Answer:
Δ h = 2958.9 kJ/kg
Explanation:
initial state
T₁ = 50⁰C
P₁ = 1 bar
from steam table , Enthalpy at 50⁰C and 1 bar
h₁ = 208.785 kJ/kg
final state
T₂ = 350⁰C
P₂ = 5 bar
from steam table , Enthalpy at 350⁰C and 5 bar
h₂ = 3167.75 kJ/kg
change in entalpy for the process
Δ h = h₂ - h₁
= 3167.75 - 208.785
Δ h = 2958.965 kJ/kg
the change in enthalpy is equal to Δ h = 2958.9 kJ/kg
