# Change in Entropy for Mixing of Two Liquids

1. Oct 2, 2012

### Alvin92SD

1. The problem statement, all variables and given/known data
You mix 1 liter of water at 20°C with 1 liter of water at 80°C. Set up an equation to find the change in entropy for one of the volumes of liquid in terms of initial temperature (T1) and the temperature after the two volumes of water mixed (T2)

2. Relevant equations
ΔS=∫(dQ/T) from state A to state B

3. The attempt at a solution
Honestly I don't even know where to start or if I am using the right formula. I tried imagining a process where the volume of water is increased at constant temperature and then the temperature is increased from T1 to T2. I would be grateful if someone pointed me in the right direction.

2. Oct 3, 2012

### clamtrox

This is a bit tricky problem; it would be much more clear if you were dealing with ideal gases instead. Now you just need to ignore the increase of entropy coming from the mixing of the two liquids.

Perhaps you can assume that the volume of the liquid does not change when it's mixed and cooled/heated. Then you can write
$$dS = \left(\frac{\partial S}{\partial T}\right)_V dT +\left(\frac{\partial S}{\partial V}\right)_T dV = C_V \frac{dT}{T}$$