# Entropy and such

1. May 8, 2005

### Jayhawk1

Here is the question:

8) [1.0/3.0] 1.6 kg of water at 34oC is mixed with 2.0 kg of water at 58o. in a well insulated container. a) What is the final temperature of the water (in degrees Celsius)? b) What is the approximate change of entropy? (Use the average temperature for the entropy calculation.)

Part a was easy enough... I got 47.3 oC

I've tried calculating part b in multiple ways, and I am getting a very small number- which is what my answer is actually suppose to be- BUT - it is not right. Any help?? Thanks.

2. May 8, 2005

### Richter915

hmm...well you know that deltaH = qt and qt/t = delta S...give that a try?

3. May 8, 2005

### Jayhawk1

Nothing I've trired has worked... I keep coming up with a negative number... I need some definitive direction.

Last edited: May 8, 2005
4. May 8, 2005

### OlderDan

$$\Delta S = \frac{\Delta Q}{T}$$

S is entropy, Q is heat, T is temperature. You are asked for an approximate value for the change in entropy using an average termperature. I interpret that to mean you should use the average temperature of the warm water to calculate its entropy loss and the average temperature of the cool water to calculate its entropy gain. There will be a net gain.

5. May 8, 2005

### Jayhawk1

Still not working... what else should I try??

6. May 9, 2005

### Jayhawk1

Can someone be a little more specific.... I am getting really frustrated. This is the last question on the last homework assignment for the year and I keep getting it wrong. Not to mention I don't have that many tries on it left. Please help me!!!!!!!!!

7. May 9, 2005

### OlderDan

How about showing us exactly what you did. I thought the last thing I told you was quite specific. What did you get for the amount of heat transferred from the warm water to the cold water, the entropy lost by the warm water, and the entropy gained by the cold water. If you are still getting negative answers for the net entropy change, you are doing something wrong in your calculations.

8. May 9, 2005

### ZapperZ

Staff Emeritus
You cannot continue complaining that you are not getting the right answer. The way you have gone about with this appears as if you want someone to do the work for you.

Show EXACTLY what you did to get the wrong answer. Only then can someone point out what you did wrong. You learn MORE from mistakes like this than from the ones you did correctly.

Zz.

9. Jun 3, 2005

### squib

Same problem for me. I've tried getting Q by computing specific heat * g * delta T, but no luck with that.

10. Jun 3, 2005

### OlderDan

What answer are you getting, how are you getting it, and what (if you know) are you supposed to get?

11. Jun 4, 2005

### Andrew Mason

Since the temperature changes as the heat is absorbed/lost, you have to integrate.

see: https://www.physicsforums.com/showthread.php?t=77857

AM

12. Jun 4, 2005

### OlderDan

Except that the problem says
These vague statements like "I've tried everything and can't get the answer" are pretty useless for figuring out where the difficulty lies.

13. Jun 4, 2005

### Andrew Mason

Ok. Then the problem Jayhawk must be having is in the determination of average temperatures for the cool and warm water. Jayhawk, what are you using for average temperatures?

AM

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