# Example Problem with Complete Solution

8A-3 : Entropy Production of Mixing Two Liquids at Different Temperatures 8 pts
The initial and final states of a sealed, insulated, rigid tank are shown below.Each side of the tank contains a different incompressible liquid at a different temperature, T1 and T2. The mass of liquid initally on each side of the tank is the same: m1 = m2 = m/2. The barrier between the two sides of the tank is removed and the two liquids mix and eventually reach the final equilibrium state.
Assume each liquid has a constant heat capacity and there are no thermal effects due to the mixing of the fluids.
a.) Show that Sgen is given by the following equation: b.) Show that Sgen must be positive.

Read : For part (a) Perform an entropy balance to determine an equation for Sgen. Then perform an energy balance to determine an expression for the final temperature and substitute the expression into Sgen and simplify.
Given: Initial State: Final State :
Chamber 1 : T1 Chamber 1 : Teq
Chamber 2 : T2 Chamber 2 : Teq
Incompressible fluids with CP = CV = C.
Find: Part (a) Show that the amount of entropy generated is: Eqn 1
Part (b) Demonstrate that Sgen must be positive.
Diagram: The diagram in the problem statement is adequate.
Assumptions: 1 - The system consists of the total mass of liquid in the entire tank.
2 - The system is isolated (adiabatic and closed).
3 - The liquid is incompressible with constant specific heat, C.
4 - No work crosses the sytem boundary.
Equations / Data / Solve:
Part a.) Let's begin with the defintion of entropy generation: Eqn 2
We can solve Eqn 2 for Sgen : Eqn 3 Since the system is isolated,
there is
no heat transferred: Eqn 4
We can use Eqn 4 to simplify Eqn 3, yielding : Eqn 5
The change in the entropy of the system is : Eqn 6 Eqn 7
We can rearrange Eqn 7 to show that the total change in entropy for the system is the sum of the changes in entropy of each of the two fluids. Eqn 8
The entropy change for an incompressible fluid depends only on temperature. Eqn 9
Because the heat capacity in this problem is a constant,
it is relatively easy to
integrate Eqn 9 to get: Eqn 10
Next, apply Eqn 10 to determine the entropy change of each fluid in this process and substitute the result into Eqn 8 : Eqn 11
Properties of logarithms let us rearrange Eqn 11 to : Eqn 12
Combining Eqn 12 with Eqn 5 gives us : Eqn 13
To complete this derivation, we must eliminate Tfinal from Eqn 13.  We can determine Tfinal in terms of T1 and T2 by applying the 1st Law to this process. Eqn 14
No work or heat crosses the system boundary, so Eqn 14 becomes : Eqn 15
Now, use the constant specific heat of
the
incompressible fluid to determine DU : Eqn 16 Eqn 17
Now, solve Eqn 17 for Tfinal : Eqn 18
Now, we can use Eqn 18 to eliminate Tfinal from Eqn 13 : Eqn 19
Simplify Eqn 19 algebraically : Eqn 20
Finally : Eqn 21
Part b.) Entropy generation is non-negative when : Eqn 22
The  values of m and C must be positive so,
Sgen is non-negative when : Eqn 23
Simplify Eqn 23 by algebraic manipulation,
as follows : Eqn 24
Squaring both sides of Eqn 24 yields : Eqn 25
( This is OK because T1 > 0 K and T2 > 0 K )
Expand the left-hand side of Eqn 25 : Eqn 26 Eqn 27
Finally, we get : Eqn 28
The inequality in Eqn 28 is satisfied for either T1 > T2 or T2 > T1.
The equality in Eqn 28 is satisfied only when T1 = T2.
Verify: The assumptions made in this solution cannot be verified with the given information.
Answers : Part a.) Part b.)  when :
which is ALWAYS true ! 