# The Basics of Thermodynamics

1. Jul 8, 2011

### Jagella

I have some questions about thermodynamics.

When liquid water is vaporized at constant pressure, the internal energy of the system may increase by fewer calories than used to vaporize the water. Is the difference in calories applied to work of some sort done by the system, or is the heat lost to the environment? I say that the heat is lost to the environment according to the Second Law of Thermodynamics.

Is it necessary to consider molecular energy when using the internal energy function? I say no: internal energy is a function of the sum of the change in heat energy and internal energy in a system.

The ocean has internal energy. Is it possible to economically extract this energy to power a ship? I say that it's not possible because you would need to use more energy to extract the ocean's energy than the energy extracted.

According to one source, an isothermal process is a thermodynamic process in which the temperature of a system and its internal energy remain constant, and all energy heat put into such a system is converted to work energy. Does this process violate the Second Law of Thermodynamics? At first I thought that it must violate the Second Law because the Second Law states that heat energy cannot be completely converted into useful work. After some reflection, I think that this paradox may be resolved by noting that the Second Law states that heat extracted from a reservoir cannot be completely converted to work. The heat from the reservoir would raise the system's temperature and its internal energy, and the system could not be isothermal.

The last issue is especially confusing.

Jagella

2. Jul 8, 2011

### Andrew Mason

Energy is conserved. So if the heat flow into the gas does not equal the change in internal energy of the gas, then either heat must flow out of the gas or it must do work: $\Delta Q = \Delta U + W$ where W is the work done by the gas on the surroundings. This does not really involve the second law.
Not quite. Internal energy pretty much consists of the molecular energy of the substance. The net heat flow into a gas less the work done by the gas must equal the change in internal energy of the gas.
The second law applies here. You cannot have a process in which the sole result is heat flow from colder to a warmer body. So in order to extract work from the ocean you have to have somewhere colder to have the heat flow to. Theoretically you could run a ship by the work produced by a heat engine operating between warm surface water and cold deep water.
Heat flows into a gas and the gas expands doing work. For an ideal gas, the work done is equal to the heat flow into the gas. You are wondering whether this violates the second law because heat flow is being converted completely into work. Good question.

In order for heat to flow into the gas and be converted entirely into work, there has to be an ever-increasing volume. Perhaps the statement of the second law by Kelvin should have been drafted to make it clear that an isothermal expansion does not violate Kelvin's statement of the second law.

AM

3. Jul 8, 2011

### Jagella

I thought that the “missing” heat may have been converted to work, but I wasn't sure what work it may have done.

I see. The book I'm using isn't clear on what energy internal energy consists of. If the burner on my stove boils water in a pan with no work output (ΔW = 0), then the resulting isocharic process is essentially the heat energy from the burner increasing the molecular energy in the water. Is that correct?

Of course. A refrigerator causes heat to flow from colder to warmer spaces, but it must do work on the system to maintain such a flow. Is that the gist of what you're saying?

So increasing gas volume is necessary for an isothermal process to be possible? If volume remains constant, would the Second Law then prohibit heat energy being entirely converted to work?

I noticed the apparent contradiction right away. I thought that the First Law was not quite correct in the case of an isothermal process and that the Second Law was then formulated to plug up this “hole” in the First Law.

Thanks a lot for the help.

Jagella

4. Jul 9, 2011

### Andrew Mason

If water is boiled in an open pan a change of state from liquid to gas occurs. The resulting vapour expands against atmospheric pressure. This is work. The water vapour molecules have to expand their volume thereby doing work on the atmosphere.
Yes. And the work required to maintain that heat flow is at least as much as the work you would obtain when running a heat engine in the reverse direction. So, for example, if you had a ship that took sea water in, cooled it and then ran a heat engine between the sea water and the cooled water, the work that you would get out of the heat engine could not be any greater than the work required to run the cooling part.

Yes. The Carnot engine uses an isothermal expansion followed by adiabatic expansion. The difference between the two parts is that in isothermal part heat flow is replacing the loss of internal energy as work is done whereas in the adiabatic part there is no heat flow. So the work done is actually greater than the total heat flow in the forward part of the cycle.

But to restore the system to its original state (ie so that the only change is heat flow from the hot reservoir - to comply with Kelvin's statement of the second law) the gas must be compressed - first isothermally so heat will flow out of the gas and then adiabatically to restore the original temperature. The result is that heat must flow out of the gas, so not all the heat flow into the gas can be converted to work.
It is not that the first law had a "hole" in it. It just did not cover this aspect. The flow of heat from cold to hot does not violate the first law since energy is conserved. But since this kind of heat flow never occurs by itself, there must be a second law stating that this cannot happen. It was not until the molecular kinetic theory was developed that the second law was fully understood.

AM

5. Jul 9, 2011

### Jagella

OK. I thought that may have been the case. It seems to me that almost all thermodynamic processes do work of some kind. Is that correct?

The work done during these two “stages” of the Carnot cycle gets its energy from the heat into the system plus internal energy. Correct?

During these two stages work is done on the system that offsets the work done by the system during the first two stages resulting in net work = 0. Am I right?

What puzzles me is how is the Carnot engine the most efficient engine, at least conceptually. Is its efficiency “ideal” in that it only loses heat during its third, isothermal stage? Is that heat loss minimized?

Thanks a lot, Andrew. Your help is appreciated.

Jagella

6. Jul 10, 2011

### Andrew Mason

Only if the volume expands against external pressure.
Correct.
No. If that was the case, the engine would do no work. The work done in isothermal compression at the lower temperature, is less than the work done in isothermal expansion. The adiabatic compression also requires less work than the adiabatic expansion because the initial and final volumes are less. So the engine definitely does net work.

AM

7. Jul 11, 2011

### Jagella

Thanks for the correction. According to the book I'm studying, the work done by the system should equal the calories of heat energy into the system less the calories of heat lost. Another way to look at it is that the net work done by the system equals the work done by the system during the isothermal-expansion process plus the work done on the system during the isothermal-compression process. Since the second work quantity has a negative value, the net work done by the system is less than the work done during the isothermal-expansion process.

Thanks a lot for the help.

Jagella