Answer
$Q_{conv}=22kW$
Work Step by Step
First we calculate the area of the surface:
$A=2m*4m=8m^2$
Knowing that the heat transfer in this case is by convection:
$Q_{conv}=hA\Delta T=55\frac{W}{m^2*^{\circ}C}*8m^2*(80^{\circ}C-30^{\circ}C)=22kW$
