Let's break down the problem step by step. ## **Given Data** - **Initial State:** - Pressure, \( P_1 = 200 \) kPa - Temperature, \( T_1 = 300^\circ \)C - **Intermediate State:** - Water cools at constant pressure (200 kPa) until it becomes a saturated vapor (\( x=1 \)), and the piston hits the stops. - **Final State:** - Water continues to cool at constant volume until the pressure drops to \( P_3 = 100 \) kPa. ## **Step 1: Find Initial State Properties (\( u_1 \))** At \( P_1 = 200 \) kPa and \( T_1 = 300^\circ \)C: - From steam tables: - At 200 kPa, the saturation temperature \( T_{sat} \approx 120.23^\circ \)C. - Since \( T_1 = 300^\circ \)C \( > T_{sat} \), the water is superheated. - Superheated steam tables at \( P = 200 \) kPa, \( T = 300^\circ \)C: - **Internal energy \( u_1 \) ≈ 2725.5 kJ/kg** --- ## **Step 2: Intermediate State (\( u_2 \)), Saturated Vapor at 200 kPa** - At 200 kPa, saturated vapor (\( x=1 \)): - From steam tables: - \( u_g \) (internal energy of vapor at 200 kPa) \( = 2506.5 \) kJ/kg So, \( u_2 = 2506.5 \) kJ/kg --- ## **Step 3: Final State (\( u_3 \)), Pressure Drops to 100 kPa at Constant Volume** - From Step 2, the specific volume at the end of step 1 (when the piston hits the stops) is: - \( v_2 = v_g \) at 200 kPa \( = .88578 \) m³/kg - At 100 kPa (final state), the specific volume remains \( v_3 = v_2 = .88578 \) m³/kg. At 100 kPa: - \( v_f = .001043 \) m³/kg - \( v_g = 1.694 \) m³/kg Since \( v_f < v_3 < v_g \), the final state is a mixture. ### **Quality at final state:** \[ x_3 = \frac{v_3 - v_f}{v_g - v_f} = \frac{.88578 - .001043}{1.694 - .001043} = \frac{.884737}{1.692957} \approx .523 \] ### **Internal energy at final state:** \[ u_3 = u_f + x_3 (u_g - u_f) \] From steam tables at 100 kPa: - \( u_f = 419.04 \) kJ/kg - \( u_g = 2506.5 \) kJ/kg \[ u_3 = 419.04 + .523 \times (2506.5 - 419.04) \] \[ u_3 = 419.04 + .523 \times 2087.46 \] \[ u_3 = 419.04 + 1091.77 \] \[ u_3 \approx 151.81 \text{ kJ/kg} \] --- ## **Step 4: Overall Change in Internal Energy** \[ \Delta u = u_3 - u_1 = 151.81 - 2725.5 = -1214.7 \text{ kJ/kg} \] --- ## **Summary Table** | State | P (kPa) | T (°C) | Quality | v (m³/kg) | u (kJ/kg) | |-------|---------|--------|---------|-----------|-------------| | 1 | 200 | 300 | - | - | 2725.5 | | 2 | 200 | 120.23 | 1. | .88578 | 2506.5 | | 3 | 100 | ? | .523 | .88578 | 151.81 | --- ## **Final Answer** \[ \boxed{ \text{The overall change in internal energy per unit mass is} \ \Delta u = -1214.7 \text{ kJ/kg} } \]