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fluidistic

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## Homework Statement

Hi guys. I'm basically clueless on what's going on, I get 2 different results while I should get the same.

The problem reads "Each of two bodies has a heat capacity at constant volume given by ##C_v=nc_v=A+2BT## where A and B are known constants (both positive).

If the 2 bodies are initially at temperatures ##T_1=200K## and ##T_2=400K## and if a reversible work source is available, what are the maximum and minimum final common temperatures to which the 2 bodies can be brought? What is the maximum amount of work that can be transferred to the reversible work source? It is to be assumed throughout that each body is maintained at constant volume.

## Homework Equations

1st law of Thermodynamics: ##dU=\delta Q + \delta W## where U is the internal energy of the system I'm considering, Q is the heat going toward the system and W is the work done on the system.

Others.

## The Attempt at a Solution

First I want to calculate the minimum final common temperature. It occurs when the work done BY the system is maximum and it corresponds to a zero change in entropy in the whole system. I.e. ##\Delta S_{\text{Total}}=\Delta S_1+\Delta S_2 + \underbrace {\Delta S _{\text{heat engine}}}_{\text{=0 because the engine works with cycles}}=\Delta S_1+ \Delta S_2##.

Now I use the equation ##dS=\frac{dQ}{T}## and that ##dQ=nc_vdT## for both bodies. I get that ##\Delta S= \int _{T_1}^{T_f} nc_v \frac{dT}{T}+\int _{T_2}^{T_f} nc_v \frac{dT}{T}=nA\ln \left ( \frac{T_f^2}{T_1T_2} \right ) +2nB[2T_f-(T_1+T_2)]## where I skipped some arithmetical steps. Now the minimum ##T_f## can be found by setting ##\Delta S=0## and isolating ##T_f## (which I don't think is possible, so a numerical method should be used). So far so good I've virtually determined the ##T_f## that they ask me and it cannot be written in terms of elemental functions.

Now if I use a different way to solve the problem I get a different answer.

If I consider the whole system 2 bodies+heat engine as my system, I get that ##\Delta U=W## where W is the work done on the system. And delta U is the change in internal energy of the 2 bodies + heat engine but the change in internal energy of the heat engine is 0 because it works with cycles.

So my goal is to get the work done by the system, in other words, ##-W##.

The change in internal energy of the first body is worth ##dU_1=dQ_1+\underbrace {dW_1}_{=0 \text{ because V=constant}}=nc_vdT_1 \Rightarrow \Delta U_1=n[A(T_f-T_1)+B(T_f^2-T_1^2)]##. In the same fashion I get ##\Delta U_2##. So ##\Delta U=\Delta U_1+\Delta U_2 =n[2AT_f-A(T_1+T_2)+2BT_f ^2 - B(T_1^2+T_2^2)]## which is worth the work done on the system (because of the 1st Law of Thermodynamics and using the fact that the whole system is isolated and therefore Q=0).

Therefore the work done by the system is ##-W=W'=n[B(T_1^2+T_2^2)-2BT_f^2+A(T_1+T_2)-2AT_f]##.( The maximum amount of work is done when ##T_f## is minimum so if I could isolate ##T_f## above and plug it here I'd answer the question of the problem.)

Here I can think of W' as a function of ##T_f## only. In such case I want to see what value of ##T_f## extremizes W'. Derivating W' with respect to ##T_f## and setting that expression equal to 0, I found ##T_f=\frac{A}{2B}## and the second derivatives gives me ##-4B<0## thus W' is maximum and therefore ##T_f## corresponds to a minimum. But it is not equal to the previous one I had found! I really don't understand what's going on. I would like someone to comment on this.

To continue the problem, if I set ##W'=0## (where the prime denotes the work done BY the system, not the derivative!) I should get ##T_f## maximum because no work is being done by the heat engine and so the heat flows from body 1 to body 2 until both temperatures are equal. In that case I get ##T_{fmax}=-\frac{A}{B}+\sqrt{\left ( \frac{A}{B} \right ) ^2 +2 [T_1^2+T_2^2 + \frac{A(T_1+T_2)}{B}] }##. I wonder if this result is good.

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