# Finding the final temperature

1. Jun 2, 2013

### Saitama

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Let the specific heat capacity of each object be $C$, then
$$C(T-200)=C(400-T)+C(400-T)$$, where T is the final temperature.
But solving this gives T=333.33 K which is wrong.

Any help is appreciated. Thanks!

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2. Jun 2, 2013

### Staff: Mentor

Contemplate upon the second law of thermodynamics.

3. Jun 2, 2013

### Saitama

I am not too much familiar with second law of thermodynamics and I did not expect that this question would deal with it. Do I have to deal with entropy here?

4. Jun 2, 2013

### Mute

By what kind of process are the objects exchanging heat?

5. Jun 2, 2013

6. Jun 2, 2013

### Staff: Mentor

The problem says nothing about how they exchange heat (radiation, conduction, convection,...). But it does say that they form an isolated system. That means no heat escape, no external forces or energies.

Can you quote a statement of the second law of thermodynamics? Hint: If heat moves between the given objects, will the heat flow spontaneously from an object of lower temperature to one of higher temperature?

7. Jun 2, 2013

### Mute

If the objects are touching, they would be transferring heat by conduction, no? The problem says the objects are isolated from the rest of the universe, so I doubt they're radiating heat.

So, your objects are isolated. I guess what I was wondering was, are the objects allowed to expand/contract when heated?

I'm not sure if these questions are pointing you in the right direction - I just find it odd that the question asks for the "highest possible temperature one of them can reach". To me that suggests there may be a process of heat exchange different than the one you calculated in which the boxes can achieve a higher temperature than 333.33 K.

e: I'll let gneill handle this. Seems to have a better idea of how to solve the problem than I do, and I won't have much time to think about it anyways.

8. Jun 2, 2013

### Saitama

From Wikipedia: "The second law of thermodynamics states that the entropy of an isolated system never decreases, because isolated systems spontaneously evolve towards thermodynamic equilibrium—the state of maximum entropy."

No, the process is not spontaneous. But I am still unsure about what equations to start with.

9. Jun 2, 2013

### Staff: Mentor

Note the phrase, "isolated systems spontaneously evolve towards thermodynamic equilibrium". That means heat, if it moves, moves from things of higher temperature to things of lower temperature. When everything has the same temperature then the system is in equilibrium.

Here we have three objects with starting temperatures. Which of the objects can achieve a higher temperature if heat always moves from higher temp to lower temp? Can the highest available temp increase?
No equations are required! Just the second law and its consequences.

10. Jun 3, 2013

### haruspex

I agree with Mute, there's something fishy about this question. On the face of it, the answer is trivial (yes?), but I suspect the question intends that the objects may form a heat engine. Thus, conservation of energy and non-reduction of entropy are the only two constraints.

11. Jun 3, 2013

### Saitama

The highest available temperature is 400 K and it increases only if the heat flows from lower temperature to higher but that is not possible (or spontaneous).

How does the system forms a heat engine?

12. Jun 3, 2013

### haruspex

In principle, the flow of heat from one of the hotter bodies to the cooler body can be used to drive some heat up the gradient to the other hot body. I know it says there are just these three bodies, nothing else, but perhaps they are in the form of a heat engine. The details don't matter.
If we assume that the only constraints are conservation of energy and non-reduction of total entropy it will be possible to calculate a max temp (and it will be > 400K). I gather that you will then be able to check if it's the accepted answer.

13. Jun 3, 2013

### Saitama

Non-reduction of entropy?

Something like initial entropy=final entropy? If so, how am I supposed to calculate the initial and final entropy?

14. Jun 3, 2013

### haruspex

Let the bodies be A, B (each 400K) and C. Suppose we arrange a small quantity of heat ΔQ to flow from A to C and use that to pump a further ΔR from A to B. Assume some arbitrary specific heat for all three. Write down the equations for no net change in entropy. This should lead to a differential equation.

Edit: For the purposes of the differential equation, you can treat one body in isolation to determine the change in entropy as its temperature changes from T1 to T2. Or you may already know a formula for that.

Last edited: Jun 3, 2013
15. Jun 4, 2013

### Saitama

I don't think I still get it but here is what I get:
$$dS=\frac{dQ}{T}$$
$dQ=CdT$
$$\Delta S=C\ln\left(\frac{T_2}{T_1}\right)$$

16. Jun 4, 2013

### haruspex

Yes. Now, if the three temperatures change from TA0, TB0, TC0 to TA1, TB1, TC1, what is the total change in entropy? What other equation can you write relating those 6 variables (since the bodies have the same S.H and mass)?

17. Jun 4, 2013

### Saitama

Total change in entropy:
$$\Delta S_{total}=C\ln\left(\frac{T_{A1}}{T_{A0}}\right)+C\ln\left(\frac{T_{B1}}{T_{B0}}\right)+C\ln\left(\frac{T_{C1}}{T_{C0}}\right)$$

I have no idea about the other equations. :uhh:

Sorry if I am missing something obvious. I do know the second law of thermodynamics but this section did not get much attention in my course, we only touched it during the Carnot cycle. Sorry again. :(

18. Jun 4, 2013

### haruspex

Good. Now, for max final temp, you can assume entropy stays constant. That will greatly simplify the above equation. You also know that total energy is constant. Express that using the same six variables. You know the initial temperatures, so it leaves you with three unknowns and two equations. We need one more equation.
Suppose that we start with A and B at 400K and C at 200K, and we use A to heat B and C. When we can pump no more heat into B, what can you say about the temperatures of A and C?

19. Jun 4, 2013

### Saitama

As the entropy stays constant
$$T_{A1}\cdot T_{B1}\cdot T_{C1}=T_{A0}\cdot T_{B0}\cdot T_{C0}$$

For energy conservation, I don't think that the following equation is correct,
$$C(T_{A1}-T_{A0}+C(T_{B1}-T_{B0})+C(T_{C1}-T_{C0})=0$$
I really have no idea about the third equation. When A heats B and C, the temperature of A decreases and that of C increases.

20. Jun 4, 2013

### haruspex

Apart from a missing parenthesis in the second, both equations look right to me. You can cancel the C in the second and rearrange more like the first equation.
Yes, but remember its the down gradient flow from A to C that we're using to pump heat into B. That can only continue until... ?