Obtaining work from two bodies by a heat engine

[Parzeevahl]

Homework Statement

Two identical bodies of constant heat capacity Cp at temperatures T1 and T2 respectively are used as reservoirs for a heat engine. If the bodies remain at constant pressure, show that the amount of work obtainable is
W = Cp * (T1 + T2 - 2 * Tf)
where Tf is the final temperature attained by both bodies. Show that if the most efficient engine is used, then Tf^2 = T1 * T2.

Relevant Equations

W = Cp * dT

Here's my attempt for the first part:

For the first body, the work obtained is
$W_1 = C_P (T_1 - T_f)$
while for the second body, it is
$W_2 = C_P(T_2 - T_f).$
So the net work obtained is the sum of these two:
$W = W_1 + W_2 = C_P (T_1 + T_2 - 2 T_f)$
and that proves the first part. However, here's my problem with the second part: The same heat engine is doing all the job, i.e., heating the two bodies. Now, the question asks me to consider the most efficient engine, in which case the efficiency is
$1 - \frac{T_{final}}{T_{initial}}.$
There are two heatings here, for which we find the efficiencies to be
$1 - \frac{T_f}{T_1}$
and
$1 - \frac{T_f}{T_2}.$
But the heat engine is the same so they should be the same, which means
$T_1 = T_2$
but that was not the case originally! Where have I gone wrong?

 

[haruspex]

Not sure where you are getting that equation for the "most efficient " condition from, but I would say it does not apply in this context.
Instead, use that maximum efficiency corresponds to no entropy gain.

 

[Chestermiller]

You are going wrong is implicitly assuming in that all the heat transfer takes place at T1 and T2. That's what your two efficiency equations assume. But, in this case, heat transfer from the hot body takes place at all temperatures between T1 and Tf, and heat transfer to the colder body takes place at all temperatures between T2 and Tf. Haruspex is correct in saying that the maximum efficiency corresponds to no entropy change for the combination of hot and cold bodies.