Question

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 1.05 kg of water decreased from 121 °C to 43.5 °C.

Property	Value	Units
Melting point	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

The number of moles of water is: 1.05 kgx 1000g 1 kg Number of moles =- 18 g/mol = 58.33mol The energy released during the change in temperature from 121°C to 43.5°C is: Total energy released = 91 +92 +93 = nCp (8)AT, + NAH cond+nC (1)AT, = nC, (g)AT, +n(-AH.) +nC (1)AT, The value of g, is: 9, = 58.33molx 33.6J/mol. °Cx(100°C-121°C) 1kJ =-41157.65 J x 1000 J =-41.16kJ The value of q, is: q2 = 58.33 molx-40.67 kJ/mol =-2372.28 kJ

The value of q, is: 92 = 58.33molx 75.3J/mol. °CX(43.5°C-100°C) = -248162.07Jx_1kJ 1000 J = -248.16 kJ The total energy released is: Total energy released = 91 +92 +93 = -41.16 kJ-2372.28 kJ - 248.16kJ = -2661.60 kJ Total energy released is –2661.60kJ.