Explanation:
The given data is as follows.
Thickness (dx) = 0.87 m, thermal conductivity (k) = 13 W/m-K
Surface area (A) = 5 [tex]m^{2}[/tex], [tex]T_{1} = 14^{o}C[/tex]
[tex]T_{2} = 93^{o}C[/tex]
According to Fourier's law,
Q = [tex]-kA \frac{dT}{dx}[/tex]
Hence, putting the given values into the above formula as follows.
Q = [tex]-kA \frac{dT}{dx}[/tex]
= [tex]-13 W/m-k \times 5 m^{2} \times \frac{(14 - 93)}{0.87 m}[/tex]
= 5902.298 W
Therefore, we can conclude that the rate of heat transfer is 5902.298 W.
