Question

Entropy change when supercooled water freezes

It is possible, with care, to cool water below its freezing point without causing the water to freeze. This is referred to as supercooling. Suppose that you did this, and had 115 g of supercooled water at -6.35 ^oC (6.35 ^oC below the freezing point) in a perfectly insulated container.

1. Now suppose that the supercooled water suddenly and spontaneously changes to a mixture of ice and water (that is, it partially freezes). Assume a constant volume, a constant pressure of 1 atm, and no gain or loss of heat to or from the container. How much of the water freezes?

Specific Heat of water = 4.184 JK^-1g^-1
Heat of fusion of ice = 333 Jg^-1

Mass of ice formed =  g

2. What entropy change occurs during the process described above?

Entropy change =  JK^-1

Answer 1

1.
Let m gram of water freezes.
Heat required to take water from -6.35 oC to 0 oC.
Q= MC*(T2-T1)
   =115*4.184*(0+6.35)
   =3055.366 J

This heat is used to convert water to ice:
m = Q/L
    = 3055.366 / 333
    = 9.2 gm

Answer: 9.2 gm

2.
Total change in Entropy=Change in entropy from heating the water +change in entropy from the water freezing

Change in entropy from heating the water = M*C*ln(T2/T1)
           = 115*4.184*ln((0+273.15)/(-6.35+273.15))
           =115*4.184*ln(273.15 / 266.8)
          = 11.3 J/K

change in entropy from the water freezing = Q/T
                                           =3055.366/273.15
                                            = 11.2 J/K
Total entropy change = 11.3+11.2 = 22.5 J/K