Question

# Based on the thermodynamic properties provided for water, determine the energy change when the temperature of...

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

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