Question

4. Steam is condensed at 100^oC to water which is cooled to 0^oC and frozen to ice. What are the molar enthalpy and entropy changes for this process? Assume that the average specific heat of water is 4.2JK^-1 g^-1 and the enthalpy of vaporization at the boiling point and the enthalpy of fusion at the freezing point are 2258.1 and 333.5 J g^-1 respectively.

(Note: Convert Jg^-1 to J mol^-1)

Answer 1

The changes that take place are

H2O (g) ------------> H2O(l) --------> H2O (l) -----------> H2O (s)

100C q1 100C q2 0C q3 0C The total heat for the change = q1 + q2 + q3

Let us assume we hav etaken 1 mole of water (18g/mol)

thus

q1 = mass x enthalpy of vaporization

= 18g x 2258.1 J /g

= 40645.8J

q2 = mass x specific heat of water x difference in temperature

= 18g x 4.2J/K.g x 100

= 7560 J

q3 = mass x enthalpy of fusion

= 18g x 33.5J/g

= 603J

Thus the total heat absorbed per mole(18g) of water =

=40645.8J +7560 J + 603 J

= 48808. 8 J

= 48.808 kJ

The entropy of process = heat / temperature

= 48808 J / 273K

= 178.78 J/K