Question

How much heat energy, in kilojoules, is required to convert 46.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ? Express your answer to three significant figures and include the appropriate units.

The constants for H2O are shown here:

Specific heat of ice: sice=2.09 J/(g⋅∘C)

Specific heat of liquid water: swater=4.18 J/(g⋅∘C)

Enthalpy of fusion (H2O(s)→H2O(l)): ΔHfus=334 J/g

Enthalpy of vaporization (H2O(l)→H2O(g)): ΔHvap=2250 J/g

Answer 1

Ti = -18.0 oC

Tf = 25.0 oC

here

Cs = 2.09 J/g.oC

Heat required to convert solid from -18.0 oC to 0.0 oC

Q1 = m*Cs*(Tf-Ti)

= 46 g * 2.09 J/g.oC *(0--18) oC

= 1730.52 J

delta Hfusion = 334 J/g

Heat required to convert solid to liquid at 0.0 oC

Q2 = m*delta Hfusion

= 46.0g *334.0 J/g

= 15364 J

Cl = 4.18 J/g.oC

Heat required to convert liquid from 0.0 oC to 25.0 oC

Q3 = m*Cl*(Tf-Ti)

= 46 g * 4.18 J/g.oC *(25-0) oC

= 4807 J

Total heat required = Q1 + Q2 + Q3

= 1730.52 J + 15364 J + 4807 J

= 21901 J

= 21.9 KJ

Answer: 21.9 KJ