Answer
$Q_{t}=50.71W$
Work Step by Step
First we calculate the area of the spherical ball:
$A=\pi D^2=\pi*(0.09m)^2=0.025447m^2$
Knowing that the heat transfer in this case is by convection and radiation:
$Q_{conv}=hA\Delta T=15\frac{W}{m^2*^{\circ}C}*0.025447m^2*(110^{\circ}C-20^{\circ}C)=34.353W$
$Q_{rad}=\varepsilon \sigma A\Delta T^4=0.8*5.67*10^-8\frac{W}{m^2*K^4}*0.025447m^2*[(110+273.15)^4-(20+273.15)^4]K^4=16.352W$
Then:
$Q_{t}=Q_{conv}+Q_{rad}=34.353W+16.352W=50.71W$
