Answer
e) $65^{\circ}C$
Work Step by Step
First we calculate the area of the board:
$A=0.1m*0.2m=0.02m^2$
Knowing that the heat transfer in this case is by convection:
$Q_{conv}=hA\Delta T$
$\Delta T=\frac{Q_{conv}}{hA}=\frac{100*0.08W}{10\frac{W}{m^2*^{\circ}C}*0.02m^2}=40^{\circ}C$
$T_{s}=25^{\circ}C+40^{\circ}C=65^{\circ}C$
