Problems and Solutions on Thermodynamics and Statistical Mechanics

Thermodynamics

(c) The change in entropy is

115

As=-= 1'4363 lo3 = 5.26 cal/K.

T 273

1117
10 kg of water at 20°C is converted to ice at -10°C by being put in
contact with a reservoir at -10°C. This process takes place at constant pres-
sure and the heat capacities at constant pressure of water and ice are 4180
and 2090 J/kg deg respectively. The heat of fusion of ice is 3.34~ lo5 J/kg.
Calculate the change in entropy of the universe.
( was co nsin)

Solution:
The conversion of water at 2OoC to ice at -10°C consists of the fol-
lowing processes. Water at 20" c -% water at ooc -+ ice at ooc 5 ice at
-lO°C, where a and c are processes giving out heat with decreases of en-
tropy and b is the process of condensation of water giving off the latent heat
with a decrease of entropy also. As the processes take place at constant
pressure, the changes of entropy are

b

273

AS, = /,,, FdT = MC,ln (g) = -2955 J/K ,

As2 = -- IQI = -^10 x 3.34 x^105 = -1.2234 x lo4 J/K ,

To 273

263

AS3 = l:r Mt 273

2dT = MC,ln - = -757 J/K.

In the processes, the increase of entropy of the reservoir due to the

absorbed heat is

10 x (4180 x 20 + 3.34 x lo5 + 2090 x 10)

263

AS, =

= 16673 J/K.

Thus, the total change of entropy of the whole system is

AS = AS, + AS2 + AS, + AS, = 727 J/K.