2 body engine, final equilibrium temperate and work produced

[mrmerchant786]

Homework Statement

2 bodies with contant heat capacity C and 2C, are initially at temp T and 2T. if a heat engine executing a reversible carnot cycle operates between the two bodies until their temperatures are equal, what is the final temp of the bodies and how much work is preformed by the heat engine

Homework Equations

W=Qη dQ=CdT

The Attempt at a Solution

q2 = C2 * (Tf-T2)
q1 = C1 * (Tf-T1)
So far this is where I'm at, I don't know how to proceed further to find the equilibrium temperature and then to fine the work produced

 

[Chestermiller]

Suppose T* is the final temperature. In terms of T*, what is the change in entropy of each of the two reservoirs. Since you are operating using a Carnot cycle, the change in entropy of the working fluid is equal to zero. What does that tell you about the total of the entropy changes of the two reservoirs?

 

[mrmerchant786]

Suppose T* is the final temperature. In terms of T*, what is the change in entropy of each of the two reservoirs. Since you are operating using a Carnot cycle, the change in entropy of the working fluid is equal to zero. What does that tell you about the total of the entropy changes of the two reservoirs?

will it be zero?
so s1 +s2 = 0

 

[Chestermiller]

will it be zero?
so s1 +s2 = 0

Yes. $\Delta S_1+\Delta S_2=0$

 

[Chestermiller]

So, have you solved it yet?