Question

Calculate the enthalpy change associated with the conversion of 25.0 grams of ice at -4.00 C to water vapor at 110.0 C. The specific heats of ice, water, and steam are 2.09 J/g-K, and 1.84 J/g-K, respectively. For H20, Delta Hfus = 6.01 kJ/mol and Delta Hvap = 40.67 kJ/mol.

Answer 1

Ti = -4.0
Tf = 110.0
here
Cs = 2.09 J/g.oC

Heat required to convert solid from -4.0 oC to 0.0 oC
Q1 = m*Cs*(Tf-Ti)
= 25 g * 2.09 J/g.oC *(0--4) oC
= 209 J

Lf = 6.01KJ/mol =
6010 J/mol


Lets convert mass to mol
Molar mass of H2O = 18.016 g/mol
number of mol
n= mass/molar mass
= 25.0/18.016
= 1.387655 mol

Heat required to convert solid to liquid at 0.0 oC
Q2 = n*Lf
= 1.387655 mol *6010 J/mol
= 8340 J

Cl = 4.18 J/g.oC

Heat required to convert liquid from 0.0 oC to 100.0 oC
Q3 = m*Cl*(Tf-Ti)
= 25 g * 4.18 J/g.oC *(100-0) oC
= 10450 J

Lv = 40.67KJ/mol =
40670J/mol


Lets convert mass to mol
Molar mass of H2O = 18.016 g/mol
number of mol
n= mass/molar mass
= 25.0/18.016
= 1.387655 mol

Heat required to convert liquid to gas at 100.0 oC
Q4 = n*Lv
= 1.387655 mol *40670 J/mol
= 56436 J

Cg = 1.84 J/goC

Heat required to convert vapour from 100.0 oC to 110.0 oC
Q5 = m*Cg*(Tf-Ti)
= 25 g * 1.84 J/goC *(110-100) oC
= 460 J

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 209 J + 8340 J + 10450 J + 56436 J + 460 J
= 75895 J
= 75.9 KJ
Answer: 75.9 KJ