Question

30. (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 120m2 and their inside surface is at 18.0ºC, while their outside surface is at 5.00ºC. (b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction?

Answer 1

(a) 960 W

(b) 960

Step-by-step explanation:

d = 13 cm = 0.13 m

K = 2 times thermal conductivity of glass wool

K = 2 x 0.040 = 0.080 W/mK

A = 120 m^2

T1 = 18°C

T2 = 5°C

(a) The rate of conduction is given by

`H = (KA\left ( T_(1)-T_(2) \right ))/(d)`

`H = (0.080* 120\left ( 18-5 \right ))/(0.13)`

H = 960 W

(b)

Number of heaters = total power / power of one heater

Number of heaters = 960 / 1 = 960