Explanation:
The given reaction is as follows.
[tex]H_{2}O(s) \rightleftharpoons H_{2}O(l)[/tex]
Hence, using standard values heat of fusion will be calculated as follows.
[tex]\Delta H_{fus} = \Delta H^{o}_{f}(H_{2}O(l)) - \Delta H^{o}_{f}(H_{2}O(s))[/tex]
= -285.8 kJ/mol - (-291.8 kJ/mol)
= 6 kJ/mol
Therefore, heat released from beverage will be calculated as follows.
q = [tex]m \times C \times \Delta T[/tex]
= [tex]355 g \times 4.18 J/g^{o}C \times (0 - 25)^{o}C[/tex]
= -37133 J
or, = -37.133 kJ (as 1 J = 0.001 kJ)
Now, heat gained by ice = - heat released by the beverage
= - (-37.133 kJ)
= 37.133 kJ
Therefore, calculate the mass of ice as follows.
Mass of ice = [tex]37.133 kJ \times \frac{\text{1 mol ice}}{6 kJ} \times \frac{18 g}{1 mol}[/tex]
= 111.40 g
Thus, we can conclude that mass of ice is 111.40 g.
