Answer
$Q_{conv}=7.422kW$
Work Step by Step
First we calculate the lateral area of the pipe:
$A_{l}=\pi DL=\pi*0.07m*18m=3.958407m^2$
Knowing that the heat transfer in this case is by convection:
$Q_{conv}=hA\Delta T=25\frac{W}{m^2*^{\circ}C}*3.958407m^2*(80^{\circ}C-5^{\circ}C)=7.422kW$
