Question

# Based on the thermodynamic properties provided for water, determine the energy change when the temperature of...

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

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

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