Answer:
T surface = 3.9°C
Explanation:
given data
emissivity 0.6
absorptivity = 0.2
solar radiation is incident rate = 1200 W/m²
solution
we get here surface temperature by equality of emitted and absorbed heat rate that is
Q (absorbed) = Q (heat ) .................1
α Qinc = [tex]\epsilon * \sigma *A*T^4(surface)[/tex]
T surface = [tex]\sqrt[4]{\frac{\alpha Qinc}{\epsilon *\sigma * A} }[/tex] ..........................2
put here value and we get
T surface = [tex]\sqrt[4]{\frac{0.2*1000}{0.6*5.67**10^{-8}} }[/tex]
T surface = 276.9 K
T surface = 3.9°C