Let's break down the problem **step-by-step**: --- ## **Given Data** - Mass of water, \( m = 2\, \text{kg} \) - Initial pressure, \( P_1 = 10\, \text{bar} \) - Initial temperature, \( T_1 = 200^\circ \text{C} \) - Heat added, \( Q = 174\, \text{kJ} \) - Process: **Constant pressure** (isobaric) --- ## **Step 1: Find Initial Properties** - Use steam tables for water at **10 bar** and **200°C**. ### At 10 bar: - Saturation temperature, \( T_{sat} = 179.9^\circ \text{C} \) - Since \( T_1 = 200^\circ \text{C} > T_{sat} \), water is **superheated**. #### From steam tables (superheated, 10 bar, 200°C): - \( h_1 = 2855.\, \text{kJ/kg} \) - \( v_1 = .1944\, \text{m}^3/\text{kg} \) --- ## **Step 2: Energy Balance** For a closed system undergoing a constant pressure process: \[ Q = m(h_2 - h_1) \] Given \( Q = 174\, \text{kJ} \), \( m = 2\, \text{kg} \), \( h_1 = 2855.\, \text{kJ/kg} \): \[ 174 = 2(h_2 - 2855.) \] \[ h_2 = \frac{174}{2} + 2855. = 870 + 2855. = 3725.\, \text{kJ/kg} \] --- ## **Step 3: Find Final State (\( h_2, P_2 \))** - \( h_2 = 3725.\, \text{kJ/kg} \) - \( P_2 = P_1 = 10\, \text{bar} \) Check the **superheated steam tables** for \( h_2 \) at 10 bar: - At 10 bar, 400°C: \( h = 3264.4\, \text{kJ/kg} \) - At 10 bar, 600°C: \( h = 3762.8\, \text{kJ/kg} \) \( h_2 = 3725.\, \text{kJ/kg} \) is between these values. **Interpolate to find \( T_2 \):** \[ T_2 = 400 + \frac{3725. - 3264.4}{3762.8 - 3264.4} \times (600 - 400) \] \[ = 400 + \frac{460.6}{498.4} \times 200 \] \[ = 400 + .9243 \times 200 \] \[ = 400 + 184.86 \approx 585^\circ\text{C} \] --- ## **Step 4: Find Work Done** At constant pressure, \[ W = P(V_2 - V_1) \] \[ V = m \times v \] From steam tables at 10 bar and ~585°C: - \( v_2 \approx .523\, \text{m}^3/\text{kg} \) (interpolated) \[ V_1 = 2 \times .1944 = .3888\, \text{m}^3 \] \[ V_2 = 2 \times .523 = 1.046\, \text{m}^3 \] \[ W = P (V_2 - V_1) = (10 \times 10^5\, \text{N/m}^2) \times (1.046 - .3888) \] \[ = 1,000,000 \times .6572 = 657,200\, \text{J} = 657.2\, \text{kJ} \] --- ## **Step 5: Plots (Sketches)** ### **P-v Diagram** - A horizontal line at \( P = 10 \) bar, - Moves from \( v_1 = .1944 \) to \( v_2 = .523 \) m³/kg. ### **T-v Diagram** - A curve at increasing T, - Moves from (200°C, .1944) to (585°C, .523). --- ## **Final Answers** ### **(a) Final Temperature** \[ \boxed{T_2 \approx 585^\circ \text{C}} \] ### **(b) Work Done** \[ \boxed{W = 657.2\, \text{kJ}} \] --- ## **Summary Table** | State | P (bar) | T (°C) | h (kJ/kg) | v (m³/kg) | V (m³) | |-------|---------|--------|-----------|-----------|---------| | 1 | 10 | 200 | 2855. | .1944 | .3888 | | 2 | 10 | 585 | 3725. | .523 | 1.046 | --- ## **Sketches** ### **P-v Plot** ``` P | _________ | / |____/___________________ v .1944 .523 ``` ### **T-v Plot** ``` T | / | / |____/___________________ v .1944 .523 ``` --- If you need the actual steam table values or more details on interpolation, just let me know!