14.30 - Conduction Part A Calculate the rate of heat conduction through house wa

Question

14.30 - Conduction Part A Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The surface area of the walls is 330 m2 and their inside surface is at 18.0°C, while their outside surface is at 5.00°c. Submit Answer Tries 0/5 Part B How many 1.0-kw room heaters would be needed to (approximately) balance the heat transfer due to conduction? Round your answer up to the nearest integer. Submit Answer Tries 0/5

Explanation / Answer

a)

Use eqn,

P = Q/t = [kA(T2-T1)]/d

Where

k=Thermal conductivity = 2*0.03 = 0.06 W.m-2.k-1

A= Surface area = 330 m2

d=width = 0.13m

(T2-T1) = temperature difference = 18 – 5 = 13

Plugging values,

P = Q/t = [kA(T2-T1)]/d

P = [0.06*330*13]/0.13

P = 1980 W

b)

No of heaters required = P/1000 = 1980/1000 = 2.0 heaters

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