Energy can be removed from water as heat at and even below the normal freezing p

Question

Energy can be removed from water as heat at and even below the normal freezing point (0.0°C at atmospheric pressure) without causing the water to freeze; the water is then said to be supercooled. Suppose a 3.00 g water drop is supercooled until its temperature is that of the surrounding air, which is at 5.40°C. The drop then suddenly and irreversibly freezes, transferring energy to the air as heat. What is the entropy change for the drop? (Hint: Use a three-step reversible process as if the water were taken through the normal freezing point.) The specific heat of ice is 2220 J/kg · K; the specific heat of liquid water is 4190 J/kg · K; and the heat of fusion of water is 333 kJ/kg.

Explanation / Answer

here,

mass of water , m = 3 gm

initial temprature , Ti = -5.4 deg C

energy , E = 3 *(4.186 * 5.4 + 334)

E = 1069.81 J

Temprature , Ti = 273 - 5.4 K

Ti = 267.6 K

entropy change for the drop = energy/T

entropy = 1069.81 / 267.6

entropy = 4 J/K

the change in entropy of the drop is 4 J/K

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