Answer:
650.65 K or 377.5°C
Explanation:
Area = A = 10 m²
Thickness of wall = L = 2.5 cm = 2.5×10⁻² m
Inner surface temperature of wall = [tex]T_i[/tex] = 415°C = 688.15 K
Outer surface temperature of wall = [tex]T_o[/tex]
Heat loss through the wall = 3 kW = 3×10³ W
Thermal conductivity of wall = k = 0.2 W/m K
Assumptions made here as follows
Considering the above assumptions we use the following formula
[tex]Q=\frac {T_i-T_o}{\frac{L}{kA}}\\\Rightarrow T_o=T_i-\frac {QL}{kA}\\\Rightarrow T_o=688.15-\frac {3\times 10^{3}\times 2.5\times 10^{-2}}{0.2\times 10}\\\Rightarrow T_o=650.65~K[/tex]
∴ The temperature of the outer surface of the wall is 650.65 K or 377.5°C