Answer
d) $3451W$
Work Step by Step
In the convection case:
$Q_{conv}=hA\Delta T=12\frac{W}{m^2*^{\circ}C}*3m^2*(80^{\circ}C-25^{\circ}C)=1980W$
In the radiation case:
$Q_{rad}=\varepsilon \sigma A\Delta T^4=1*5.67*10^-8\frac{W}{m^2K^4}*3m^2*[(80+273)^4-(15+273)^4]K^4=1471W$
Then the tital rate of heat loss is:
$Q_{t}=Q_{rad}+Q_{conv}=1471W+1980W=3451W$
