To solve for the outlet velocity, we need to use the conservation of mass and conservation of energy equations:
Conservation of mass: m1 = m2
where m1 is the mass flow rate at the inlet and m2 is the mass flow rate at the outlet
Conservation of energy: (m1 * u1 * cp) + (m1 * h1) = (m2 * u2 * cp) + (m2 * h2)
where u1 and u2 are the velocities at the inlet and outlet. h1 and h2 are the enthalpy values at the inlet and outlet.
Therefore, we can solve for u2, the velocity at the outlet:
u2 = ((m1*u1*cp) + (m1*h1) - (m2*h2)) / (m2*cp)
Plugging in the given values:
u2 = ((3.8 kg/s*265 m/s*1001 J/kg-K) + (3.8 kg/s*1520.4 kJ/kg) - (3.8 kg/s*1537.1 kJ/kg)) / (3.8 kg/s*1001 J/kg-K)
u2 = 253.4 m/s
For the second part of your question, the passive heating of a fluid in a steady-flow control volume means that Δh > 0. Since the flow is steady, there is no change in kinetic or potential energy, and therefore the total enthalpy change must be greater than zero.