Question

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

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.750 kg of water decreased from 115 °C to 43.0 °C.

Answer 1

mass of water = 0.750 kg = 750 g

moles of water = 750 / 18.02 = 41.63 mol

115o C ---------- 100 oC   ----------- 43 oC

Q1 = m Cp dT

     = 41.63 x 33.6 x (115 - 100)

      = 20982.51 J

Q2 = n x delta Hvap

     = 41.63 x 40.67

     = 1693.09 kJ

Q3 = n Cp dT

     = 41.63 x 75.3 x (100 - 43)

    = 178680 J

total heat = Q1 + Q2 + Q3

                = 1892.75 kJ

Energy change = - 1.89 x 10^3 kJ