Work or Heat transfer to system?

[CAF123]

Homework Statement

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?

Homework Equations

First Law of thermodynamics

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?

 

[rude man]

Homework Statement

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?

Homework Equations

First Law of thermodynamics

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.

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.

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.

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.

 

[TSny]

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?

The Attempt at a Solution

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.

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.

 

[CAF123]

Hi rude man,

The heat is produced by friction with the container sides and I guess internal friction as well.

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?

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.

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.

 

[rude man]

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.

Yeah, but the k.e. disappears!

W is defined as work done by the system (water and container) on the environment.

No, on the system.

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.

Well, then, I wonder how you explain that the internal energy of the water increased?
ΔU = δQ - δW, δQ = 0 so what does that leave for accounting for ΔU ? Not the k.e. which has gone poof!

EDIT: Think of the work as the product of torque and total angle of rotation. Torque is finite because of the friction between the water's edge and the wall. The rotational k.e. is thus changed to work by W = ∫τdθ, τ = τ(t) = torque.

 

[rude man]

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?

Correct.

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.

Yes it is. Absolutely correct.

@Chet - join in, willya?

 

[CAF123]

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.

Does this not answer your below question?

Well, then, I wonder how you explain that the internal energy of the water increased?

 

[rude man]

Does this not answer your below question?

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).

 

[TSny]

Well, then, I wonder how you explain that the internal energy of the water increased?
ΔU = δQ - δW, δQ = 0 so what does that leave for accounting for ΔU ? Not the k.e. which has gone poof!

EDIT: Think of the work as the product of torque and total angle of rotation. Torque is finite because of the friction between the water's edge and the wall. The rotational k.e. is thus changed to work by W = ∫τdθ, τ = τ(t) = torque.

CAF123 has answered this.

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.

 

[rude man]

CAF123 has answered this.

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.

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

 

[TSny]

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

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

 

[rude man]

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

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.

 

[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.

 

[CAF123]

Hi TSny,

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.

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.

 

[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

 

[Chestermiller]

Correct.


Yes it is. Absolutely correct.

@Chet - join in, willya?

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

 

[rude man]

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.

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.

 

[ehild]

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.

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

 

[Chestermiller]

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.

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

Chet

 

[rude man]

@Chet, thanks for chiming in.
rudy

 

[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.

Thanks everyone for your responses.

 

[ehild]

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).

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

 

[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?
W_{surr on system} = -W_{system 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.

 

[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.)

 

[ehild]

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?

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

 

[rude man]

You start with a system in equilibrium (no fluid rotation). Then, work is done by the environment on the system to set the fluid in motion.

You cannot start with a system that is not in equilibrium; it's not subject to thermodynamic analysis; it does not possesses thermodynamic coordinates.

Therefore, work was done on the system.

The system consists of the water at rest, and the container. The k.e. of the water represents an external agent performing work on the system.

 

[CAF123]

Think of the work as the product of torque and total angle of rotation. Torque is finite because of the friction between the water's edge and the wall. The rotational k.e. is thus changed to work by W = ∫τdθ, τ = τ(t) = torque.

The system consists of the water at rest, and the container. The k.e. of the water represents an external agent performing work on the system.

Are the above comments referring to the time when the stirrer is still acting to rotate the water? While the stirrer is doing work, I believe the first law is: $$\Delta U = -\text{bulk}(KE) + W,$$however I don't see how that equation makes sense. Certainly the internal energy of the water is increasing as we do work, but that eqn also suggests that the internal energy also increases as the bulk KE decreases. But the bulk KE is not decreasing when the stirrer is still in motion.

Perhaps I have misinterpreted the eqn, because you said in the quote above that the bulk k.e is a source of work.

Thanks.

 

[rude man]

Are the above comments referring to the time when the stirrer is still acting to rotate the water? While the stirrer is doing work, I believe the first law is: $$\Delta U = -\text{bulk}(KE) + W,$$however I don't see how that equation makes sense. Certainly the internal energy of the water is increasing as we do work, but that eqn also suggests that the internal energy also increases as the bulk KE decreases. But the bulk KE is not decreasing when the stirrer is still in motion.

The problem deals with the situation when the stirring has stopped. The equation makes sense once the stirring has stopped, the lid is emplaced on the container, and the k.e. is allowed to be dissipated into increasing the internal energy U. The sum of k.e and ΔU = 0.

We are bickering only abot whether work was done on the system after the lid is closed. As I said, we're dealing in semantics.

To me your equation is not a thermodynamic equation since the system so described is not in equilibrium, with no fixed thermodynamic coordinates (p,V,T).

So the k.e. is to be regarded as an external agent doing work on the system. The k.e. is clearly produced by external work. The system is the quiescent water and the container, not the rotating water and the container.

If you were given the above equation including k.e. this is something I have never encountered in a thermodynamics text. I have never seen the 1st law written in any way other than dU = δQ - δW (except chemsits like to use +δW) , but your instructor is the boss and in your shoes I would probably and reluctantly accept that no work was done to the system by an external agent once the stirring has stopped and the container is sealed and allowed to come to rest. The k.e. is changed to ΔU by friction and clearly work is done on the system by the diminishing k.e. (torque times angular displacement, intermolecular friction, whatever), but this work is not external work. The first law for an adiabatic system stipulates that external work is changed to internal energy on a 1 to 1 basis.

I have to follow my textbook which is by the undisputed expert thermodynamicist Prof. Mark Zemansky, who clearly states that work is done on this system. I myself am a dumb electrical engineer!

 

[CAF123]

So the k.e. is to be regarded as an external agent doing work on the system. The k.e. is clearly produced by external work. The system is the quiescent water and the container, not the rotating water and the container.

I think I can say since the only two things that can cause the internal energy to increase are work on system and heat transfer to system, then in the eqn $\Delta U = -\text{bulk}(KE) + W$, the first term on the right should be regarded as work. (it is not heat). Can you or someone else tell me if this is a correct interpretation?

If you were given the above equation including k.e. this is something I have never encountered in a thermodynamics text.

Oh no, I am just playing about with the problem trying to understand the various steps in the process.

I have never seen the 1st law written in any way other than dU = δQ - δW

If there is bulk movement, then the full form of the first law includes bulk changes in the kinetic and potential energy of the substance.

Thanks.

 

[rude man]

I think I can say since the only two things that can cause the internal energy to increase are work on system and heat transfer to system, then in the eqn $\Delta U = -\text{bulk}(KE) + W$, the first term on the right should be regarded as work. (it is not heat). Can you or someone else tell me if this is a correct interpretation?

That's my view. But not everyone's on PF & maybe your instructor/textbook.

 

[ehild]

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.

Your formula misses Δ-s. $$\Delta U + \text{bulk}(ΔKE + ΔPE) = Q + W$$. The work W done on the system by external forces + the heat Q transferred to the system changes its total energy , mechanical +internal. That is conservation of energy.

If a system is not in equilibrium, like the whirling water in the tank, it can do internal work, that is, its parts do work on other parts. But internal work is not subject of Thermodynamics. It investigates equilibrium states, not the process of transfer from one equilibrium state to the other one.

If the system is isolated, no work is done and no heat is added. Then $$\Delta U + \text{bulk}(ΔKE + ΔPE) =0$$. With PE also zero, $$\Delta U = -\text{bulk}(ΔKE) $$.

When the rotation stopped, ΔKE<0 and the system got into thermal equilibrium, its internal energy being greater than it was before by the original kinetic energy. But that does not mean work on or by the environment.

I think I can say since the only two things that can cause the internal energy to increase are work on system and heat transfer to system, then in the eqn ΔU=−bulk(KE)+W, the first term on the right should be regarded as work. (it is not heat). Can you or someone else tell me if this is a correct interpretation?

Your problem is that you use two different equations for the same thing. One is conservation of energy:

$$\Delta U + \text{bulk}(ΔKE +Δ PE) = Q + W$$,

the other is the First Law of Thermodynamics, stating that the change of the internal energy between two equilibrium states of a system is $$\Delta U = Q + W$$.

The stirring put the system into a non-equilibrium state, and the work done changed the total energy, mechanical and internal. The following irreversible process led the system to an other equilibrium state. Such process can be treated by the methods of Non-Equilibrium Thermodynamics. http://en.wikipedia.org/wiki/Non-equilibrium_thermodynamics. Question b) refers to this process, asking the work done by an isolated system on the environment. Isolated means no interaction between the system and its surroundings - no work done by the system or on the system.

ehild

 

[CAF123]

When the rotation stopped, ΔKE<0 and the system got into thermal equilibrium, its internal energy being greater than it was before by the original kinetic energy. But that does not mean work on or by the environment.

If I understand you correctly, then your view is different from rude man's view?

Your problem is that you use two different equations for the same thing. One is conservation of energy:

$$\Delta U + \text{bulk}(ΔKE +Δ PE) = Q + W$$,

the other is the First Law of Thermodynamics, stating that the change of the internal energy between two equilibrium states of a system is $$\Delta U = Q + W$$.

I see, my book actually says that the first form above is the generalised version of ΔU = Q + W. So the form ΔU + bulk(ΔKE) = Q + W means transfers of energy in the form of work and heat to system increases the total energy of the system. This means one need not consider,in the eqn ΔU = -bulk(ΔKE) + W, that the first term is necessarily work done on the system.

The stirring put the system into a non-equilibrium state, and the work done changed the total energy, mechanical and internal. The following irreversible process led the system to an other equilibrium state. Such process can be treated by the methods of Non-Equilibrium Thermodynamics. http://en.wikipedia.org/wiki/Non-equilibrium_thermodynamics. Question b) refers to this process, asking the work done by an isolated system on the environment. Isolated means no interaction between the system and its surroundings - no work done by the system or on the system.

ehild

 

[ehild]

If I understand you correctly, then your view is different from rude man's view?

I don't agree with the following sentence:

The system consists of the water at rest, and the container. The k.e. of the water represents an external agent performing work on the system.

The kinetic energy does not perform work.

I see, my book actually says that the first form above is the generalised version of ΔU = Q + W. So the form ΔU + bulk(ΔKE) = Q + W means transfers of energy in the form of work and heat to system increases the total energy of the system. This means one need not consider,in the eqn ΔU = -bulk(ΔKE) + W, that the first term is necessarily work done on the system.

ΔU + bulk(ΔKE) = Q + W is valid in general, meaning conservation of energy. But the system in the state when it has KE of an ordered motion is not in thermodynamic equilibrium. The process afterwards, during which the bulk KE transforms into internal energy (that is KE of disordered motion of molecules), falls out of the frames of Thermodynamics. ΔU = Q + W, the First Law is not valid for it.

Anything happens inside, the work and heat from outside of the isolated system is zero.

ehild

 

[Chestermiller]

Smith and Van Ness, "Introduction to Chemical Engineering Thermodynamics," state the equation for the first law of thermodynamics as:
$$ΔU+ΔE_k+ΔE_p=Q-W$$
They then state that "Closed systems often undergo processes that cause no changes in external potential or kinetic energy, but only changes in internal energy. For such processes, Eq. (2.3) reduces to $ΔU=Q-W$."

 

[rude man]

Smith and Van Ness, "Introduction to Chemical Engineering Thermodynamics," state the equation for the first law of thermodynamics as:
$$ΔU+ΔE_k+ΔE_p=Q-W$$
They then state that "Closed systems often undergo processes that cause no changes in external potential or kinetic energy, but only changes in internal energy. For such processes, Eq. (2.3) reduces to $ΔU=Q-W$."

Uh-oh, there's that W again!