Question

The temperatures on the inside and outside surface of a wall are 20°C and 5°C respectively. The wall is 8 m wide, 4 m high, and 0.2 m thick.

The thermal conductivity is 0.3 W/m.K.
1
The heat flux through the wall is negative
W/m2. Use two significant figures in your answer.

Answer 1

Heat Flux = - 22.5 W/m²

Step-by-step explanation:

We can use Fourier's Law of Heat Conduction:

q = -kAΔT/t

q/A = -KΔT/t

where,

q/A = Heat Flux = ?

K = Thermal Conductivity = 0.3 W/m.K

ΔT = Change in Temperature = 20°C - 5°C = 15K

t = Thickness = 0.2 m

Therefore,

Heat Flux = -(0.3 W/m.k)(15 k)/(0.2 m)

Heat Flux = - 22.5 W/m²

Negative sign shows the heat flowing out of the system.