An ice cube tray contains enough water at \(22.0{ }^{\circ} \mathrm{C}\) to make 18 ice cubes that each have 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 \(\mathrm{H}_{2} \mathrm{O}(I)\) 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 energy for cooling the water from 22.0 °C to 0 °C

First, we have to calculate the energy required to lower the temperature of the water from 22.0 °C to 0 °C. The formula to calculate the energy is: \(Q = mcΔT\) Where: - Q is the energy required (Joules) - m is the mass of the water (grams) - c is the specific heat capacity of liquid water (4.18 J/g· °C) - ΔT is the temperature change (from 22.0 °C to 0 °C) Total mass of the water is given as 18 ice cubes each having a mass of 30.0 g, so the total mass is 18 * 30.0 g = 540 g. Now we can substitute the values and find the energy for cooling the water: \(Q_{cooling} = (540 \mathrm{~g}) (4.18 \mathrm{~J / g \cdot ^{\circ}C})(0 - 22)\)

02

Calculate the energy for converting liquid water to ice

Next, we need to calculate the energy for converting water (at 0 °C) to ice. We will use the enthalpy of fusion value: Enthalpy of fusion = 6.02 kJ/mol The molar mass of water is around 18 g/mol. We have 540 g of water which is equivalent to \(\frac{540g}{18g/mol} = 30 mol\). Now we can calculate the energy required for converting water to ice: \(Q_{fusion} = (30 \mathrm{~mol})(6.02 \mathrm{~kJ/mol})\) Convert the energy to Joules: \(Q_{fusion} = (30 \mathrm{~mol})(6.02 \mathrm{~kJ/mol}) \times 1000 \mathrm{~J/kJ}\)

03

Calculate the energy for cooling the ice from 0 °C to -5.0 °C

In this step, we will calculate the energy needed to lower the temperature of the ice from 0 °C to -5.0 °C. Using the formula \(Q = mcΔT\), we can calculate the energy required: Here, c is the specific heat capacity of ice (2.03 J/g· °C). \(Q_{cooling\_ice} = (540 \mathrm{~g}) (2.03 \mathrm{~J / g \cdot ^{\circ}C})(-5)\)

04

Calculate the total energy

Now, combining the energy required from steps 1, 2, and 3, we can calculate the total energy: \(Q_{total} = Q_{cooling} + Q_{fusion} + Q_{cooling\_ice}\)

05

Calculate the mass of CF2Cl2 needed to provide the total energy

We are given the heat of vaporization of CF2Cl2 as 158 J/g. Let m be the mass of CF2Cl2 required to provide the total energy (Q_total). Then: \(Q_{total} = (m \mathrm{~g})(158 \mathrm{~J/g})\) Now we can calculate the mass of CF2Cl2 needed: \(m = \frac{Q_{total}}{158 \mathrm{~J/g}}\) Calculate m using the Q_total value found in Step 4 to get the final answer.

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