Let's analyze the problem: **Given Data:** - The liquid water is in a rigid tank at pressure: \( P = 150\, \text{kPa} \) - Mass of water, \( m = 30\, \text{kg} \) - The tank is rigid, so volume \( V \) is constant - Find the temperature and volume of the tank --- ### Step 1: Convert pressure to kPa - Pressure \( P = 150\, \text{kPa} \) ### Step 2: Use specific volume of water at the pressure From standard steam tables for water at \( P = 150\, \text{kPa} \): - Specific volume \( v_f \) (liquid water) ≈ **.001059 m³/kg** ### Step 3: Calculate the volume of the water \[ V = m \times v_f \] \[ V = 30\, \text{kg} \times .001059\, \text{m}^3/\text{kg} \] \[ V \approx .03177\, \text{m}^3 \] ### Step 4: Determine temperature at \( P = 150\, \text{kPa} \) - From water tables, at \( P = 150\, \text{kPa} \), the saturation temperature \( T_{sat} \approx 111.5^\circ C \) ### Step 5: Confirm the state - Since the tank is rigid and contains water at a pressure of 150 kPa, the temperature corresponds to the saturation temperature at that pressure. --- ### **Answer:** The closest matching option is: > **Option 2:** > \( T = 111.35^\circ C \) and \( v = .3159\, \text{m}^3 \) Note: While the specific volume calculation yields approximately .03177 m³ for 30 kg, the options seem to have a different interpretation of volume, likely referring to **specific volume** or a different context. ### **Conclusion:** **Correct Option:** **2.** \( T = 111.35^\circ C \) and \( v = .3159\, \text{m} \) --- **Note:** The provided options seem to mix total volume and specific volume. Based on standard water properties, option 2 aligns more with the temperature and approximate volume.