[jaysup_2006]

I have another problem where I am not sure about the equation:

Calculate the entropy change for a process in which 2.09 mol of liquid water at 0°C is mixed with 1.05 mol of water at 100.°C in a perfectly insulated container. (Assume that the molar heat capacity of water is constant at 75.3 J/(Kmol).)

I think I find my q first..q=massx heat capacityx detla t.
Not sure though, any tips??? Thanks

[allanf]

I'm not really sure.  My back of the napkin calculation would start off sort-of like this:

Entropy change is a state-function, which (among other things) means that the change in entropy from one state to another doesn't depend on how you get from point A to point B.  So an equivalent way of thinking of this problem is that we have a thermally insulated container with the 0C water on one side of a divider and the 100C water on the other (this being an imaginary wall).  We let the two compartments come to thermal equilibrium, then we remove the divider and let them mix.  OK?

The total entropy of this process then is the sum of the change in entropy for the 0C water heating up plus the 100C water cooling down plus the entropy of them mixing.  I believe since both sides are water, the entropy of mixing is 0.  So we can ignore this.

The entropy change from heating or cooling something is n*Cv*ln(Tf /Ti), where Tf is the final temperature and Ti the inital one (ln is the natural log btw)

So if you can figure out what the equilibrium temperature will be, then you can just add together the entropies of heating up the 0C water and cooling down the 100C water.

Hope that helps.

[jaysup_2006]

It kinda helps, but its still a little confusing. Is there any way you could break it down further, if not, its alright.

[allanf]

Let's see if I can make it clearer.

For simplicity I'm going to call the ~2mol of 0C water "A" and the ~1mol of 100C water "B".

From my previous post, I'm saying the following is true:
S_total = S_A + S_B

This equation follows from what I said earlier.  Taking A and B, putting them next to each other until they come to the same temperature, then mixing them is equivalent to taking the cold A and hot B and mixing them straight away (and I'm conjecturing that mixing water with water has no associated change in entropy)

In this equation S_X is the entropy change from heating or cooling, which I'm saying is n*C_v*ln(T_f/T_i)

So, the total entropy should be:
S_total = n_A*C_v*ln(T_f/T_A) + n_B*C_v*ln(T_f/T_B)

Where does T_f come from?  Well, you should have already done stuff on thermal equilibrium, the only heat being transferred is from B to A, from the 100C water to the 0C water.  So to get Tf set q_A=q_B and solve for the final temperature.

Just so you know where I got that n*C_v*ln(Tf/Ti), I remembered the definition of entropy and integrated in the following way (this forum really needs LaTeX support!):

So you might want to double-check that this is the right equation (I may have messed up my maths)

[jaysup_2006]

OK, I had worked out an equation before actually posting my question(i should have mentioned it, sorry) it was (2.09mol)(75.3 j)(ln 373/273) divided by (1.05mol)(75.3 j)(ln 373/273). I understand you said it was an addition of both. So i have to change that. Wouldnt the temperature become halfway after mixing them?..like 50..also doesnt the cold water now have the final temp as the hot water(reason I say this is because I did a lab similar, and after mixing hot and cold water, I was told the cold water has the same temp as the hot water).

[allanf]

Yes, the hot and cold water will have the same final temperature.  I wouldn't think it would be half-way (50C) because there is twice as much cold water as hot.  I seem to think it would be the average temperature so (I'm just rounding the moles to 2 and 1 so I can do this in my head): (2*0C + 1*100C)/(2+1) = 100/3 ~ 33C

So the natural log for the 0C water should be ln(306/273) and for 100C it should be ln(306/373).  Try it then and see if it makes sense:  the entropy should be positive (i.e. this is a spontaneous process, warm water doesn't spontaneously separate into regions hot and cold), and it should be small (1 or 2 kJ likely).

[jaysup_2006]

cool, ok i did the calulations again(not completely, i only have my cellphone at this time), but my new equation looks like this: (2.09mol)(75.3J)(ln(306.43949/273) + (1.05mol)(75.3J)(ln(306.43949/273)...I think this is the right equation. I did have a couple questions though:

1) Its supposed to be addition, right?
2) You said the entropy should be small, I do agree, however you said in kJ..the answer requires it to be in J/mol...is any conversion required(I dont think so because the heat capacity is already in Joules right)

You've been such a big help..thanks.

[allanf]

Yeah its supposed to be addition.  You're right it should be Joules not Kilojoules.  The only conversion is a factor of 1000 anyways (yay metric!).

Your equation looks right to me.  Its addition since entropy, like energy in general, is additive.