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Answer :
To find the amount of heat that must be removed, follow these steps:
1. First, determine the number of moles of water. The number of moles is given by the formula
[tex]$$
n = \frac{m}{M},
$$[/tex]
where [tex]$m$[/tex] is the mass of water and [tex]$M$[/tex] is the molar mass of water. For 40.0 g of water and [tex]$M \approx 18.015\, \text{g/mol}$[/tex],
[tex]$$
n = \frac{40.0\, \text{g}}{18.015\, \text{g/mol}} \approx 2.22\, \text{mol}.
$$[/tex]
2. Next, calculate the heat that must be removed using the molar heat of fusion. The heat removed is given by
[tex]$$
q = n \cdot \Delta H_{\text{fus}},
$$[/tex]
where [tex]$\Delta H_{\text{fus}} = 6.01\, \text{kJ/mol}$[/tex]. Substituting the values,
[tex]$$
q \approx 2.22\, \text{mol} \times 6.01\, \text{kJ/mol} \approx 13.34\, \text{kJ}.
$$[/tex]
Thus, approximately [tex]$13.4\, \text{kJ}$[/tex] of heat must be removed to freeze the tray of ice cubes.
The correct answer is [tex]$\boxed{13.4\, \text{kJ}}$[/tex].
1. First, determine the number of moles of water. The number of moles is given by the formula
[tex]$$
n = \frac{m}{M},
$$[/tex]
where [tex]$m$[/tex] is the mass of water and [tex]$M$[/tex] is the molar mass of water. For 40.0 g of water and [tex]$M \approx 18.015\, \text{g/mol}$[/tex],
[tex]$$
n = \frac{40.0\, \text{g}}{18.015\, \text{g/mol}} \approx 2.22\, \text{mol}.
$$[/tex]
2. Next, calculate the heat that must be removed using the molar heat of fusion. The heat removed is given by
[tex]$$
q = n \cdot \Delta H_{\text{fus}},
$$[/tex]
where [tex]$\Delta H_{\text{fus}} = 6.01\, \text{kJ/mol}$[/tex]. Substituting the values,
[tex]$$
q \approx 2.22\, \text{mol} \times 6.01\, \text{kJ/mol} \approx 13.34\, \text{kJ}.
$$[/tex]
Thus, approximately [tex]$13.4\, \text{kJ}$[/tex] of heat must be removed to freeze the tray of ice cubes.
The correct answer is [tex]$\boxed{13.4\, \text{kJ}}$[/tex].
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