductivities of the insulation and plywood are, respectively, 0.030 and 0.080 J/(smoC), and the area of the wall is 35 m2. Find the heat conducted through the wall in one hour (a) with the insulation and (b) without the insulation.Explanation / Answer
The thickness of the plywood L1 = 0.019m
The thickness of the insulation L2 = 0.076m
The temperature difference T = 25 -4 = 21C
The thnermal conductivities k1 = 0.08 J/smC
and the thermal conductivity of insulation k2 = 0.03 J/smC
The area of the wall A = 35m^2
Let the thickess of the wall be L
the thermal conductivity of the wall be k
Therefore the heat conducted in 1hr
P = AT /[(L1/k1) + (L2/k2) + (L3/k3) ]
Therefore the heat in 1hour is
Q = (35)(21)(3600s) / [(0.019/0.08)+(0.076/0.03) + (L/k)]
Without Insulation is
Q/t = AT /[(L1/k1) + (L3/k3) ]
Therefore the energy
Q = (35)(21)(3600s) / [(0.019/0.08) + (L/k)]
if we know the thickness of the wall we can determin the heat energy
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