Change in temperature for a system with entropy change

[MalcolmMck]

Homework Statement

I am Pretty Lost with this problem...[/B]
A 2.45-kg aluminum pan at 155∘C is plunged into 3.58 kg of water. If the entropy change of the system is 162 J/K, what was the initial temperature of the water? NOTE:We did not receive a Tf for the system.

Homework Equations

Δs=mCln(T1/T2)
Δs=∫(1/T)dQ
Q=mcΔT

The Attempt at a Solution

Here's what I have so far:
The change in entropy of the system is equal to the change in entropy of the aluminum plus the change in entropy of the water. This gave me the equation:
Δs=MaCa(ln(Tf/Tia))+MwCw(ln(Tf/Tiw))
where...
Ma and Mw are the masses of aluminum and water respectively
Ca and Cw are the specific heats of aluminum and water respectively
Tf is the final equilibrium temperature for the system\
Tia is the initial temperature of the aluminum
Tiw is the desired answer, the initial temperature of the water

I don't know how to move on without the Tf of the system. My attempts at solving for Tf led to something very messy. Any help is greatly appreciated!

 

[BvU]

Is there some other quantity in this scenario that is a conserved quantity ? Perhaps that way you can come to another equation so you end up with just as many equations as unknowns ...

 

[MalcolmMck]

As the entropy of the system is changing, the only thing that I could think of that would be conserved is energy. The first law of thermodynamics states that ΔU=W+Q where U is the internal energy. Also the equation relating T,U, and s is 1/T=Δs/ΔU. I'm not quite sure how I would use these two to produce another useful equation or incorporate conservation.

 

[BvU]

Perhaps you are allowed to look up the c_p ?

 

[MalcolmMck]

Yes we are allowed to look up the specific heats. For aluminum Ca is 0.900 J/gK and for water, Cw is 4.184 J/gK. However, plugging these into my original equation still leaves Tiw and Tf unknown