Adiabatic mixing of liquids


> 
> One mole of benzene at $\pu{20^\circ C}$ is adiabatically mixed with 2 moles of toluene at $\pu{40^\circ C}$. The heat capacity of benzene is $\pu{84 J K-1 mol-1}$ and the heat capacity of toluene is $\pu{96 J K-1 mol-1}$. If benzene and toluene form an ideal solution, compute the entropy change for the process. 
> 
> 
> 


I tried to use


$$\Delta S = n C \ln{(T\_\mathrm f/T\_\mathrm i)}$$


where $C$ is the heat capacity and $T\_\mathrm f$ and $T\_\mathrm i$ are the final and initial temperatures of the liquids, respectively. However, I am unsure of what the final equilibrium temperature of the mixture comes to.




Let's say the final temperature is $T$. Also we assume the system is isolated and the process is adiabatic, as you said. Also all energy that is released by toluene or gained by benzene is heat.


The *absolute* amount of heat then released by toluene equals to the *absolute* amount of heat gained by benzene:


$|Q\_\mathrm t| = |Q\_\mathrm b|$


Considering signs, it will look like this:


$Q\_\mathrm t = -Q\_\mathrm b$


By definition, $Q = nc\Delta T$. Thus,


$Q\_\mathrm t = 2\cdot96\cdot(40-T)$, and


$Q\_\mathrm b = 1\cdot84\cdot(20-T)$


Plug in the latter two equations into the second one, and here's what you get:


$2\cdot96\cdot(40-T) = -1\cdot84\cdot(20-T)$


Now just solve it for $T$. And plug that in your equation as $T\_\mathrm f$.


Note that it's not really important to do the temperature in kelvin, since we are dealing with changes in temperature. And yeap, your $T$ should be between $20\ \mathrm{^\circ C}$ and $40\ \mathrm{^\circ C}$, because there are such things as thermodynamics laws. So if it's not, you know for sure you did something not really correct.


 -1