Answer
$\mathcal{V}_1=825\ ft/min$
$T_2=64.0°F$
Work Step by Step
For the velocity:
$\dot{V}_1=A_1\mathcal{V}_1$
Given $A_1=\frac{\pi}{4}D_1^2,\ D_1=10\ in,\ \dot{V}_1=450\ ft³/min$
$\mathcal{V}_1=825\ ft/min$
The mass flowrate is given by:
$\dot{m}=\frac{P_1\dot{V}_1}{RT_1}$
and since $P_1=15\ psia,\ R=0.3704\ psia.ft³/lbm.°R,\ T_1=50°F=510°R$:
$\dot{m}=0.595\ lbm/s$
From the energy balance:
$\dot{Q}+\dot{m}h_1=\dot{m}h_2$
$\dot{Q}=\dot{m}c_p(T_2-T_1)$
In the case of $\dot{Q}=2\ Btu/s,\ c_p=0.240\ Btu/lbm.°F$
$T_2=64.0°F$
