Respuesta :
Explanation:
The heat transferred through conduction = heat transfer through radiation + heat transfer through convection
It is given that thermal conductivity of wall is 1 W/(m.K)
Boltzmann constant ([tex]\sigma_{e}[/tex]) = [tex]5.68 \times 10^{-8} W/T^{4} m^{2}[/tex]
Formula to calculate the inner temperature of the furnace is as follows.
kA(T - 80) = [tex]\sigma_{e} \times A \times (80 + 273)^{4} - (25 + 273)^{4} + hA(80 - 25)[/tex]
It is known that for black body emissitivity = 1
Now putting the given values into the above formula as follows.
kA(T - 80) = [tex]\sigma_{e} \times A \times (80 + 273)^{4} - (25 + 273)^{4} + hA(80 - 25)[/tex]
[tex]1 \times A(T - 80)[/tex] = A ([tex]\sigma_{e} \times (80 + 273)^{4} - (25 + 273)^{4} + h(80 - 25)[/tex])
T - 80 = [tex]5.68 \times 10^{-8} \times [(353)^{4} - (298)^{4}] + 1 \times 55[/tex]
T = [tex]1338.25^{o}C[/tex]
Thus, we can conclude that the inner temperature of the furnace wall is [tex]1338.25^{o}C[/tex].
