Calculate max temp reached, change in internal energy, heat transfer. When vol=0.2m^3, at 1.2 bars, 60°c compressed 12bars temp 270°c at constant vol. P becomes twice at the end(cp=1.072kj,R=294j/kgk?

Given data:
Initial volume (V1) = 0.2 m^3
Initial pressure (P1) = 1.2 bars = 1.2 * 10^5 Pa
Initial temperature (T1) = 60°C = 333 K
Final pressure (P2) = 12 bars = 12 * 10^5 Pa
Final temperature (T2) = 270°C = 543 K
Specific heat at constant pressure (cp) = 1.072 kJ/kgK = 1072 J/kgK
Gas constant (R) = 294 J/kgK

1. Calculate the maximum temperature reached during compression:
Using the ideal gas law, we can calculate the initial and final number of moles of gas:
n1 = (P1 * V1) / (R * T1)
n2 = (P2 * V1) / (R * T2)

Since the volume is constant during the process, the number of moles remains constant:
n1 = n2

Therefore, (P1 * V1) / (R * T1) = (P2 * V1) / (R * T2)
Solving for T2:
T2 = (P2 * T1) / P1
T2 = (12 * 333) / 1.2
T2 = 3330 K

Therefore, the maximum temperature reached during compression is 3330 K.

2. Calculate the change in internal energy:
The change in internal energy (ΔU) can be calculated using the first law of thermodynamics:
ΔU = n * cp * (T2 - T1)
ΔU = n * 1072 * (3330 - 333)
ΔU = n * 1072 * 2997
ΔU = 3 * 1072 * 2997
ΔU = 3.22 * 10^6 J

Therefore, the change in internal energy is 3.22 * 10^6 J.

3. Calculate the heat transfer:
The heat transfer (Q) during the process can be calculated using the equation:
Q = ΔU + W
Where W is the work done during the process. Since the volume is constant, no work is done (W = 0).

Therefore, Q = ΔU
Q = 3.22 * 10^6 J

Therefore, the heat transfer during the process is 3.22 * 10^6 J.