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 total mass of water

To find the total mass of water, multiply the mass of each ice cube (30.0 g) by the number of cubes (18): \(30.0\,\mathrm{g} \times 18 = 540\,\mathrm{g}\). Next, we need to calculate the heat required to lower the temperature of the water to its freezing point (0°C).

02

Calculate heat to lower water temperature to 0°C

Using the heat capacity for liquid water and the formula \(q = mcΔT\), we have: \(q_1 = (540\,\mathrm{g})(4.18\,\mathrm{J}/\mathrm{g} \cdot{ } ^{\circ}\mathrm{C})(22.0 - 0)\,\mathrm{J} = 49{,}514\,\mathrm{J}\). Now, we will calculate the heat needed to change the water from liquid to solid at 0°C using the enthalpy of fusion.

03

Calculate heat to change water to ice

First, we need to convert the mass of water to moles. We know that the molar mass of water is \(18.015\,\mathrm{g}/\mathrm{mol}\). So the number of moles is \(\frac{540\,\mathrm{g}}{18.015\,\mathrm{g}/\mathrm{mol}} = 30\,\mathrm{mol}\). Next, we multiply the moles by the enthalpy of fusion per mole: \(q_2 = 30\,\mathrm{mol} \times 6.02\,\mathrm{kJ}/\mathrm{mol} = 180.6\,\mathrm{k}\mathrm{J}\). Converting this to joules, we get \(180{,}600\,\mathrm{J}\). Afterward, we need to calculate the heat necessary to lower the ice temperature to -5°C.

04

Calculate heat to lower ice temperature to -5°C

Using the heat capacity for solid water and the formula \(q = mcΔT\), we have: \(q_3 = (540\,\mathrm{g})(2.03\,\mathrm{J}/\mathrm{g} \cdot{ } ^{\circ}\mathrm{C})(0 - (-5))\,\mathrm{J} = 5{,}481\,\mathrm{J}\). Now, by adding all the heats, we can find the total heat removed from the water.

05

Calculate total heat removed from water

The total heat removed from the water is: \(q_\text{total} = q_1 + q_2 + q_3 = 49{,}514\,\mathrm{J} + 180{,}600\,\mathrm{J} + 5{,}481\,\mathrm{J} = 235{,}595\,\mathrm{J}\). Finally, we will determine the mass of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) needed.

06

Calculate mass of \(CF_2Cl_2\) needed

We know that the heat of vaporization for \(\mathrm{CF}_{2}\mathrm{Cl}_{2}\) is \(158 \mathrm{J} / \mathrm{g}\). To find the mass, we divide the total heat by the heat of vaporization: \(m_\text{\(CF_2\)}\) = \(\frac{235{,}595\,\mathrm{J}}{158\,\mathrm{J}/\mathrm{g}} = 1491.51\,\mathrm{g}\). Therefore, around \(1491.51\,\mathrm{g}\) of \(\mathrm{CF}_{2} \mathrm{Cl}_{2}\) must be vaporized to convert all the water at \(22.0^{\circ}\mathrm{C}\) to ice at \(-5.0^{\circ}\mathrm{C}\).

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