Work or Heat transfer to system?

[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!