Question

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.750 kg of water decreased from 111 °C to 51.0 °C.

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

Answer 1

Amount of energy change ,

Q = heat change for conversion of vapour at 111 ^oC to vapour at 100 ^oC + heat change for conversion of steam at 100 ^oC to water at 100 ^oC + heat change for conversion of water at 100^oC to water at 51.0 ^oC

Amount of heat released , Q = mcdt + mL + mc'dt

                                              = m(cdt + L + c'dt' )

Where

m = mass of steam = 0.750kg x(1000g/kg) = 750 g

c = Specific heat of steam = 33.6 J/mol-K x (1mol/18g) =1.87 J/g degree C

c' = Specific heat of water = 75.3 J/mol-K x (1mol/18g) = 4.183 J/g degree C

L = Heat of Vaporization of water = 40.67 kJ/mol x ( 1000J/kJ) x (1 mol/18 g )= 2259.4 J/g

dt = 111-100 = 11^oC

dt' = 100 -51.0=49.0^oC

Plug the values we get Q = m(cdt + L + c'dt ' ) = 1863.7x10³ J = 1863.7 kJ