What is the final temperature of the water if 35.1 kj of energy are added to 100.0 g of ice at -4.5° c?

Because we don't know the final state, let's test if it goes through melting stage.

Q = mΔHfus, where ΔHfus for water is 333 kJ/kg

Q = 333 kJ/kg(0.1 kg) = 33.3 kJ

So, it means the final state is liquid water. The remaining Q would now be: 35.1 - 33.3:
Q = mCp,iceΔT + mCp,waterΔT
Q = (0.1 kg)(2.108 kJ/kg°C)(0 - -4.5) + 0.1 kg(4.187 kJ/kg°C)(T - 0) = 35.1 - 33.3
Solving for T,
T = 2.03°C