Help me solve this??

Question

How much heat is required to warm 32.07g of ice (solid water) at -10.48°C to liquid water at 21.49°C ?

Specific Heat (ice) = 37.6 J/K*mol; Specific Heat (water) = 75.3 J/K*mol; Specific Heat (steam) = 35.8 J/K*mol; .

ΔHfusion (ice) = 6.01 kJ/mol; ΔHvaporization (water) = 40.67 kJ/mol; ΔHsublimation (ice) = 46.7 kJ/mol.

Answer 1

q=m⋅c⋅ΔT

Using the specific heat capacity of ice as 2.09 J/g*C

q= 32.07g * 2.09J/g*C * [0-(-10.48)]C

q= 702.44 J

q=m⋅ΔHfus

ΔHfus = 334 J/g

q= 32.07g * 334J/g = 10711.4 J

q=m⋅c⋅ΔT

Using the specific heat capacity of water as 4.18 J/g*C

q= 32.07g * 4.18J/g*C * [21.49-(0)]C

q= 2880.79 J

Total heat will require adding all three together:

702.44 + 10711.4 + 2880.79 = 14294.63 J

Round to 3 sig figs and express in kJ:

14.3 kJ