Question

Based on the thermodynamic properties provided for water, determine the amount of energy needed for 2.60 kg of water to go from -4.00 °C to 74.0 °C.

Property	Value	Units
Melting point	0.0	°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 = -4.0 oC

Tf = 74.0 oC

here

Cs = 37.1 J/mol.oC

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 2600.0/18.016

= 144.3162 mol

Heat required to convert solid from -4.0 oC to 0.0 oC

Q1 = n*Cs*(Tf-Ti)

= 144.3162 mol * 37.1s J/mol.oC *(0--4) oC

= 21416.5187 J

Lf = 6.01KJ/mol =

6010J/mol

Heat required to convert solid to liquid at 0.0 oC

Q2 = n*Lf

= 144.3162 mol *6010 J/mol

= 867340.1421 J

Cl = 75.3 J/mol.oC

Heat required to convert liquid from 0.0 oC to 74.0 oC

Q3 = n*Cl*(Tf-Ti)

= 144.3162 mol * 75.3l J/mol.oC *(74-0) oC

= 804158.5258 J

Total heat required = Q1 + Q2 + Q3

= 21416.5187 J + 867340.1421 J + 804158.5258 J

= 1692915 J

= 1693 KJ

Answer: 1693 KJ