Answer:
q=471.19kj.kg
Explanation:water constant volume = C
First state
Quantity X1=1
Pressure=1 bar
Second state
T2=400C
Required heat transferred q in kj.kg
Assumption:
System is at equilibrium
Kinetic and Potential energy are neglected
Therefore, from 1st law of thermodynamics q - w=∆u. . . . 1
From steam table A
U1=Ug=2553.6kj.kg
U2=Vg=1625m³.kg
V1=V2=4625m³.kg
From steam Table B
Px= 7 bar
Vx=4667kj.kg
Ux=3026.6kj.kg
Py= 10 bar
Vy= 3066kj.kg
Uy=2957.5kj.kg
Assuming a Linear interpolation
Ux-V2/Ux-Vy=Vx-U2/Ux-Vy
U2 =Vx-{vx-vy}× vx-v2/Ux-Vy
U2 = 3026.6 - (3026.6-2957.3)× 4667-4625/4667-3066
=3024.79kj.kg
Therefore, from q-w=∆u
q=3024.79-253.6
=471.19kj.kg
Where u = internal energy
V= specific volume
