An ice cube tray contains enough water at \(22.0^{\circ} \mathrm{C}\) to make 18 ice cubes that each has a mass of 30.0 \(\mathrm{g} .\) The tray is placed in a freezer that uses \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) as a refrigerant. The heat of vaporization of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) is 158 $\mathrm{J} / \mathrm{g}\( . What mass of \)\mathrm{CF}_{2} \mathrm{Cl}_{2}$ must be vaporized in the refrigeration cycle to convert all the water at $22.0^{\circ} \mathrm{C}\( to ice at \)-5.0^{\circ} \mathrm{C} ?$ The heat capacities for \(\mathrm{H}_{2} \mathrm{O}(s)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) are 2.03 \(\mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) and 4.18 $\mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}$ , respectively, and the enthalpy of fusion for ice is 6.02 \(\mathrm{kJ} / \mathrm{mol} .\)

Step by step solution

01

Calculate the total mass of the water

We know that the ice cube tray can make 18 ice cubes, each with a mass of 30.0 g. So, the total mass of the water can be calculated as: Total mass of water = Number of ice cubes × Mass of each ice cube Total mass of water = 18 × 30.0 g = 540.0 g

02

Calculate the heat needed to cool the water from 22.0°C to 0°C

To calculate the heat needed to cool down the water to 0°C, we will use the formula: Heat (Q) = Mass (m) × Specific Heat Capacity (c) × Temperature difference (ΔT) Given that the specific heat capacity of water (H2O(l)) is 4.18 J/(g·°C): Q1 = 540.0 g × 4.18 J/(g·°C) × (0 - 22.0)°C Q1 = 540.0 g × 4.18 J/(g·°C) × -22.0°C Q1 = -50,173.6 J; we take the absolute value, as we are considering only the magnitude of heat exchange: Q1 = 50,173.6 J

03

Calculate the heat needed to convert the water at 0°C into ice at 0°C

For freezing the water at 0°C to ice at 0°C, we will use the formula: Q2 = mass × (ΔHf / molar mass of H2O) where ΔHf is the enthalpy of fusion, and the molar mass of H2O is 18.015 g/mol. In this case, ΔHf = 6.02 kJ/mol. Q2 = 540.0 g × (6.02 kJ/mol) / 18.015 g/mol Q2 = 540.0 g × (6020 J/mol) / 18.015 g/mol Q2 = 180,569.1 J

04

Calculate the heat needed to cool the ice from 0°C to -5.0°C

To calculate the heat needed to cool down the ice from 0°C to -5.0°C, we will use the formula: Heat (Q) = Mass (m) × Specific Heat Capacity (c) × Temperature difference (ΔT) Given that the specific heat capacity of ice (H2O(s)) is 2.03 J/(g·°C): Q3 = 540.0 g × 2.03 J/(g·°C) × (0 - (-5.0))°C Q3 = 540.0 g × 2.03 J/(g·°C) × 5.0°C Q3 = 5,490.7 J

05

Calculate the mass of refrigerant needed

Now, we will sum up the heats calculated in steps 2, 3, and 4, and divide the total heat by the heat of vaporization of CF2Cl2 to find the mass of refrigerant needed: Total heat (Q_total) = Q1 + Q2 + Q3 Q_total = 50,173.6 J + 180,569.1 J + 5,490.7 J Q_total = 236,233.4 J Next, we can find the mass of CF2Cl2 required using the heat of vaporization (Lv = 158 J/g): Mass of CF2Cl2 = Q_total / Lv Mass of CF2Cl2 = 236,233.4 J / 158 J/g Mass of CF2Cl2 = 1,494.6 g Therefore, 1,494.6 grams of CF2Cl2 must be vaporized in the refrigeration cycle to convert all the water at 22.0°C to ice at -5.0°C.

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