Answer
$2.29\times10^{-3}m^{2}$
Work Step by Step
Here we use equation 14.8 $m=\frac{(DA\Delta C)t}{L}$ to find the cross-sectional area of the pipe.
$m=\frac{(DA\Delta C)t}{L}=>A=\frac{mL}{D\Delta Ct}$
Let's plug known values into this equation.
$A=\frac{(9\times10^{-4}kg)(1.5\space m)}{(2.1\times10^{-5}m^{2}/s)(0.65\space kg/m^{3}-0\space kg/m^{3})(4.32\times10^{4}s)}=2.29\times10^{-3}m^{2}$
