Question

Based on the thermodynamic properties provided for water, determine the amount of energy needed for 2.60...

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

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