Answer
$T_{s}=32.2^{\circ}C$
Work Step by Step
As the problem says:
$Q_{absorbed}=Q_{conv}$
$Q_{absorbed}=\alpha Q_{solar}=0.8*450\frac{W}{m^2}=360\frac{W}{m^2}$
$Q_{conv}=hA\Delta T$
And: $\frac{Q_{conv}}{A}=h\Delta T=360\frac{W}{m^2}$
Solving for $\Delta T$:
$\Delta T=\frac{360\frac{W}{m^2}}{h}=\frac{360\frac{W}{m^2}}{50\frac{W}{m^2*^{\circ}C}}=7.2^{\circ}C$
Taking $T_{2}$ as the temperature of the surface and $T_{1}$ as the air temperature::
$\Delta T=T_{2}-T_{1}$
$T_{2}= 7.2^{\circ}C+25^{\circ}C=32.2^{\circ}C$
