Answer
$\nu_2=40.7\ m/s$
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 $h_2=h_1+c_p(T_2-T_1),\ c_p=1.007\ kJ/kg.K,\ T_2=90°C,\ T_1=30°C,\ \nu_1=350\ m/s$:
we can solve for the outlet velocity $\nu_2=40.7\ m/s$
