Question

The constants for H2O are shown here: Specific heat of ice: sice=2.09 J/(g⋅∘C) Specific heat of...

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

How much heat energy, in kilojoules, is required to convert 36.0 g of ice at −18.0 ∘C to water at 25.0 ∘C ?

Homework Answers

Answer #1

Ti = -18.0

Tf = 25.0

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)

= 36 g * 2.09 J/g.oC *(0--18) oC

= 1354.32 J

Lf = 334.0 J/g

Heat required to convert solid to liquid at 0.0 oC

Q2 = m*Lf

= 36.0g *334.0 J/g

= 12024 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)

= 36 g * 4.18 J/g.oC *(25-0) oC

= 3762 J

Total heat required = Q1 + Q2 + Q3

= 1354.32 J + 12024 J + 3762 J

= 17140 J

= 17.1 KJ

Answer: 17.1 KJ

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