Calculate the amount of heat, in kJ, needed to convert 15.0 g of ice at -21 C to liquid water at 75 C.
Specific heat of ice = 2.10 J/(g C); heat of fusion of ice = 334 J/g; Specific heat of water = 4.18 J/(g C)
2.10 X 15.0 X 21 = 661.5 correct sig figs 660
334 X 15.0 = 5010 correct sig figs 5010
4.18 X 15.0 X 75.0 = 4702.5 correct sig figs 4700 (first zero is significant)
660 + 5010 + 4700 = 10370 % 1000 = 10.37 kJ