Internal energy of a container filled with water & steam

[etotheipi]

Homework Statement

A container has its bottom half filled with water and the top half filled with steam, both of which are at 100 degrees celsius. The surroundings are kept at 100 degrees celsius and heat can flow through the walls of the container.

If a massless piston at the top of the container is slowly pushed down so that the volume of steam is halved, what is the net heat added to the system?

Relevant Equations

Ideal gas equations, P-V Work

I've worked out that half of the steam must condense (since the pressure of the steam needs to remain constant for the forces on the massless piston to balance). Also, if the total volume of the container is $V$, the work done on the system equals $\frac{P_{atm}V}{4}$. When half of the steam condenses, the heat released to the system is $Q_{1} = \frac{m_{steam}l}{2}$. Since I don't know what the heat exchanged between the container and surroundings is, I'll call this extra contribution $Q_{2}$.

From the first law, $\Delta U = Q_{T} + W = \frac{m_{steam}l}{2} + Q_{2} + \frac{P_{atm}V}{4}$.

In the solutions, the final line is to set $\Delta U = 0$. From this, if we were to substitute in numerical values, we could find out $Q_{2}$.

I don't, however, understand why $\Delta U = 0$. I know that if the system contained only gas, the internal energy depends only on $n_{gas}$ and $T$ so at constant temperature $\Delta U$ is necessarily zero. I believe it is similar for incompressible liquids, that internal energy only depends on $V_{liquid}$ and $T$ (i.e. if we have more liquid at the same temperature, the internal energy has to be greater). However, in this system we have some liquid and some gas, and the relative amounts are also changing.

So why does the condition that $\Delta U = 0$ still hold?

 

[TSny]

There might be different ways to think about this. But if you define your "system" to be all of the H₂O molecules, then there will be a nonzero change in the internal energy of the system. Some of the molecules that were initially at 100 ^oC in the gas phase end up at 100 ^oC in the liquid phase. The internal energy of these molecules in the liquid phase is less than the internal energy of these molecules when they were in the gas phase.

It seems to me that the quantity $\frac{m_{steam}l}{2}$ is a representation of the loss in the internal energy of the molecules that condensed from steam to water. So I would say that for the system, $\Delta U = -\frac{m_{steam}l}{2}$. Thus, I would not take $\frac{m_{steam}l}{2}$ to be some sort of "heat released to the system". [EDIT: As pointed out by @Chestermiller below, this isn't correct. The latent heat is a change in enthalpy, not internal energy.]

When you set up the first law, $\Delta U = W + Q$ , the $Q$ on the right is the net heat added to the system. This $Q$ is the quantity that you are asked to calculate.

 

[etotheipi]

It seems to me that the quantity $\frac{m_{steam}l}{2}$ is a representation of the loss in the internal energy of the molecules that condensed from steam to water. So I would say that for the system, $\Delta U = -\frac{m_{steam}l}{2}$. Thus, I would not take $\frac{m_{steam}l}{2}$ to be some sort of "heat released to the system".

I wonder, is the latent heat actually defined in terms of the internal energy (i.e. $\Delta U = \pm ml$?), instead of heat? In the case that no external work is done, this would reduce to the more common $\Delta Q = ml$ however in examples such as the one in the problem it would mean we end up with $$\Delta U = -ml = Q + W$$ which is in the form that we need?

 

[Chestermiller]

This is a constant pressure process. So $$\Delta U=Q-P\Delta V$$or $$Q=\Delta H$$So the problem boils down to determining the change in enthalpy of the container contents. The only thing that changes enthalpy within the container is the steam that condenses; the enthalpy of the liquid initially present does not change, nor does the enthalpy of the vapor that does not condense.

 

[etotheipi]

This is a constant pressure process. So $$\Delta U=Q-P\Delta V$$or $$Q=\Delta H$$So the problem boils down to determining the change in enthalpy of the container contents. The only thing that changes enthalpy within the container is the steam that condenses; the enthalpy of the liquid initially present does not change, nor does the enthalpy of the vapor that does not condense.

Hi Chestermiller,

I saw in one of your other posts that we actually define $\Delta U = C_{v} \Delta T$ and $\Delta H = C_{p} \Delta T$ instead of using $Q$. Is it also the case that we define $\Delta U = ml$ instead of $Q$? Otherwise I'm not sure how we can also deal with work done.

Also, I'm not too sure how to calculate $\Delta H$ for the condensation of steam, apart from maybe looking up the value.

 

[Chestermiller]

Hi Chester,

I saw in one of your other posts that we actually define $\Delta U = C_{v} \Delta T$ and $\Delta H = C_{p} \Delta T$ instead of using $Q$. Is it also the case that we define $\Delta U = ml$ instead of $Q$?

Also, I'm not too sure how to calculate $\Delta H$ for the condensation of steam.

Those equations involving temperature change only apply if there is no change of phase, which is not the case in this problem. The "heat of vaporization" $l$ is synonymous with the change in enthalpy between saturated liquid and saturated vapor.

 

[etotheipi]

Those equations involving temperature change only apply if there is no change of phase, which is not the case in this problem. The "heat of vaporization" $l$ is synonymous with the change in enthalpy between saturated liquid and saturated vapor.

The thing I'm struggling to understand is that I have learned the equation $Q = ml$, which seems to imply that on condensation of some mass $m$ of steam we have evolved some quantity $Q$ of heat within our container.

However, it appears one valid way of solving the problem is to instead set $\Delta U = - ml$. I don't see how these two concepts are linked in this manner.

 

[Chestermiller]

The thing I'm struggling to understand is that I have learned the equation $Q = ml$, which seems to imply that on condensation of some mass $m$ of steam we have evolved some quantity $Q$ of heat within our container.

However, it appears one valid way of solving the problem is to instead set $\Delta U = - ml$. I don't see how these two concepts are linked in this manner.

Steam is condensing, not evaporating. So heat has to be removed. So $$Q=\Delta H=-ml$$where l is the heat of vaporization. The change in internal energy of the system is then $$\Delta U=Q-mP(v_{liquid}-v_{vapor})=-ml+mP(v_{vapor}-v_{liquid})$$where $v_{vapor}$ is the specific volume of the vapor and $v_{liquid}$ is the specific volume of the liquid.

 

[TSny]

Thank goodness for @Chestermiller! I stand corrected. The latent heat equals the change in enthalpy of the condensing vapor.

For the rest of this thread, I demote myself to a fellow learner. Let me try a little more and see if I can dig myself into an even deeper hole.

@etotheipi, don't accept any of the following if it doesn't sound right.

It helps me to have a picture:

Part I is the initial part of the vapor that doesn't condense in the process.
Part II is the vapor that does condense
Part III is the initial water

During the process, II is converted from vapor to water.

I take my system to be I+II+III. The first law applied to the system is

$\Delta U_{sys} = Q_{sys} + W_{sys}$ [Eq. 1]

Here, $Q_{sys}$ is the net heat transfer between the system and the environment and represents the quantity asked for in the problem.

Now, parts I and III do not change their internal energy during the process. So, $\Delta U_{sys} = \Delta U_{II}$.

Also, I believe that $W_{sys} = -P\Delta V_{sys} = -P \Delta V_{II} = W_{II}$. So, $W_{sys} = W_{II}$. Thus, Eq. 1 becomes

$\Delta U_{II} = Q_{sys} + W_{II}$ [Eq. 2]

or

$Q_{sys} = \Delta U_{II} - W_{II}$ [Eq. 3]

By the first law, the right side represents the heat gained by II during the process. But we know that the amount of heat gained by II equals $-m_{\small II}l$. (The minus sign is because II loses the "latent heat" as it condenses.)

So, $Q_{sys} = -m_{\small II}l = -\frac{m_s l}{2}$, where $m_s$ is the original mass of steam in the container.

Now, this doesn't agree with the answer suggested in the first post:

From the first law, $\Delta U = Q_{T} + W = \frac{m_{steam}l}{2} + Q_{2} + \frac{P_{atm}V}{4}$.

In the solutions, the final line is to set $\Delta U = 0$. From this, if we were to substitute in numerical values, we could find out $Q_{2}$.

 

[etotheipi]

So, $Q_{sys} = -m_{\small II}l = -\frac{m_s l}{2}$, where $m_s$ is the original mass of steam in the container.

Now, this doesn't agree with the answer suggested in the first post:

Your reasoning seems very airtight, and after a little more thought it does seem intuitively that $-ml$ should be the change in heat of the system (if the container and surroundings are both at the same temperature/in thermal equilibrium, why should any heat transfer between them apart from the latent heat from within the system?).

The difficulty in this question seems to be in careful definition of the system/subsystems.

 

[TSny]

Your reasoning seems very airtight, and after a little more thought it does seem intuitively that $-ml$ should be the change in heat of the system (if the container and surroundings are both at the same temperature/in thermal equilibrium, why should any heat transfer between them apart from the latent heat from within the system?).

It finally became fairly intuitive to me also, but it took quite a while for me to get there.

The difficulty in this question seems to be in careful definition of the system/subsystems.

Yes. If you take just parts I and III as "the system", then it's easy to see that $\Delta U = 0$ and $W = 0$ for this system. Hence, also, $Q = 0$ for this system via the first law. That means that all of the heat lost by part II (i.e., $ml$) must show up as heat transferred to the environment outside the container.

 

[Chestermiller]

$-ml$ should be the change in heat of the system

This phrase has no meaning in thermodynamics. Heat is not a property of a system, so it can't be something that changes in a system. Heat is the amount of thermal energy transferred from the surroundings to a system through its boundary with the surroundings. The things that change are the internal energy and the enthalpy of the system.

the latent heat from within the system

This is another phrase that has no meaning in thermodynamics. Latent heat is the amount of heat that has to be transferred between the surroundings and the system to change a unit mass of the liquid to vapor at constant temperature and pressure. A. system does not have latent heat within it.

 

[etotheipi]

This phrase has no meaning in thermodynamics. Heat is not a property of a system, so it can't be something that changes in a system. Heat is the amount of thermal energy transferred from the surroundings to a system through its boundary with the surroundings. The things that change are the internal energy and the enthalpy of the system.

This is another phrase that has no meaning in thermodynamics. Latent heat is the amount of heat that has to be transferred between the surroundings and the system to change a unit mass of the liquid to vapor at constant temperature and pressure. A. system does not have latent heat within it.

Yes you're quite right about both of these, I was being a little (or more than a little) sloppy.