Question

Based on the thermodynamic properties provided for water, determine the amount of energy released for 160.0 g of water to go from 83.0 °C to -17.5 °C.

Property	Value	Units
Melting point	0.0	°C
Boiling point	100.0	°C
ΔHfusΔHfus	6.01	kJ/mol
ΔHvapΔHvap	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 = 83.0 oC

Tf = -17.5 oC

Cl = 75.3 J/mol.oC

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 83.0/18.016

= 4.607 mol

Heat released to convert liquid from 83.0 oC to 0.0 oC

Q1 = n*Cl*(Ti-Tf)

= 4.607 mol * 75.3 J/mol.oC *(83-0) oC

= 28793.3892 J

Lf = 6.01KJ/mol =

6010J/mol

Heat released to convert liquid to solid at 0.0 oC

Q2 = n*Lf

= 4.607 mol *6010 J/mol

= 27688.1661 J

Cs = 37.1 J/mol.oC

Heat released to convert solid from 0.0 oC to -17.5 oC

Q3 = n*Cs*(Ti-Tf)

= 4.607 mol * 37.1 J/mol.oC *(0--17.5) oC

= 2991.1051 J

Total heat released = Q1 + Q2 + Q3

= 28793.3892 J + 27688.1661 J + 2991.1051 J

= 59472.6604 J

= 59.5 KJ

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

Answer: - 59.5 KJ