Question

40. The molar entropy of ice at 0 ◦C is 160.2 J/(mole K). Using Lvaporization =

40680 J/mole, Lmelting = 6006J/mole, find the molar entropy of water (a) as

a liquid at 0 ◦C, (b) as a liquid at 100 ◦C, (c) as a vapor at 100 ◦C, (d) in ice

melt at 0 ◦C that is 40% ice and 60% liquid.

Answer 1

1.molar entroy of liquid at 0 deg.c= heat of melting/ Temperature ( o deg.c) in K= 6006 /273= 22 J/mole.K

2. first the water need to be converted from liquid at 0 deg.c to liquid at 100 deg.c. entropy change during the process

= Cpliquid* ln (T2/T1)= 75.4 * ln (100+273)/(0+273)= 23.53 J/mole.K

total entropy of liquid at 100 deg.c= 22+23.53= 45.53 J/mole.K

3. at 100 deg.c, liquid water becomes vapor by supplying latent heat, entropy change = 40680/373 =109 J/mole.K

total entropy change = 45.53+109= 154.53 J/mole.K

4. for converting ice to 60% liquid, heat to be supplied = 0.6*6006= 3604 J

entropy change= 3604/273= 13.2 J/mole.K