How much heat must be removed to freeze 92 moles of water at its freezing point? The heat of crystallization of water equals -6.02kJ/mol.

Answer 1

Hi, Carmen,

The value of -6.02kJ/mol is negative, so that means energy is being removed. It is the "Heat of Crystallization," which is a term that refers to solidification. In the case of water, it is the heat that needs to be removed in order to freeze (crystallize) it.

The units are helpful. -6.02 kJ/mole is the amount of energy removed to freeze each mole of water. So the calculation becomes:

92 moles H2O * -6.02 kJ/mole = -554 kJ (the moles cancel out).

The phrase "at its freezing point" simply means the water is already at 0 degrees C, so there is no other energy that needs to be removed other than the crystallization energy. So 554 kJ is removed to freeze the water (550 kJ if you use significant digits).

Extra credit

If, for example, the water started at 10 deg C, we would need to cool it to 0 deg C before using the heat of crystallization. The specific heat of liquid water is 75.4 J/moleC, or 0.0754 kJ/moleC. We would need to remove heat to get to 0 C before using the heat of crystallization. Note the huge difference in the molar heat capacities! Water molecules resist being told to freeze, so to speak. If we do this calculation:

92 moles (0C - 10C)*0.0754 kJ/moleC = -69 kJ is the heat removed to cool 92 moles of water from 10 to 0 deg C.

Add the two to get the total energy that needs to be removed to take 92 moles of water at 10 deg C to a large ice cube:

-554 kJ

-69 kJ

-623 kJ

For fun, let's calculate the size of this ice, assuming we make it into a cube. Calculate 92 moles of water as grams, using the molar mass of water (18 g/mole)

92 moles * 18g/mole = 1656 g (ignoring sig figs).

The density of ice at 0 C is 0.9340 g/cm^3. Liquid water density is 1 g/cm^3, so that is why ice floats - it is less dense.

Next: 1656 g = 1773 cm^3 of ice is formed from 92 moles of water.

0.9340 g/cm^3

There are 16.39 cm^3 per in^3, so

1773 cm^3 / (16.39 cm^3/in^3) = 108.2 in^3

Assume that this is in the shape of a cube. The volume of a cube is X^3, where X is one of the dimensions.

Set X^3 = 108.2 in^3, then X = 4.8 inches.

We have a cube of ice 4.8" x 4.8" x 4.8" = 108.2 in^3

Fit that in your glass!

I hope this helps,

Bob

Answer 2

At the freezing point, changing liquid to solid represents a phase change, and there is NO CHANGE IN TEMP.

q = m∆Hfusion

q = heat = ?

m = mass = 92 moles

∆Hfusion = 6.02kJ/mol

q = (92 mol)(6.02 kJ/mol)

q = 553.8 kJ = 550 kJ (3 sig.figs.) of heat must be removed (sign would be negative)