Answer:
T = 212.8125°C
Explanation:
Given
radius of the wire, [tex]r_{0}[/tex] = 5 mm 0.005 m
heat generated, g = 5 x [tex]10^{7}[/tex] W/[tex]m^{3}[/tex]
outer surface temperature, [tex]T_{S}[/tex] = 180°C
Thermal conductivity, k = 8 W / m-k
Now maximum temperature occurs at the center of the wire
that is at r=0,
Therefore, [tex]T_{o}=T_{S}+\frac{g\times r_{o}^{2}}{4\times k}[/tex]
[tex]T_{o}=180+\frac{5\times 10^{7}\times 0.005^{2}}{4\times 8}[/tex]
[tex]T_{o}=219.0625[/tex]°C
Therefore, temperature at r = 2 mm
[tex]\frac{T-T_{S}}{T_{O}-T_{S}}= 1-\left (\frac{r}{r_{O}} \right )^{2}[/tex]
[tex]\frac{T-180}{219.0625-180}= 1-\left (\frac{2}{5} \right )^{2}[/tex]
Therefore, T = 212.8125°C