# Entropy Change of Supercooled Water Freezing Spontaneously

1. Jan 12, 2009

### sanitykey

1. The problem statement, all variables and given/known data

A 100 g mass of supercooled water at -2ºC in thermal contact with surroundings also at -2ºC
freezes spontaneously.
Calculate the entropy change of the universe assuming that the surroundings act as a large
temperature reservoir. [The specific heat capacity of ice is 2090 J kg-1 K-1, the specific heat
capacity of water is 4183 J Kg-1 K-1, and the latent heat of fusion of water is 334.7 kJ kg-1.]

Is this process reversible?

2. Relevant equations

Change in entropy of the universe = change in entropy of the system + change in entropy of the surroundings

dS = dQ/T (for a reversible process)

3. The attempt at a solution

latent heat of fusion = l = 334700J/Kg
T2 = freezing temperature of water = 273K
T1 = -2 degrees Celcius = 271K
mass = m = 0.1Kg

Change in entropy of the system = (-)m*l/T2 = (0.1*334700)/273 = -122.60J/K

Change in entropy of the surroundings = (+)m*l/T1 = (0.1*334700)/271 = 123.51J/K

Change in entropy of the universe = 123.51 -122.60 = 0.9J/K

The process is irreversible

I have a feeling this is wrong because i haven't used the heat capacities of ice and water (not sure how to) any help would be appreciated, thanks.

2. Jan 12, 2009

### Andrew Mason

Why would the final temperature be 273K rather than 271 K?

If the liquid water at -2C (271K) turns to ice it must release heat into the surroundings (33.47 KJ). The surroundings are at the same temperature, 271K, and do not change temperature as a result of the change of state of the water and the release of this heat. So how does any part of this occur at a temperature other than 271K?

AM

3. Jan 12, 2009

### sanitykey

Thanks for your response, i'm still not quite sure how to work this out though. The reason i used 273K was i was following a (what i thought to be) similar example i found at:

(the green box about 1/3 down the page)

If i use 271K for both processes then would that make the change in entropy of the universe 0? Would that make it a reversible process?

How do i include the heat capacities? I mean if the temperature of the water/ice isn't actually changing i don't see how i can use them?

Any help is much appreciated.

4. Jan 12, 2009

### Mapes

You need to apply the equations to a reversible process, which spontaneous freezing isn't. An appropriate process would be:

1) Take the supercooled water up to 0°C reversibly
2) Freeze the water (at 1 atm, 0°C is the only temperature for which freezing is reversible)
3) Take the ice back down to -2°C reversibly

The outcome is the same, but the whole process is now reversible. The heating and cooling steps are where the heat capacities come in. You can find more information on this technique in most thermo texts.

5. Jan 13, 2009

### Andrew Mason

This is correct. The change in entropy is defined as the integral of dQ/T for the reversible path. Since ice will never melt at -2C in these surroundings, the spontaneous freezing of supercooled water at -2C is not reversible.

The path Mapes has given can be reversible (for example the water is heated and the ice is then cooled using a Carnot heat pump between the water and the surroundings). The change in entropy of the water is:

$$\Delta S_{water} = \int_{-2}^{0}mCdT/T + \left{(}-\frac{Q_{fusion}}{T_{0}}\right{)} + \int_{0}^{-2}mCdT/T = -Q_{fusion}/T_{0} = -33.47/273$$

since the integrals cancel each other. Similarly, the change in entropy of the surroundings (temperature constant at 271K) is:

$$\Delta S_{surr} = mC\Delta T/T + Q_{fusion}/T + mC(-\Delta T)/T = Q_{fusion}/T = 33.47/271$$

So the total change is:

$$\Delta S = \Delta S_{water} + \Delta S_{surr} = -33.47/273 + 33.47/271 > 0$$

Since the change in entropy of the universe is positive, the process is not reversible.

AM

6. Jan 27, 2009

### sanitykey

I got caught up in exams recently and didn't get a chance to respond to this but it's a perfect explanation thanks!