# Step-by-Step Solution Let's break down the problem using the information given and thermodynamic principles. --- ## **Given Data** - **Initial Volume, \( V_1 = 1.3 \, \text{m}^3 \)** - **Initial State:** Saturated vapor (steam) at \( T_1 = 185^\circ\text{C} \) - **Final State:** Saturated liquid at \( T_2 = 185^\circ\text{C} \) (since the process is isothermal) - **Device:** Frictionless piston-cylinder (constant pressure during phase change) - **Process:** Isothermal condensation (from saturated vapor to saturated liquid) --- ## **Step 1: Find Properties at 185°C** From steam tables for **saturated water at \( T = 185^\circ\text{C} \):** - **Saturation pressure, \( P_{\text{sat}} \)** - **Specific volume of saturated vapor, \( v_g \)** - **Specific volume of saturated liquid, \( v_f \)** - **Specific internal energy of vapor, \( u_g \)** - **Specific internal energy of liquid, \( u_f \)** - **Specific enthalpy of vapor, \( h_g \)** - **Specific enthalpy of liquid, \( h_f \)** **At \( T = 185^\circ\text{C} \):** From steam tables (approximate values, check your textbook for exact values): - \( P_{\text{sat}} = 1.1557 \) MPa - \( v_g = .15841 \) m³/kg - \( v_f = .001127 \) m³/kg - \( u_g = 255.4 \) kJ/kg - \( u_f = 780.22 \) kJ/kg - \( h_g = 2778.1 \) kJ/kg - \( h_f = 781.06 \) kJ/kg --- ## **Step 2: (a) Find the Mass of the Water** \[ \text{Initial state:} \quad V_1 = m \cdot v_g \] \[ m = \frac{V_1}{v_g} = \frac{1.3}{.15841} = 8.206 \text{ kg} \] --- ## **Step 3: (b) Initial Pressure in the Tank** \[ P_1 = P_{\text{sat}} \text{ at } 185^\circ\text{C} = 1.1557 \text{ MPa} \] --- ## **Step 4: (c) Final Pressure and Volume of the Water** - **Final State:** Saturated liquid at \( 185^\circ\text{C} \) - **Pressure:** \( P_2 = P_{\text{sat}} = 1.1557 \) MPa (same as initial; isothermal at saturation) - **Volume:** \( V_2 = m \cdot v_f = 8.206 \times .001127 = .00925 \text{ m}^3 \) --- ## **Step 5: (d) Change in Internal Energy** \[ \Delta U = m(u_f - u_g) = 8.206 \times (780.22 - 255.4) \] \[ \Delta U = 8.206 \times (-177.18) = -14,525 \text{ kJ} \] --- ## **Step 6: (e) Work Done On the Water** \[ W_{on} = -\int_{V_1}^{V_2} P_{ext} dV \] Since isothermal and at constant pressure: \[ W_{on} = -P_{ext} (V_2 - V_1) \] \[ W_{on} = -1.1557 \times 10^6 \times (.00925 - 1.3) \] \[ = -1.1557 \times 10^6 \times (-1.29075) \] \[ = 1,492,214 \text{ J} = 1,492.2 \text{ kJ} \] --- ## **Step 7: (f) Amount of Heat Removed** \[ Q_{out} = \Delta U + W_{by} \] But here, \( W_{by} = -W_{on} \). So, \[ Q_{out} = \Delta U + (-W_{on}) \] \[ Q_{out} = -14,525 \text{ kJ} + 1,492.2 \text{ kJ} = -13,032.8 \text{ kJ} \] --- ## **Step 8: (g) Change in Enthalpy** \[ \Delta H = m(h_f - h_g) \] \[ \Delta H = 8.206 \times (781.06 - 2778.1) \] \[ \Delta H = 8.206 \times (-1997.04) = -16,387 \text{ kJ} \] --- ## **Summary Table** | Part | Quantity | Formula/Value | Answer | |------|---------------------------|----------------------------|---------| | (a) | Mass | \( m = \frac{V_1}{v_g} \) | **8.21 kg** | | (b) | Initial pressure | \( P_{\text{sat}} \) | **1.16 MPa** | | (c) | Final pressure, volume | \( P_2 = P_{\text{sat}} \), \( V_2 = m v_f \) | **1.16 MPa**, **.00925 m³** | | (d) | Change in internal energy | \( m(u_f-u_g) \) | **-14,525 kJ** | | (e) | Work done on water | \( -P(V_2-V_1) \) | **1,492 kJ** | | (f) | Heat removed | \( \Delta U + W_{by} \) | **-13,033 kJ** | | (g) | Change in enthalpy | \( m(h_f-h_g) \) | **-16,387 kJ** | --- ### **Final Answers (Rounded as Appropriate)** **(a) Mass of water:** \( \boxed{8.21\,\text{kg}} \) **(b) Initial pressure:** \( \boxed{1.16\,\text{MPa}} \) **(c) Final pressure and volume:** \( \boxed{1.16\,\text{MPa}}, \boxed{.00925\,\text{m}^3} \) **(d) Change in internal energy:** \( \boxed{-14,525\,\text{kJ}} \) **(e) Work done on water:** \( \boxed{1,492\,\text{kJ}} \) **(f) Heat removed:** \( \boxed{-13,033\,\text{kJ}} \) **(g) Change in enthalpy:** \( \boxed{-16,387\,\text{kJ}} \) --- **Note:** - All property values are from steam tables for 185°C. - Negative sign in heat/energy means heat is removed or internal energy decreases. - Check your steam tables for more precise values if required.