Absolutely! Let’s break down the **thermodynamics problem** step by step. You’re dealing with a piston-cylinder device containing water, and you’re asked about changes when heating it in two different ways. --- # **Step-by-Step Explanation** ## **Given Data** - Mass of water, \( m = 3\, \text{kg} \) - Initial state: **Saturated liquid** at \( T_1 = 250^\circ \text{C} \) - **Device is frictionless** (no energy loss) - **No other substances** (pure water) - **Piston-cylinder**: Pressure inside = Pressure outside (constant pressure for isobaric process) --- ## **Part 1: Volume Triples, Entirely Becomes Vapor** ### **(a) Initial Volume of the Tank** - **Saturated liquid at 250°C:** Use steam tables to find the **specific volume** of saturated liquid at \( 250^\circ \text{C} \) (\( v_f \)): - \( v_{f,250^\circ C} \approx 0.001252\, \text{m}^3/\text{kg} \) (from steam tables) - **Total initial volume:** \[ V_1 = m \times v_f = 3\, \text{kg} \times 0.001252\, \text{m}^3/\text{kg} = 0.003756\, \text{m}^3 \] --- ### **(b) Final Temperature and Pressure** - **Volume triples:** \( V_2 = 3 V_1 \) - **All water is now saturated vapor:** At final state: \( x_2 = 1 \) (saturated vapor quality) - **At final state:** \( V_2 = m \times v_g \) (where \( v_g \) is specific volume of saturated vapor at final T/P) Set up: \[ V_2 = 3 V_1 = m \times v_g^\prime \] \[ v_g^\prime = \frac{V_2}{m} = \frac{3 V_1}{m} = 3 \times v_{f,250^\circ C} \] \[ v_g^\prime = 3 \times 0.001252 = 0.003756\, \text{m}^3/\text{kg} \] - **Now, find T/P where \( v_g \approx 0.003756 \) in steam tables.** This is typically at a much lower pressure than at 250°C, so you’d interpolate to find the exact T and P where saturated vapor has this specific volume. --- ### **(c) Change in Internal Energy (\( \Delta U \))** - **Initial:** Internal energy of saturated liquid at 250°C: \( u_{f,250^\circ C} \) (from steam tables) - **Final:** Internal energy of saturated vapor at the final state: \( u_{g,final} \) - **Change:** \[ \Delta U = m \left[ u_{g,final} - u_{f,250^\circ C} \right] \] --- ## **Part 2: Isobaric Heating (Constant Pressure)** ### **Now, the process is at constant pressure, starting with saturated liquid at 250°C (find \( P_{sat,250^\circ C} \)).** ### **(d) Final Temperature and Volume** - **Final state:** All water is saturated vapor at the same pressure as the initial state. - **Pressure:** \( P_{sat,250^\circ C} \) (from steam tables, about 3975 kPa) - **Temperature:** Still 250°C (because at constant pressure, phase change occurs at constant temperature) - **Final volume:** \[ V_2 = m \times v_{g,250^\circ C} \] --- ### **(e) Change in Internal Energy** \[ \Delta U = m \left[ u_{g,250^\circ C} - u_{f,250^\circ C} \right] \] --- ### **(f) Work Done by the Water** - **At constant pressure:** \[ W = P \Delta V = P \left( V_2 - V_1 \right) \] - \( P \) in Pa, \( V \) in \( \text{m}^3 \) --- ### **(g) Total Heat Added (\( Q \))** - **From the first law:** \[ Q = \Delta U + W \] --- ## **Summary Table** | Quantity | Formula/Approach | Where to Find Data | |---------------|------------------------------------------|----------------------------| | Specific volume (\( v \)) | Steam Tables at given T/P | Steam Tables | | Internal energy (\( u \)) | Steam Tables at given T/P | Steam Tables | | Volume (\( V \)) | \( V = m \times v \) | Calculation | | Pressure (\( P \)) | \( P_{sat,250^\circ C} \) | Steam Tables | | Temperature (\( T \)) | Given or from Steam Tables | Steam Tables | | Work (\( W \)) | \( P (V_2 - V_1) \) | Calculation | | Heat (\( Q \)) | \( Q = \Delta U + W \) | Calculation | --- ## **Next Steps** 1. **Get values** from steam tables for 250°C: - \( v_{f,250^\circ C} \), \( v_{g,250^\circ C} \) - \( u_{f,250^\circ C} \), \( u_{g,250^\circ C} \) - \( P_{sat,250^\circ C} \) 2. **Plug into formulas above** to calculate each answer. --- ### **Need help with the actual steam table values or with plugging numbers in?** Let me know, and I can walk you through the calculations with the numbers!