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

`\dot{Q} =1.5\ W`

Step-by-step 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 * 10^(-3)\ W.m^(-1).K^(-1)`

Now, according to Fourier's Law of conduction:

`\dot{Q} =k.A.(dT)/(dx)`

putting the respective values:

`\dot{Q} =26.24 * 10^(-3)* 24* (5)/(2.1)`

`\dot{Q} =1.5\ W`