Question

Calculate the rate of heat conduction (in W) through house walls that are 14.5 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 145 m² and their inside surface is at 21.5°C, while their outside surface is at 5.00°C.

How many 1 kW room heaters would be needed to balance the heat transfer due to conduction? (Round your answer to the next whole integer.)

Answer 1

(a) The rate of heat conduction through house walls which will be given as :

using a formula, we have

Q / t = k A (T₂ - T₁) / d

where, k = average thermal conductivity twice that of glass wool = 2 (0.042 J/s.m ⁰C)

A = surface area of the walls = 145 m²

d = thickness = 0.145 m

then, we get

Q / t = 2 (0.042 J/s.m ⁰C) (145 m²) [(21.5 - 5) ⁰C] / (0.145m)

Q / t = 1386 W

Q / t = 1.386 x 10³ W

(b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction?

Two    (1-kW room heaters)