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