To solve this problem, we can use the First Law of Thermodynamics, which states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W):
ΔU = Q - W
We can assume that the process is quasi-static and reversible, so there is no entropy change (ΔS = 0). Therefore, the heat added to the system is simply the heat transfer (Q) and can be calculated using the specific heat capacity of R-134a at constant pressure (cp):
Q = m * cp * ΔT
where m is the mass of R-134a, cp is the specific heat capacity at constant pressure, and ΔT is the temperature change.
To calculate the work done by the refrigerant, we can use the following equation:
W = -P * ΔV
where P is the pressure and ΔV is the change in volume.
First, we need to find the initial and final states of the refrigerant:
State 1: P1 = 600 kPa, T1 = 150°C, V1 = 400.3 m³
State 2: P2 = 600 kPa, T2 = -30°C, V2 = 0.1 m³
We can use the R-134a tables to find the specific volume and specific internal energy at each state:
State 1: v1 = 0.1532 m³/kg, u1 = 783.1 kJ/kg
State 2: v2 = 0.0044 m³/kg, u2 = 131.3 kJ/kg
The mass of R-134a can be calculated from the initial volume and specific volume:
m = V1 / v1 = 400.3 m³ / 0.1532 m³/kg = 2614.4 kg
The heat transfer can be calculated as:
Q = m * cp * ΔT = m * cp * (T2 - T1) = 2614.4 kg * 1.14 kJ/kg·K * (-30°C - 150°C) = -573876.48 kJ
Note that the negative sign indicates that heat is being removed from the system (the refrigerant is being cooled).
The work done by the refrigerant can be calculated as:
W = -P * ΔV = P1 * (V1 - V2) = 600 kPa * (400.3 m³ - 0.1 m³) = 240120 J
Note that we converted the pressure from kPa to Pa and the volume from m³ to m³ to obtain the correct units for work (Joules).
Therefore, the transfer of heat from the refrigerant during this process is -573876.48 kJ, and the work produced by the refrigerant is 240120 J.