# Finding entropy involving metal in water

1. Apr 12, 2010

### aal0315

1. The problem statement, all variables and given/known data
A piece of metal at 80C is placed into 1.2L of water at 72C. This thermally isolated system reaches a final temperature of 76C. Estimate the overall change of entropy for this system.

2. Relevant equations
deltaS = Q/T
Q=m*c*deltaT

3. The attempt at a solution
I think that i have to find Q for metal and water separately. correct? and if so i dont know how to find Q for the metal because i dont have its mass or what kind of metal it is to know its heat capacity.
does anyone have any suggestions?

2. Apr 14, 2010

### aal0315

Please, can anybody help me with this question?

3. Apr 14, 2010

### AtticusFinch

The amount of heat that leaves the metal is equal to the amount of heat that goes into the water.

Last edited: Apr 14, 2010
4. Apr 14, 2010

### iRaid

I don't think it's possible without the metal :|

5. Apr 14, 2010

### aal0315

But don't I still need to know the type of metal or mass of metal to find the Q in order to find the deltaS for the metal?

And I dont even know how to find out the water part because what is 1.2L in kg?

6. Apr 14, 2010

### AtticusFinch

Well you can find the mass of the water using the density of the water. But I just noticed this process isn't isothermal so I do think you are lacking information here.

7. Apr 15, 2010

### iRaid

That's what I thought too.

8. Apr 18, 2010

### aal0315

there has to be a way to answer this question .. it is part of an assignment that I have to hand in, so any other suggestions?

9. Apr 18, 2010

### AtticusFinch

Thinking about it some more there is a solution.

You should be able to show using a formula from your book that

$$\Delta S = m_w c_w ln(\frac{349}{345}) + m_m c_m ln(\frac{349}{353})$$

Also, since heat in equals heat out

$$m_w c_w (349-345) = -m_m c_m (349-353)$$

That's all you need to solve this problem.

10. Apr 18, 2010

### aal0315

thank you for the help!

11. Jun 14, 2010

### mlostrac

I am having this same problem. What's the ln mean?