Question

A rigid tank initially contains \(1.4-\mathrm{kg}\) saturated liquid water at \(200^{\circ} \mathrm{C}\). At this state, 25 percent of the volume is occupied by water and the rest by air. Now heat is supplied to the water until the tank contains saturated vapor only. Determine \((a)\) the volume of the \(\tan \mathrm{k},(b)\) the final temperature and pressure, and \((c)\) the internal energy change of the water.

Short answer

Answer: The final temperature and pressure of the water are approximately 150.5°C and 477.56 kPa, respectively.

Step-by-step solution

To find the initial volume of the water, we will use the steam table to look up the specific volume \(v_{f}\) of saturated liquid water at 200°C. Using the steam table, we can find that \(v_{f} = 1.0603 \times10^{-3} \, \frac{\mathrm{m}^3}{\mathrm{kg}}\). #Step 2: Calculate the volume of the water#

Now that we know the specific volume, we can find the volume \(V_{w}\) of the water by multiplying it with the mass of the water, \(m\). \(V_{w} = m \times v_{f} = 1.4\,\mathrm{kg} \times 1.0603 \times10^{-3} \, \frac{\mathrm{m}^3}{\mathrm{kg}} = 1.4844 \times10^{-3} \,\mathrm{m}^{3}\) #Step 3: Calculate the volume of the tank#

Since the water occupies 25% of the tank, we can find the volume of the tank by dividing the volume of the water by 25%. \(V_{\mathrm{tank}} = \frac{V_{w}}{0.25} = \frac{1.4844 \times10^{-3} \,\mathrm{m}^{3}}{0.25} = 5.9376 \times10^{-3} \, \mathrm{m}^{3}\) #Step 4: Determine the final state of the water#

When the water becomes saturated vapor, the volume of the water will be equal to the volume of the tank. Thus, we can find the specific volume of the saturated vapor, \(v_{g}\). \(v_{g} = \frac{V_{\mathrm{tank}}}{m} = \frac{5.9376 \times10^{-3} \,\mathrm{m}^{3}}{1.4\,\mathrm{kg}} = 4.2411 \times10^{-3} \, \frac{\mathrm{m}^3}{\mathrm{kg}}\) #Step 5: Find the final temperature and pressure using steam table#

We will now consult the steam table again to find the temperature and pressure corresponding to this specific volume \(v_{g}\) for saturated vapor. The steam table reveals that the final temperature \(T_{2}\) is approximately \(150.5^{\circ}\mathrm{C}\) and the final pressure is approximately \(477.56\,\mathrm{kPa}\). #Step 6: Calculate the internal energy change of the water#

Next, we will apply the first law of thermodynamics to calculate the internal energy change of the water. From the steam table, we can find the initial internal energy, \(u_{1}\), as well as the final internal energy, \(u_{2}\). \(u_{1} = 884.2 \, \frac{\mathrm{kJ}}{\mathrm{kg}}\) for the initial state and \(u_{2} = 2564.5\, \frac{\mathrm{kJ}}{\mathrm{kg}}\) for the final state. To find the internal energy change, \(\Delta u\), we can subtract the initial internal energy from the final internal energy multiplied by the mass of the water. \(\Delta u = m \times (u_{2} - u_{1}) = 1.4\,\mathrm{kg} \times (2564.5 - 884.2) \,\frac{\mathrm{kJ}}{\mathrm{kg}} = 2352.6 \, \mathrm{kJ}\) In summary, the volume of the tank is approximately \(5.9376 \times10^{-3}\,\mathrm{m}^3\), the final temperature and pressure are approximately \(150.5^{\circ}\mathrm{C}\) and \(477.56\,\mathrm{kPa}\) respectively, and the internal energy change of the water is approximately \(2352.6\,\mathrm{kJ}\).