Why heat and work?

Question

In thermodynamics we study energy from a heat and work transfer perspective.

Confusion:

If we take the example of an electric-engine consuming electric energy, creating a momentum/rotation and "losing" heat in the process.

Why INTERNAL ENERGY = GIVEN HEAT + GENERATED WORK?

Why not -->

INTERNAL ENERGY = INPUT - OUTPUT, and letting energies be part of the income and the outcome?

or -->

INTERNAL ENERGY = TOTAL WORK THAT THE SYSTEM CAN MAKE

Entropy would be easier to understand and more useful... like, "the amount of energy that is not productive because of a loss in frictions and stuff". Instead of going from HEAT/TEMPERATURE and micro-macro states that are abstract and lead to confusing/non-realistic endings like: "the entropy of the universe is always increasing".

Enthalpy is also like a wrapper for internal energy that was made up later. We just say... okay take INTERNAL ENERGY and also add the pV amount and call it ENTHALPY. Why not saying --> Internal Energy is the whole amount of energy (including everything: velocity, gravitational force, pression, temperature... etc). Why sticking to old concepts that were made-up on the go?

I really like thermodynamics because it is beautiful, but... I want to understand what makes the concepts of HEAT and WORK be distinct? If an engine converts electricity into momentum and friction (a form of work), why is HEAT not just a subpart of WORK?

In less words:

Why do we separate HEAT from WORK, if HEAT is also a form of WORK, just at a smaller level than work, but... still work?

Answer 1

But why don't we use "input" and "output" energy?

Because that isn't a unique distinction. For example, systems can lose or gain kinetic energy. If I tell you the system's energy changed due to an input of energy, then you aren't able to tell me what that energy is, how it was transferred, etc.

However, heat and work are unique. Heat is energy transfer specifically due to a temperature difference. Work is energy transfer due to forces acting over some distance (not attributed to the random motion of molecules). To link it to the above paragraph, systems can both input and output both work and heat, so your proposed "input-output" classification just isn't unique enough to be useful.

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Something that might also be tripping you up is the conflict between the microscopic and the macroscopic. The whole point of concept of heat is for systems where we cannot keep track of all molecules of the system. If we could, then we would just consider the work each particle does on all other particles. We could then track energy transfers due to all of this work, and then we would be good to go. Unfortunately, we cannot do this. Therefore, we bring in concepts like heat, temperature, and entropy, so that we can talk about the "most likely" behavior of the system. Objects at a "high temperature" are more likely to transfer energy to "low temperature" objects. We do not need to talk about individual molecules to understand and apply this. We call this energy transfer "heat" to differentiate it from other mechanisms of energy transfer (work) that are not due to the above statistical properties.

Answer 2

We need to distinguish between heat and work in order to state the Second Law of Thermodynamics. There are a number of ways to state the Second Law and all of them dpend critically on this distiction. Most notably

• Clausius Statement: Heat can sponaniously flow from a body with a higher temperature to a lower temperature but not the reverse
• Kelvin's Statement: No process can have its only outcome be the complete conversion of heat into work
• The Increase in Entropy: The total entropy of a closed system can never decrease (Note that entropy is defined in terms of heat transfer along a reversible path)
• The Clausius inequality: $$\int \frac{d Q}{T} \le \int \frac{d Q_\mathrm{rev}}{T} = \Delta S $$

There are a number of other statements, all of which also make use of the distinction between heat and work, though possibly in a somewhat indirect fashion.

This relationship between the distiction between heat and work and the second law suggests that we may view this distinction in terms of entropy. In the reversible case the definition of entropy tells us that reversible heat transfere entropy from one body to another, while reversible work does not. In the irreversible case, roughly speaking the same thing happens, but extra entropy can be produced along the way.

Answer 3

If we take the example of an electric-engine consuming electric energy, creating a momentum/rotation and "losing" heat in the process.

Just to be clear, momentum is not energy. The electric motor takes in electrical energy, converts it and transfers part of it as mechanical work to whatever it is connected to, and loses part of it mainly in the form of heat due to friction and $I^{2}R$ heat losses in the motor itself.

Why INTERNAL ENERGY = GIVEN HEAT + GENERATED WORK?

The first term should be change in internal energy. Then the expression would read:

Change in internal energy = Heat added to system minus work done by system. Or in equation form

$$\Delta U=Q-W$$

Why not --> INTERNAL ENERGY == INPUT - OUTPUT, and letting energies be part of the income and the outcome?

You can think the change in internal energy in a similar way if you think about it as an energy bank account. I would use the term "expense" rather than "outcome". Using the first law equation, $\Delta U$ is the change in your energy account balance. In thermodynamics the only two types of deposits and withdrawals you are able to make to the account are in the form of heat $Q$ and work $W$.

Now let's apply the first law to your electric motor. Considering the motor as the "system":

$$\Delta U_{motor}=Q-W$$

The internal energy of the electric motor itself does not change since after it performs its task it stops and cools back down to the original temperature of its environment. Therefore

$$\Delta U_{motor}=0$$

And

$$Q=-W$$

There are two forms of work involved. The electrical energy input to the electric motor is considered to be electrical work, and we will call it $W_{elect}$. It is an energy input (income) to the motor. The motor converts part of this energy input and transfers it out to perform mechanical work on whatever it is connected to. This is an energy output (expense). We can call it $W_{mech}$.

No heat is transferred to the motor. But we do have heat transfer out of the motor, the aforementioned heat losses (an expense). We can call this heat, $Q_{loss}$.

Putting this together we have

$$Q_{loss}=-(W_{elec}+W_{mech})$$

By convention in the first law equation, $Q_{loss}$ is negative, $W_{elec}$ is negative and $W_{mech}$ is positive. So in terms of the magnitudes of each of $W$ and $Q$ we have

$$-Q_{loss}=+W_{elec}-W_{mech}$$

or, in terms of mechanical work done by the motor

$$W_{mech}=W_{elec}-Q_{loss}$$

The net work done by the electric motor equals its electrical energy input minus its heat losses. I should also note that there are other minor energy losses, such as sound, vibration, etc..

Entropy would be easier to understand and more useful...

More useful for what? Entropy doesn't address energy conservation in processes. It addresses the direction that processes can take. For example heat can only naturally (spontaneously) transfer from a hot object to a cold object. The reverse process never happens naturally (spontaneously) but if it could it would not violate the first law (energy conservation). But it would violate the second law.

Enthalpy is also like a wrapper for internal energy that was made up later.

Enthalpy is a property of a system. Just like internal energy. But it is a derived property from other system properties according to

$$H=U+PV$$

or

$$\Delta H=\Delta U +\Delta (PV)$$

It is simply a more convenient property to use in certain applications, because it groups together $\Delta U$ and $\Delta (PV)$ terms simplifying the analysis. For example, the change enthalpy property alone can be used in the analysis of each of the components of an ideal Rankine cycle (boiler, turbine, condenser, and pump).

Why do we separate HEAT from WORK, if HEAT is also a form of WORK, just at a smaller level than work, but... still work?

Heat is not a form of work. And work is not a form of heat. They are separate and distinct forms of energy transfer. Heat is produced as a by-product of work, but heat is not work. Work may be produced from heat (a heat engine) but work is not heat.

Hope this helps.