Question

Model the concrete slab as being surrounded on both sides (contact area 24 m2) with a 2.1-m-thick layer of air in contact with a surface that is 5.0 ∘C cooler than the concrete. At sunset, what is the rate at which the concrete loses thermal energy by conduction through the air layer?

Answer 1

Answer:

$\dot{Q} =1.5\ W$

Explanation:

Given that:

• area of concrete slab, $A=24\ m^2$

• thickness of the layer of air, $dx=2.1\ m$

• temperature difference between the air and the concrete, $dT=5\ K (\rm difference\ will\ be\ same\ in\ K\ and\ ^{\circ}C)$

We have:

• thermal conductivity of air at S.T.P., $k=26.24 \times 10^{-3}\ W.m^{-1}.K^{-1}$

Now, according to Fourier's Law of conduction:

$\dot{Q} =k.A.\frac{dT}{dx}$

putting the respective values:

$\dot{Q} =26.24 \times 10^{-3}\times 24\times \frac{5}{2.1}$

$\dot{Q} =1.5\ W$