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tekilochka [14]
3 years ago
14

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?
Physics
1 answer:
Sloan [31]3 years ago
5 0

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}

<u>Now, according to Fourier's Law of conduction:</u>

\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

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