Question

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 1.45 kg of water decreased from 117 °C to 23.5 °C.

Property	Value	Units
Melting point	0	°C
Boiling point	100.0	°C
ΔHfus	6.01	kJ/mol
ΔHvap	40.67	kJ/mol
cp (s)	37.1	J/mol · °C
cp (l)	75.3	J/mol · °C
cp (g)	33.6	J/mol · °C

Answer 1

Ti = 117.0 oC

Tf = 23.5 oC

Cg = 33.6 J/mol.oC

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 1450.0/18.016

= 80.484 mol

Heat released to convert vapour from 117.0 oC to 100.0 oC

Q1 = n*Cg*(Ti-Tf)

= 80.484 mol * 33.6 J/mol.oC *(117-100) oC

= 45972.4689 J

Lv = 40.67KJ/mol =

40670J/mol

Heat released to convert gas to liquid at 100.0 oC

Q2 = n*Lv

= 80.484 mol *40670 J/mol

= 3273284.8579 J

Cl = 75.3 J/mol.oC

Heat released to convert liquid from 100.0 oC to 23.5 oC

Q3 = n*Cl*(Ti-Tf)

= 80.484 mol * 75.3 J/mol.oC *(100-23.5) oC

= 463624.1397 J

Total heat released = Q1 + Q2 + Q3

= 45972.4689 J + 3273284.8579 J + 463624.1397 J

= 3782881 J

= 3783 KJ

Since it is heat released, please enter your answer with negative sign

Answer: - 3783 KJ