calculate the total heat needed to convert 3.035 grams of ice at -7.326C to water at 8.722x10^1C?

Question

The melting point at 1 atm 0.0C
Cwater(s)=2.09 J/g C
Cwater(l)=4.21 J/g C
delta H fus=6.02 kJ/mol
answer in joules

Answer 1

The conversion consists of 3 steps:
Ice is heated to 0 deg C
Ice is melted at 0 deg C
Water is heated to 87.2 deg C.
Step 1: 3.035*2.09*7.327 = appx 44 joules
Step 2: 3.035*6020*1/18= appx 1000 joules
Step 3: 3.035*4.21*87.2 = axxp 1100 joules.