Answer
$\nu_2=838.6\ ft/s$
Work Step by Step
The specific energy balance:
$\dot{e}_{in}=\dot{e}_{out}$
$h_1+\nu_1^2/2=h_2+\nu_2^2$
With $h_2=h_1+c_p(T_2-T_1),\ c_p=0.253\ Btu/lbm°R,\ T_2=645°F,\ T_1=700°F,\ \nu_1=80\ ft/s$:
Solving for the outlet velocity $\nu_2=838.6\ ft/s$
