Question

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 \(6.02 \mathrm{kJ} / \mathrm{mol}\).

Short answer

To convert all the water at \(22.0^\circ\text{C}\) to ice at \(-5.0^\circ\text{C}\) in the given scenario, \(1498.0\text{g}\) of \(\text{CF}_2\text{Cl}_2\) must be vaporized in the refrigeration cycle.

Step-by-step solution

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 \(= 540\text{g}\)

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 \(q1 = 540\text{g} \times 4.18\frac{\text{J}}{\text{g}\cdot^\circ\text{C}} \times (0^\circ\text{C} - 22.0^\circ\text{C})\) \(q1 = -49603.2\text{J}\)

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 \(q2 = 30\text{mol} \times 6.02 \frac{\text{kJ}}{\text{mol}}\) \(q2 = -180.6\text{kJ}\) \(q2 = -180600\text{J}\)

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 \(q3 = 540\text{g} \times 2.03\frac{\text{J}}{\text{g} \cdot^\circ\text{C}} \times (-5.0^\circ\text{C} - 0^\circ\text{C})\) \(q3 = -5478.1\text{J}\)

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 \(= q1 + q2 + q3\) \(Q = -49603.2\text{J} + (-180600\text{J}) + (-5478.1\text{J})\) \(Q = -236681.3\text{J}\)

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 \(-5.0^\circ\text{C}\).

Key concepts

Heat Capacity

Heat capacity is a measure of the amount of heat energy required to change the temperature of a substance by a certain amount. It helps us understand how different substances respond to heat. Specifically, the **specific heat capacity** is a critical concept which is defined as the amount of heat needed to raise the temperature of 1 gram of a substance by 1 degree Celsius. In this exercise,

• the specific heat capacity of water (liquid) is used as it changes temperature from \(22.0^{\circ}\text{C}\) to \(0^{\circ}\text{C}\).
• A different specific heat capacity is applied to ice as it cools from the freezing point down to \(-5.0^{\circ}\text{C}\).

This concept allows us to capture and calculate the heat exchanged during temperature changes, which is crucial for determining the total heat loss needed in the entire phase change process.

Enthalpy of Fusion

The enthalpy of fusion is the amount of energy required to change a substance from solid to liquid at its melting point, without changing its temperature. In the reverse process, it represents the energy released when a liquid becomes solid. For water, the fundamental enthalpy of fusion is used to calculate the energy involved in converting liquid water into ice at \(0^{\circ}\text{C}\).
This energy is necessary because even at the melting point, the internal structure must overcome attractive forces to switch phases.

• The enthalpy of fusion of ice is given as \(6.02 \, \text{kJ/mol}\).
• This value helps determine how much energy must be removed per mole of water as we freeze it.

Considering the enthalpy of fusion is vital for capturing the total heat loss when converting water into ice.

Vaporization

Vaporization is the process of turning a liquid into vapor, and it requires energy known as the heat of vaporization. Conversely, this energy can be released when vapor condenses back into a liquid. In the refrigeration process discussed in the exercise,

• the heat of vaporization pertains to \(\text{CF}_2\text{Cl}_2\).
• This determines how much energy is carried away by one gram of \(\text{CF}_2\text{Cl}_2\) as it vaporizes.

With this data, we can calculate how much of the refrigerant is necessary to absorb the full heat loss when cooling and freezing the water. Understanding vaporization is essential in determining the refrigerant requirements to achieve the desired temperature change.

Phase Change

Phase change refers to the transition of a substance from one state of matter to another, such as from liquid to solid or vice versa. This exercise involves several key phase changes:

1. The phase change from liquid water at \(22.0^{\circ}\text{C}\) to ice at \(0^{\circ}\text{C}\).
2. Additionally, further cooling of the solid ice from \(0^{\circ}\text{C}\) to \(-5.0^{\circ}\text{C}\).

Phase changes are accompanied by energy exchanges due to breaking or forming of molecular bonds. During these changes,

temperature remains constant at points of fusion or solidification because the energy change involves restructuring, not heating or cooling.

The accurate assessment of energy exchanges during these transitions, both isothermally and across temperature changes, helped determine the entire cooling requirement in the exercise.