needs help for my study in thermodynamics?

Question

i have questions here, i cant solve.

1 kg of water in piston cylnder assembly undergoes two processes in series from an initial state where P=3bar and T=160 C.
Process 1-2:water is compressed isothermally to a V=0.25m*3 with W=120kJ.

Process 2-3:water is heated at constant volume to final pressure of 10 bar.

how do i find the change in internal energy for the system and the amount of heat transfered. THANK YOU in advance.

Answer 1

Process 1-2: Use the 1st Law of Thermodynamics:
ΔU = ΔQ - ΔW
where U = internal energy, Q = heat, W = work done by the system

ΔQ = 0 because it was isothermal
ΔW = -120 kJ (work was done to, not by, the system)
ΔU = 0 - (-120 kJ) = 120 kJ (solution!)

Process 2-3:
ΔQ = Use PV=nRT to find the change in temperature.
ΔW = 0 because there is no work done on (other than heating) or by (no volume change) the system.

Step 1: Rewrite PV = nRT to PV/T = nR. Thus: P1V1/T1 = P2V2/T2.
Note: Now, V cancels, because it was constant. This leaves you with P1/T1 = P2/T2

Step 2: Solve to find the change in temperature (ΔT)
P1 = 3 bar
P2 = 10 bar
T1 = 160 C = 160 + 273.15 K = 433.15K

P1/T1 = P2/T2
T2 = P2T1/P1 = 10 * 433.15 / 3 = 1443.83 K

ΔT = 1443.8 K - 433.15K = 1010.68 K

Step 3: Use ΔT to find ΔQ.
ΔQ = ΔT * specific heat of water * mass
ΔQ = 1443.83 * 4180 J/(kg·K) * 1 K = 4225 kJ

Step 4: Use ΔQ and ΔW to find ΔU.
ΔU = ΔQ + ΔW
ΔU = 4225kJ - 0
ΔU = 4225kJ (solution!)