Let's solve the problem step by step. ## **Step 1: Understand the Problem** You are given: - Volume of liquid water, \( V_f = .005\,m^3 \) - Volume of water vapor, \( V_g = .81\,m^3 \) - Pressure, \( P = 600\,kPa \) - Both phases are in equilibrium (saturated mixture). - You need to **determine the total mass of the water in the system**. --- ## **Step 2: Find Specific Volumes from Steam Tables** At \( P = 600\,kPa \): From the steam tables: - Specific volume of saturated liquid, \( v_f \) at 600 kPa ≈ **.001127 m³/kg** - Specific volume of saturated vapor, \( v_g \) at 600 kPa ≈ **.3196 m³/kg** --- ## **Step 3: Calculate Mass in Each Phase** The mass in each phase is: \[ m = \frac{V}{v} \] ### **Liquid Water:** \[ m_f = \frac{V_f}{v_f} = \frac{.005}{.001127} \approx 4.44\,kg \] ### **Water Vapor:** \[ m_g = \frac{V_g}{v_g} = \frac{.81}{.3196} \approx 2.54\,kg \] --- ## **Step 4: Find Total Mass** \[ m_{total} = m_f + m_g = 4.44 + 2.54 = 6.98\,kg \] --- ## **Step 5: Final Answer** **The total mass is \( \boxed{6.98\,\text{kg}} \).** --- ### **Summary Table** | Phase | Volume (m³) | Specific Volume (m³/kg) | Mass (kg) | |-----------------|-------------|-------------------------|-----------| | Liquid (f) | .005 | .001127 | 4.44 | | Vapor (g) | .81 | .3196 | 2.54 | | **Total** | | | 6.98 |