Total Entropy of the Freezing of Water

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Ryan Fang 1D
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Joined: Sat Jul 22, 2017 3:00 am

Total Entropy of the Freezing of Water

Postby Ryan Fang 1D » Sun Feb 11, 2018 1:42 pm

Can someone explain, using calculations, how the total entropy of the universe increases during the freezing of water at 0 degrees Celsius, thus making it spontaneous? How would you calculate the change in entropy of the surroundings? I don't understand how to get a value other than 0 for the total entropy.

Jason Muljadi 2C
Posts: 62
Joined: Thu Jul 27, 2017 3:01 am

Re: Total Entropy of the Freezing of Water

Postby Jason Muljadi 2C » Sun Feb 11, 2018 2:45 pm

Generally when at equilibrium, delta Stot = 0. Keep in mind that the equation for total entropy is delta Stot = delta Ssurr + delta Ssys.

With water, you want to take into account that there is an unfavorable phase change, so you would have to take into account - delta Sfus and you have to take into consideration delta Stot as the water goes from a liquid to a solid. I'm not sure how the total entropy of the universe increased during the freezing of water when there is less disorder/freedom as a liquid goes into a solid. You'd have to specify what other things are going on. But to calculate, I would find delta S = C ln (T2/T1) if i you are given C. Or, if you are given delta H, you can technically use delta Ssurr which is equal to delta H/T. Many ways to figure this out, just manipulate the equations to your benefit.

Jason Muljadi 2C
Posts: 62
Joined: Thu Jul 27, 2017 3:01 am

Re: Total Entropy of the Freezing of Water

Postby Jason Muljadi 2C » Sun Feb 11, 2018 2:46 pm

Another thing to note is that, when there is an isothermal, reversible expansion of a gas, delta Stot = 0. When there is an isothermal, irreversible free expansion of a gas, the internal energy is 0, meaning work is equal to 0, so q is equal to 0. This means that delta S is equal to 0.

This doesn't apply to this topic, but I found these facts to be incredibly useful when doing expansion problems.


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