66. We consider a three-step reversible process as follows: the supercooled water drop (of mass m) starts

at state 1 (T

1

= 268 K),moves on to state 2 (still in liquid form but at T

2

= 273 K),freezes to

state 3 (T

3

= T

2

),and then cools down to state 4 (in solid form,with T

4

= T

1

). The change in

entropy for each of the stages is given as follows: ∆S

12

= mc

w

ln(T

2

/T

1

),∆S

23

=

−mL

F

/T

2

,and

∆S

34

= mc

I

ln(T

4

/T

3

) = mc

I

ln(T

1

/T

2

) =

−mc

I

ln(T

2

/T

1

). Thus the net entropy change for the water

drop is

∆S

=

∆S

12

+ ∆S

23

+ ∆S

34

= m(c

w

− c

I

) ln

T

2

T

1



−

mL

F

T

2

=

(1.00 g)(4.19 J/g

·K − 2.22 J/g·K) ln

273 K

268 K



−

(1.00 g)(333 J/g)

273 K

=

−1.18 J/K .