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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 :

Answer:

[tex]\dot{Q} =1.5\ W[/tex]

Explanation:

Given that:

  • area of concrete slab, [tex]A=24\ m^2[/tex]
  • thickness of the layer of air, [tex]dx=2.1\ m[/tex]
  • temperature difference between the air and the concrete, [tex]dT=5\ K (\rm difference\ will\ be\ same\ in\ K\ and\ ^{\circ}C)[/tex]

We have:

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

Now, according to Fourier's Law of conduction:

[tex]\dot{Q} =k.A.\frac{dT}{dx}[/tex]

putting the respective values:

[tex]\dot{Q} =26.24 \times 10^{-3}\times 24\times \frac{5}{2.1}[/tex]

[tex]\dot{Q} =1.5\ W[/tex]

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