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 begin, we will calculate the total mass of the water to be converted to ice by multiplying the mass of a single ice cube by the number of ice cubes: Total mass of the water = Number of ice cubes x Mass of a single ice cube Total mass of the water \(= 18 \times 30.0\text{g}\) Total mass of the water \(= 540\text{g}\)

02

Calculate Heat Loss during Cooling of Water

Now, we will calculate the amount of heat lost during the cooling of water from \(22.0^\circ\text{C}\) to \(0^\circ\text{C}\): q1 = mass x specific heat capacity of water x temperature change \(q1 = 540\text{g} \times 4.18\frac{\text{J}}{\text{g}\cdot^\circ\text{C}} \times (0^\circ\text{C} - 22.0^\circ\text{C})\) \(q1 = -49603.2\text{J}\)

03

Calculate Heat Loss during Freezing of Water

Next, we will calculate the amount of heat lost when the water freezes. To do this, we need to first find the number of moles of water and then multiply this by the enthalpy of fusion of ice. Number of moles of water \(= \frac{540\text{g}}{18.02\frac{\text{g}}{\text{mol}}}\) Number of moles of water \(= 30\text{mol}\) Now, we can calculate the heat loss during freezing: q2 = Number of moles x Enthalpy of fusion \(q2 = 30\text{mol} \times 6.02 \frac{\text{kJ}}{\text{mol}}\) \(q2 = -180.6\text{kJ}\) \(q2 = -180600\text{J}\)

04

Calculate Heat Loss during Cooling of Ice

Finally, we will calculate the amount of heat lost during the cooling of ice from \(0^\circ\text{C}\) to \(-5.0^\circ\text{C}\): q3 = mass x specific heat capacity of ice x temperature change \(q3 = 540\text{g} \times 2.03\frac{\text{J}}{\text{g} \cdot^\circ\text{C}} \times (-5.0^\circ\text{C} - 0^\circ\text{C})\) \(q3 = -5478.1\text{J}\)

05

Calculate Total Heat Loss

Now, we can find the total heat loss by summing the heat losses from step 2, step 3, and step 4. Total Heat Loss, Q \(= q1 + q2 + q3\) \(Q = -49603.2\text{J} + (-180600\text{J}) + (-5478.1\text{J})\) \(Q = -236681.3\text{J}\)

06

Calculate Mass of CF2Cl2 to be Vaporized

Finally, we will use the heat of vaporization of \(\text{CF}_2\text{Cl}_2\) to determine the mass of \(\text{CF}_2\text{Cl}_2\) that must be vaporized: Mass of \(\text{CF}_2\text{Cl}_2\) \(= \frac{-Q}{\text{Heat of vaporization}}\) \(= \frac{236681.3\text{J}}{158\frac{\text{J}}{\text{g}}}\) Mass of $\text{CF}_2\text{Cl}_2 = 1498.0\text{g}\) Therefore, \(1498.0\text{g}\) of \(\text{CF}_2\text{Cl}_2\) must be vaporized in the refrigeration cycle to convert all the water at \(22.0^\circ\text{C}\) to ice at \(-5.0^\circ\text{C}\).

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