Answer
$Q_{conv}=230.86W$
Work Step by Step
First we calculate the lateral area of the modeled cylinder:
$A_{l}=\pi Dh=\pi*0.3m*1.75m=1.649m^2$
Knowing that the heat transfer in this case is by convection:
$Q_{conv}=hA\Delta T=10\frac{W}{m^2*^{\circ}C}*1.649m^2*(34^{\circ}C-20^{\circ}C)=230.86W$
