Answer:254.7 K
Explanation:
Given
diameter(d)=0.5 m
Energy dissipate[tex](E_g)[/tex]=150 w
emissivity[tex](\epsilon )[/tex]=0.8
According to conservation of Energy Energy generation and energy dissipation at any instant is given by
[tex]-E_{out}+E_g=0[/tex]
[tex]E_g=\epsilon A_s\sigma T^4[/tex]
[tex]T=\frac{E_g}{\epsilon A_s\simga }^{\frac{1}{4}}[/tex]
[tex]T=(\frac{150}{0.8\pi (0.5)^2\times 5.67\times 10^{-8}})^{\frac{1}{4}}[/tex]
T=254.7 K