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72 3 THE SECOND AND THIRD LAWS
�erefore the overall entropy change for the system is
∆S = ∆S1 + ∆S2 + ∆S3 + ∆S4 + ∆S5
= nCp ,m(H2O(s)) ln(Tm
Ti
) + n∆fusH
−○
m
Tm
+ nCp ,m(H2O(l)) ln( Tb
Tm
)
+ n
∆vapH−○
m
Tb
+ nCp ,m(H2O(g)) ln( Tf
Tb
)
= 15.0 g
18.02 gmol−1
× (37.6 JK−1mol−1) × ln( 273.15 K
273.15 K − 12.0 K
)
+ 15.0 g
18.02 gmol−1
× 6.01 × 10
3 Jmol−1
273.15 K
+ 15.0 g
18.02 gmol−1
× (75.3 JK−1mol−1) × ln(273.15 K + 100.0 K
273.15 K
)
+ 15.0 g
18.02 gmol−1
× 40.7 × 10
3 Jmol−1
273.15 K + 100.0 K
+ 15.0 g
18.02 gmol−1
× (33.6 JK−1mol−1) × ln(273.15 K + 105.0 K
273.15 K + 100.0 K
)
= (+1.40... J K−1) + (+18.3... J K−1) + (+19.5... J K−1)
+ (+90.7... J K−1) + (0.372... J K−1)
= +130.4 JK−1 .
Solutions to problems
P3B.2 Because entropy is a state function, ∆S between the initial and �nal states is
the same irrespective of the path taken.�us the overall process can be broken
down into steps that are easier to evaluate.
First consider cooling the water at constant pressure to from the initial temper-
ature T to that of freezing, Tf . Entropy variation with temperature at constant
pressure is given by [3B.7–90], S(Tfinal) = S(Tinitial) + Cp ln (Tfinal/Tinitial).
�us the change in entropy, ∆S = S(Tfinal) − S(Tinitial), of this step is
∆S1 = Cp ln(
Tf
T
) = nCp ,m(H2O(l)) ln(Tf
T
)
Next consider the phase transition from liquid to solid at the freezing temper-
ature; note that freezing is just the opposite of fusion, thus ∆H2 = n(−∆fusH−○
m).
�e entropy change of a phase transition is given by [3B.4–89], ∆trsS = ∆trsH/Ttrs,
thus
∆S2 =
∆H2
Tf
= n−∆fusH
−○
m
Tf
�e ice is then heated to the �nal temperature, T . In analogy to the �rst step
∆S3 = nCp ,m(H2O(s)) ln( T
Tf
)