Answer: The outer surface temperature is [tex]377.5^{o}C[/tex].
Explanation:
Given: Heat = 3 kW (1 kW = 1000 W) = 3000 W
Area = 10 [tex]m^{2}[/tex]
Length = 2.5 cm (1 cm = 0.01 m) = 0.025 m
Thermal conductivity = 0.2 W/m K
Temperature (inner) = [tex]415^{o}C[/tex]
Formula used is as follows.
[tex]q = KA \frac{(t_{in} - t_{out})}{L}[/tex]
where,
K = thermal conductivity
A = area
L = length
[tex]t_{in}[/tex] = inner surface temperature
[tex]t_{out}[/tex] = outer surface temperature
Substitute the values into above formula as follows.
[tex]q = KA \frac{(t_{in} - t_{out})}{L}\\3000 W = 0.2 \times 10 \times \frac{415 - t_{out}}{0.025 m}\\t_{out} = 377.5^{o}C[/tex]
Thus, we can conclude that the outer surface temperature is [tex]377.5^{o}C[/tex].
