# Homework Help: Work or Heat transfer to system?

1. Aug 6, 2013

### CAF123

1. The problem statement, all variables and given/known data
a) Liquid is stirred at constant volume inside a container with adiabatic walls. The liquid and container are regarded as the system. Is heat being transferred to the system? Is work being done on the system?

b) Water inside a rigid cylindrical insulated tank is set into rotation and left to come to rest under the action of viscous forces. Regard the tank and water as the system. Is any work done by the system as the water comes to rest? Is there any heat flow to or from the system?

2. Relevant equations
First Law of thermodynamics

3. The attempt at a solution
I think I have the correct answers but I want to check my reasoning for each case:
a)Work is being done on the system. The stirring produces a mechanical exchange of energy to the water. No heat flow since the walls are adiabatic. I think you could also say given that the stirrer is in contact with the water, after a time, no object is hotter than the other so this may not be regarded as heat transfer.

b)'left to come to rest' implies no work is done by the water. The water comes to rest solely by the viscous forces which act as a retarding force to the rotational motion of the water. No heat transfer because the system is insulated.

Are these ok responses?

2. Aug 6, 2013

### rude man

That is right, except that the spoon would transfer heat to the water if it's hotter than the water initially, but I don't see why that should be the case. The heat is produced by friction with the container sides and I guess internal friction as well.

No. How do you account for the rise in internal energy of the water? The kinetic energy of the rotating water is changed to heat. Since there is no heat transferred the rotating water must be doing work on the water. Again, by the 1st law.

Both a and b are examples of adiabatic dissipation of work into internal energy of the system.

3. Aug 6, 2013

### TSny

I think your answer to (b) is correct: W = Q = 0.

In problems like this where there is bulk movement of the system, the first law should be stated in a general form ΔE = Q - W where E is the total energy of the system including kinetic energy of bulk motion. W is defined as work done by the system (water and container) on the environment. Since the volume of the container is constant, the system does zero work on the environment. Likewise, Q is defined as heat transferred to the system from the environment. Since the walls are adiabatic, Q = 0.

4. Aug 6, 2013

### CAF123

Hi rude man,
I presume since the heat produced by the container sides is within the container, (i.e internal to the system), we do not regard this as heat transfer to the system?

Mathematically, we have the first law as follows: $$\Delta U + \text{bulk}(KE + PE) = Q + W$$ The right hand side is zero, so this gives $\Delta U = - \text{bulk}(KE + PE)$ i.e the rotational kinetic energy of the water is converted to the internal energy of the water.
Is it what you meant?
Thanks.

5. Aug 6, 2013

### rude man

Last edited: Aug 6, 2013
6. Aug 6, 2013

### rude man

Correct.
Yes it is. Absolutely correct.

@Chet - join in, willya?

Last edited: Aug 6, 2013
7. Aug 6, 2013

8. Aug 6, 2013

### rude man

I just told you it did! That post was not directed at you.

Actually, the work is done externally on the system by spinning the water. This is analogous to boring out a cannon (if the cannon was insulated from the environment).

Last edited: Aug 6, 2013
9. Aug 6, 2013

### TSny

It's like a system where you have a closed box with adiabatic walls and fixed volume. Inside the box is a block of wood that is initially sliding along the floor of the box (call this the initial state). The block is brought to rest by friction between the block and the floor of the box (call this the final state).

We define the system to be the block + the box. No energy is transferred to the system as heat or work during the time that the system passes from the initial state to the final state. Q = W = 0. The first law applied to this system implies that the total energy E of the system remains constant. You can still have energy transformations taking place inside the system. The internal energy U of the block and walls will increase as the KE of the block decreases.

10. Aug 6, 2013

### rude man

There is work done external to the system here also. It's the work done by gravity on the block as it slides.

11. Aug 6, 2013

### TSny

I am assuming the block is sliding horizontally along the floor of the box. The work done by gravity will then be zero.

12. Aug 7, 2013

### rude man

And what impelled the block to move horizontally? It had to be accelerated via an applied force, and W = ∫F ds.

EDIT: it's occurred to me that this little exchange could go on forever. Perhaps I can truncate it by a quote from Mark W. Zemansky, Heat and Thermodynamics, 4th edition, p. 178:

"Processes Exhibiting External Mechnical Irreversibility:

Those involving the adiabatic dissipation of work into internal energy of a system, such as:

2. Coming to rest of a rotating or vibrating thermally insulated liquid.

Note the word "work" in that penultimate sentence.

Last edited: Aug 7, 2013
13. Aug 7, 2013

### CAF123

The answer in the back of my book is that Q = W = 0. It doesn't give any reasons for these answers though in case you wondered.

14. Aug 7, 2013

### CAF123

Hi TSny,
I am wondering how this conforms with what I wrote in (a). There I argued that the stirrer did mechanical work but the volume of the system was not changing. In (b), I suppose the rotating device can be regarded as a stirrer also, but you say there is no work done?

Thanks.

15. Aug 7, 2013

### ehild

Thermodynamics deals with systems in steady state. It means that all its parts are in equilibrium with the surrounding parts in the system. Only in that case are the state variables like pressure, temperature, defined. Internal energy is function of the state variables. And it is the sum of the energies of the random motion of its particles. Internal energy is changed by work done on the system or heat transfer to the system.
When stirred, external work is done on the fluid. The fluid is set into motion as a whole, and different parts move with respect to each other. It is some kind of regular motion, not random. The fluid as whole has got both momentum (angular momentum) and kinetic energy.
The system is not in equilibrium. If you divide the fluid to small parts, these parts move with respect to each other, change their shapes, and exert force to each other, exchanging momentum and energy . After the stirring stopped the energy of the whirling parts of the fluid is transformed to that of the random motion of the molecules and thermal equilibrium is established. The liquid reaches a new state, with higher internal energy as before. The change of internal energy was caused by the work of the stirrer. At the same time, the adiabatic walls prevented heat exchange.

In case b, there was work done on the system when the water was set into rotation. Till it moves on, it has angular momentum and kinetic energy as a whole, its thermodynamic state can not be specified. Isolated from the surroundings, the energy transforms to that of random motion of the constituents, thermal equilibrium is set up. You can not apply the laws of thermodynamics during the process leading to equilibrium, when the parts of the fluid interact with the other parts and with the wall, exchanging momentum and energy.

ehild

Last edited: Aug 7, 2013
16. Aug 7, 2013

### Staff: Mentor

Hi Rude man.

I agree with the analysis of part b. With regard to part a, it seems to me that the main thing that's happening is that stirrer is doing work to deform the fluid (viscous deformation), and the viscous heat generation within the system translates into an increase in internal energy. I regard the stirrer as part of the surroundings. So Q = 0 and ΔU = -W. I don't regard the interaction of the stirrer with the walls of the container as comprising the bulk of the work done by the stirrer.

Chet

17. Aug 7, 2013

### rude man

In the final analysis I suppose one could argue that this is a semantic issue.

Certainly, the moving liquid is doing work on the system, and when it stops, the internal energy has increased by that amount of work. And obviously it took work to set the liquid in motion. But i suppose one can ignore those work quantities, say that the rotating liquid represents initial kinetic energy, and that there is no work done on the system externally once the liquid is set in motion.

To me, and to famed thermodynamicist Zemansky, the k.e. is doing work on the system consisting of the liquid and the container. I will leave it at that.

18. Aug 7, 2013

### ehild

PΔV = W, work by the system is valid only in quasi-static processes, so slow ones, that the system can be considered in equilibrium. Otherwise, the pressure is not defined.
You can do work on a system in very different ways. Mechanical work, by moving a piston, but stirring a liquid is also mechanical work. In an electric field, the liquid or gas is polarized, it needs electric work done. A magnet can change the magnetic state of the system, doing magnetic work. Putting the liquid into the microwave oven, the electromagnetic field does work.
In most cases the external work is easier to calculate than the work done by the system. The process can be fast, and the state variables of the system are well defined only in equilibrium - in quasi-static processes.
Of course, the water stirred does work on the blades of the stirrer. There is force of interaction between the water and the blades, and the blades displace. There is force and displacement. But how do you know the force the water acts on the blades? At the same time, you can measure the electric power consumed by the stirrer, and you can also estimate its efficiency. It is not difficult to find the work done by the stirrer on the water.
In case b, the work had been done by setting the liquid into rotation. After that, the parts of liquid do work on each other till equilibrium is set up. There is no external work done. The energy of the rotational motion will transform into energy of random motion of molecules, that is, into internal energy.

ehild

19. Aug 7, 2013

### Staff: Mentor

From the implied problem description, I agree with your interpretation in bold.

Chet

20. Aug 7, 2013

### rude man

@Chet, thanks for chiming in.
rudy

21. Aug 8, 2013

### CAF123

Here are my thoughts:
a)The stirrer does mechanical work on the water and if we assume no bulk motion of the liquid, then this work is transferred into the internal energy of the water. We can also say that the water does work on the stirrer. So, in answering the question, work is done on the system.

b)Putting the water into motion requires work. ehild, you mentioned that the water does work on the blades of the rotary device and this also makes sense. Would this not mean though that the system does work? (If work is the work done by the system on the environment and I think it is agreed that the rotary device is part of the environment).

At the moment, I am under the impression that there is no clear cut answer to this - it all depends on what we really regard as the environment.

22. Aug 8, 2013

### ehild

Question b) says that the water is put into rotational motion and then left alone. It asks about work done by the system afterwards.
No word about a stirrer. The water can be set rotating without a stirrer. Or the stirrer can be removed. The system is a rigid container with the water whirling inside. On what can this system do work? The energy of rotation is transferred to the energy of random motion of the molecules with time.

You certainly have seen magnetic stirrers in Chemistry lab. It uses magnetic energy to to rotate a little magnetic bar inside the water. If you remove the vessel from the stirrer the stir bar will rotate for a short time, and the water would do work on it. Is the stir bar part of the system? If it is, the rotating magnetic bar creates a rotating magnetic field outside the vessel - and that needs work done by the system. In case of a metal tank, the rotating magnet creates eddy current inside the wall, heating it up, and no work on the environment. A "system" must be exactly specified when we ask about the work it does on the environment.

ehild

Last edited: Aug 8, 2013
23. Aug 8, 2013

### CAF123

Ok, so part a) was asking if there was any work done on the system. The answer is yes - the mechanical work by the stirrer.

In part b), the question asks about the work done by the system. If we assume that the water was set into rotation by a stirrer, then is it then correct to say that the system does work?
Wsurr on system = -Wsystem on surr. As in a), the stirrer does work so from that equation, I inferred that then the system also did work. (I think you have already said this - the water does work on the blades - I just want to check my understanding).

Thank you.

24. Aug 8, 2013

### TSny

In part b), the environment is doing work on the system while the stirring is going on to get the water swirling. By reaction forces, the system does negative work on the surroundings during this time.

However, the question is asking about the time period between after the stirring is stopped (but the water is still swirling) until the time the water comes to rest. During this time, no work is done on or by the system. (Here we assume the usual thermodynamic definition of work as involving transfer of energy between system and surroundings.)

25. Aug 8, 2013

### ehild

The question asks the work done by the system after the stirring has finished. With the rigid and isolated tank, the system water-tank can not do any work on the surrounding.

ehild