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 ?
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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