How much heat (kJ) is required to completely melt 34.9 grams of ice at 0 oC ?

Answer 1

It is called "Heat of Fusion" of water: 79.72 Cal/g or 333.55 J/g.
The requested answer is:
(333.55 J/g)*(34.9 g) =
---You do the math.
Please read the reference:
http://en.wikipedia.org/wiki/Heat_of_fusion