Answer:
Q=486.49 KJ/kg
Explanation:
Given that
V= 0.2 m³
At initial condition
P= 2 MPa
T=320 °C
Final condition
P= 2 MPa
T=540°C
From steam table
At P= 2 MPa and T=320 °C
h₁=3070.15 KJ/kg
At P= 2 MPa and T=540°C
h₂=3556.64 KJ/kg
So the heat transfer ,Q=h₂ - h₁
Q= 3556.64 - 3070.15 KJ/kg
Q=486.49 KJ/kg