Question

Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.350 kg of water decreased from 123 °C to 30.0 °C.

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

Answer 1

To Find: the energy change when the temperature of 0.350 kg of water decreased from 123 °C to 30.0 °C.

Solution:

Number of moles of water

= (Mass of water /molar mass of water)

= (0.350 x 10³ g)/( 18 gmol^-1)

= 19.4444 mol

This problem can be solved through following steps:

Step 1: Energy change when water vapor is cooled from 123 to saturated vapor 100

H₁ = Cp (T₁ - T₂)

=( 33.6 Jmol ^-1-1) x (100 - 123)

= - 772.8 J/mol

Step 2: Calculate enthalpy of vaporization to condence all vapor to form saturated liquid at 100 which is given in the problem.

H₂ = - 40.67 kJ/mol

Step 3: Enthaply change when liquid is cooled from saturated liquid at 100 to 30

H3 = Cp (T₁- T₂)

= 75.3 Jmol^-1-1 (30 - 100)

= - 5271 J/mol

Hence total enthalpy change H = H₁ + ₂₃ = - 46.7138 kjmol^-1

For 19.444 mol of water enthalpy change = x n = - 908.322 kJ

-ve sign indicates that heat is given by the system.