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Water has the following thermodynamic values: ΔH°fus of H2O = 6.02 kJ/mol ΔH°vap of H2O =...

Water has the following thermodynamic values: ΔH°fus of H2O = 6.02 kJ/mol ΔH°vap of H2O = 40.7 kJ/mol heat capacity of solid H2O = 2.09 J/g°C heat capacity of liquid H2O = 4.18 J/g°C heat capacity of gaseous H2O = 1.97 J/g°C How much energy (in kJ) is required to raise the temperature of 25.0 g of H2O from -101°C to 218°C? Enter your answer in units of kJ to three significant figures.

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Answer #1

Ti = -101.0 oC
Tf = 218.0 oC
here
Cs = 2.09 J/g.oC

Heat required to convert solid from -101.0 oC to 0.0 oC
Q1 = m*Cs*(Tf-Ti)
= 25 g * 2.09 J/g.oC *(0--101) oC
= 5277.25 J

Hf = 6.02KJ/mol =
6020J/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
= 1.3877 mol

Heat required to convert solid to liquid at 0.0 oC
Q2 = n*Hf
= 1.3877 mol *6020 J/mol
= 8353.6856 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

Hv = 40.7KJ/mol =
40700J/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
= 1.3877 mol

Heat required to convert liquid to gas at 100.0 oC
Q4 = n*Hv
= 1.3877 mol *40700 J/mol
= 56477.5755 J

Cg = 1.97 J/g.oC

Heat required to convert vapour from 100.0 oC to 218.0 oC
Q5 = m*Cg*(Tf-Ti)
= 25 g * 1.97 J/g.oC *(218-100) oC
= 5811.5 J

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
= 5277.25 J + 8353.6856 J + 10450 J + 56477.5755 J + 5811.5 J
= 86370 J
= 86.4 KJ
Answer: 86.4 KJ

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