Explanation:
We must convert the ice's grams into moles, and use the enthalpy of water in order to find the total amount of energy required in joules.
Formula: [tex]\[\boxed{q = n \times \Delta H}\][/tex]
[tex]\bf{Where:\begin{itemize} - \item ~\( q \) = heat\\ ~~~~ - \item \( n \) = moles\\ ~~~~ -\item \( \Delta H \) = enthalpy\\\end{itemize}}[/tex]
Converting 35 g water to moles:
[tex]35 \, \text{g} \times \left( \frac{1 \, \text{mol}}{18.02 \, \text{g}} \right) = \boxed{1.94 \, \text{mol} ~H_2O}[/tex]
The enthalpy of fusion of water is 6.01 kJ/mol, since we have both components of the equation, we can begin solving.
Solving:
[tex]q = n \times \Delta H = 1.94 \times 6.01 = \boxed{11.66 \, \text{kJ}}[/tex]
Now we need to convert it to Joules:
[tex]11.66 \, \text{kJ} \times \left( \frac{1000 \, \text{J}}{1 \, \text{kJ}} \right) = \boxed{11,660 \, \text{J}}[/tex]
We need to round to 2 significant figures since that was what was given in the problem.
[tex]\therefore \boxed{q = 12000 ~J}[/tex]
