Find the difference in entropy between a cup of water and a cup of ice

[erin88]

Find the difference in entropy between a cup of water and a cup of ice for each of these cases:
1. both are at T=0C
2. ice is at T=0C water is at 20C

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If the "water" is at 0C then it is ice right? and the difference in entropy is 0 correct?

For the second part... do you think I am supposed to find the Q for each one? since the equation for Q is Q=cmT I don't think this would make sense because the equation for entropy (S) is S=Q/T. For ice it would be 0/0 and you can't do that...

Do you think maybe he's asking for the change in entropy if you mix the two together?

Thanks.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
EDIT

I just remembered the equation is Q=cmdeltaT

So this must mean that he does want to know the change of entropy in the system if you mix the two together??

 

[dark adonis]

Find the difference in entropy between a cup of water and a cup of ice for each of these cases:
1. both are at T=0C
2. ice is at T=0C water is at 20C

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
If the "water" is at 0C then it is ice right? and the difference in entropy is 0 correct?

For the second part... do you think I am supposed to find the Q for each one? since the equation for Q is Q=cmT I don't think this would make sense because the equation for entropy (S) is S=Q/T. For ice it would be 0/0 and you can't do that...

Do you think maybe he's asking for the change in entropy if you mix the two together?

Thanks.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
EDIT

I just remembered the equation is Q=cmdeltaT

So this must mean that he does want to know the change of entropy in the system if you mix the two together??

First I'll address the conceptual problem:
Ice is not simply water at T=0C, there is change in energy associated with melting/freezing. I think this might give you a bit of a start on the first part.

The next issue to address:
We have $$Q = S T$$
The first question that crosses my mind is what are the units on T?
Should it be K? C? F? R? I think the answer to this question will relieve your divide by zero error.

 

[nlightner]

Thanks.I would like to clarify the concepts of entropy and heat (Q) in this context. Entropy is a measure of the disorder or randomness in a system, while heat (Q) is a form of energy transfer between two objects due to a temperature difference.

In the first case, both the cup of water and cup of ice are at 0°C, meaning they both have the same temperature. However, the ice has a more ordered and less random molecular structure compared to the liquid water. This results in a higher entropy for the cup of water compared to the cup of ice. The difference in entropy between the two can be calculated by finding the change in entropy (ΔS) during the phase change from ice to water. This can be done by using the equation ΔS=Q/T, where Q is the heat absorbed during the phase change and T is the temperature at which the phase change occurs.

In the second case, the ice is at 0°C and the water is at 20°C. This means that the water has a higher temperature and thus a higher kinetic energy compared to the ice. When the two are mixed together, heat (Q) will transfer from the water to the ice until they reach a common temperature. This heat transfer will result in a change in entropy, as the molecules in the ice will become more randomly arranged and the water molecules will have a lower kinetic energy. The difference in entropy between the two can again be calculated using the equation ΔS=Q/T, where Q is the heat transferred and T is the final temperature reached by the system.

In summary, the difference in entropy between a cup of water and a cup of ice depends on the temperature and phase of the two substances. In both cases, there will be a change in entropy when the two are mixed together, but the specific values will depend on the temperature and heat transfer involved.