Question

Calculate the rate of heat conduction through house walls that are 14.7 cm thick and that have an average thermal conductivity of 0.057 J/(s•m•ºC). Assume there are no windows or doors. The surface area of the walls is 136 m2 and their inside surface is at 19.9ºC , while their outside surface is at 4.9ºC .

Answer 1

Rate of heart conduction P = ds/dt = KA (T_1 - T_2)/L K = 0.0575/s - m - - c degree, A = 136 m^2 T_1 = 19.9 degree c, T_2 = 4.9 degree c L = 14.7 cm = 0.147 m P = 0.057 times 136 [19.9 - 4.9]/0.147 P = 791 J/s

Answer 2

SOLUTION :

Rate of Heat transfer through wall

= k * (A / d) * ∆ T 

where,

k = heat transfer constant though thick surface = 0.057 J/(s*m*ºC)

A = surface area  = 136 m^2 

d = thickness = 14.7 cm = 0.147 m 

∆T = Ti - T0 = 19.9 - 4.9 = 15ºC 

So, Rate of Heat transfer = 0.057 * (136 / 0.147) * 15 = 791.02 J / s (ANSWER).