Answer with Explanation:
We are given that
Irradiation,G=6300[tex]W/m^2[/tex]
Surface A absorb.[tex]G_A=5600 W/m^2[/tex]
a.[tex]G=\sigma T^4[/tex]
[tex]T^4=\frac{G}{\sigma}[/tex]
[tex]T=(\frac{G}{\sigma})^{\frac{1}{4}}[/tex]
Where [tex]\sigma =5.67\times 10^{-8}[/tex]=Boltzman's constant
[tex]T=(\frac{6300}{5.67\times 10^{-8}})^{\frac{1}{4}}=577.4 K[/tex]
Temperature of surface =577.4 K
b.[tex]E_{in}=E_{out}[/tex]
[tex]E_{emitted}=\sigma E_A T^4[/tex]
[tex]5600=5.67\times 10^{-8}E_A(577.4)^4[/tex]
[tex]E_A=\frac{5600}{5.67\times 10^{-8}\times (577.4)^4}=0.89[/tex]
Hence, the emissivity of surface =0.89
