Question

An insulated piston-cylinder device initially contains \(0.01 \mathrm{m}^{3}\) of saturated liquid-vapor mixture with a quality of 0.2 at \(120^{\circ} \mathrm{C}\). Now some ice at \(0^{\circ} \mathrm{C}\) is added to the cylinder. If the cylinder contains saturated liquid at \(120^{\circ} \mathrm{C}\) when thermal equilibrium is established, determine the amount of ice added. The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg},\) respectively.

Short answer

Answer: Approximately \(8.9462 \mathrm{kg}\) of ice is added to the piston-cylinder device.

Step-by-step solution

Determine the initial mass of the liquid-vapor mixture

For a saturated liquid-vapor mixture, the specific volume \(v\) is given by: $$ v = (1-x)v_f + xv_g $$ Where \(x\) stands for the quality, while \(v_f\) and \(v_g\) are the specific volumes of the saturated liquid and saturated vapor, respectively. We are given the quality (\(x = 0.2\)) and the total volume (\(V = 0.01 \mathrm{m}^3\)). To find \(v_f\) and \(v_g\) at \(120^{\circ}\mathrm{C}\), we can use the steam tables. At \(120^{\circ}\mathrm{C}\), from the steam tables, we have: - \(v_f = 0.00106 \mathrm{m}^3/\mathrm{kg}\) - \(v_g = 0.0993 \mathrm{m}^3/\mathrm{kg}\) Now, we can find the specific volume \(v\) of the mixture: $$ v = (1-0.2) \times 0.00106 + 0.2 \times 0.0993 = 0.0205 \mathrm{m}^3/\mathrm{kg} $$ To find the initial mass of the liquid-vapor mixture \(m_{initial}\), we can use: $$ m_{initial} = \frac{V}{v} = \frac{0.01}{0.0205} = 0.4878 \mathrm{kg} $$

Calculate the mass of liquid formed due to melting and heating of ice

To find the mass of liquid formed due to the melting of ice and heating it to \(120^{\circ}\mathrm{C}\), we can set up an energy balance equation. Let \(m_{ice}\) be the mass of ice added and \(Q_{in}\) be the energy transferred from the surroundings to the system. The energy balance equation can be written as: $$ Q_{in} = m_{ice}(hf_{ice} + c_{pw}(120 - 0)) $$ Where \(hf_{ice}\) is the heat of fusion of ice and \(c_{pw}\) is the specific heat of water. They are given: - \(hf_{ice} = 333.7 \mathrm{kJ/kg}\) - \(c_{pw} = 4.18 \mathrm{kJ/kg\cdot K}\) We can now rewrite the energy balance equation as: $$ Q_{in} = m_{ice}(333.7 + 4.18 \times 120) = 833.6m_{ice} $$

Determine the final mass of the saturated liquid

When thermal equilibrium is established, there will be saturated liquid at \(120^{\circ}\mathrm{C}\) in the piston-cylinder device. The final mass of the saturated liquid \(m_{final}\) can be obtained using the specific volume of the saturated liquid (\(v_f\)) at \(120^{\circ}\mathrm{C}\): $$ m_{final} = \frac{V}{v_f} = \frac{0.01}{0.00106} = 9.4340 \mathrm{kg} $$

Find the amount of ice added

To find the mass of ice added \(m_{ice}\), subtract the initial mass of the liquid-vapor mixture from the final mass of the saturated liquid: $$ m_{ice} = m_{final} - m_{initial} = 9.4340 - 0.4878 = 8.9462 \mathrm{kg} $$ Hence, the amount of ice added to the piston-cylinder device is approximately \(8.9462 \mathrm{kg}\).