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Vol. 9, 2024-10
Cite as: Hernandez, H. (2024). Understanding Work, Heat, and The First Law of Thermodynamics 2:
Examples. ForsChem Research Reports, 9, 2024-10, 1 - 56. Publication Date: 28/08/2024.
Understanding Work, Heat, and The First Law of Thermodynamics 2:
Examples
Hugo Hernandez
ForsChem Research, 050030 Medellin, Colombia
hugo.hernandez@forschem.org
doi:
Abstract
The First Law of Thermodynamics represents the principle of energy conservation applied to the
interaction between different macroscopic systems. The traditional mathematical description of
the First Law (e.g. ) is rather simplistic and lack universal validity, as it is only
valid when several implicit assumptions are met. For example, it only considers mechanical work
done associated with a change in volume of a system, but completely neglects other types of
work. On the other hand, it employs the concept of entropy which is not only ambiguous but
also implies only heat associated with a temperature difference, neglecting other types of heat
transfer that may take place at mesoscopic and/or microscopic levels. In addition, it does not
consider mass transfer effects. In the previous report of this series, a more general
representation of the First Law is obtained considering different conditions and different types
of interactions between the systems. In this report, the expression previously obtained is
applied to different representative examples, involving macroscopic systems with no volume
change, gas systems with volume change, and even a case where mass transfer between the
systems takes place.
Keywords
Compression, Energy Conservation, Expansion, Forces, Friction, Gravity, Heat, Interaction,
Piston, Thermodynamics, Volume Change, Work
1. Introduction
This is the second part of a series of reports discussing work, heat and the first law of
Thermodynamics. In the first part [1], the main theoretical concepts were discussed and a new
framework for understanding energy transfer between systems was presented. In this second
part, representative examples are presented and explained using the proposed framework.
Let us first recall the following basic definitions introduced in the first part:
Understanding Work, Heat, and the First
Law of Thermodynamics 2: Examples
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Work (Physics): The [theoretical] work done by a force on a body represents the
expected change in translational kinetic energy of the body if such force were acting alone
on it.
(1.1)
where represents the position of the body at time .
Apparent Work: The apparent work done by a force on a body represents the
theoretical amount of work that would be done when the force is assumed to remain
constant.
(1.2)
where represents the displacement of the body during the time interval .
Net Work: The net mechanical work done by all forces acting on a body represents the
true change in translational kinetic energy of the body.
(1.3)
where represents the number of forces acting on the body.
Work (Thermodynamics): The thermodynamic work done by a closed system 1 on a
closed system 2 is the apparent theoretical amount of energy transferred from system 1 to
system 2 when a force is exerted by system 1 on system 2, and corresponds to the
apparent mechanical work done by such force.
(1.4)
where represents the displacement of the center of mass position of system 2.
Total Work done onto a System: The total thermodynamic work done on a system by
its surrounding systems is the apparent theoretical amount of energy transferred to
the system by each system in the surroundings.
(1.5)
where represents the number of systems in the surroundings of system .
Understanding Work, Heat, and the First
Law of Thermodynamics 2: Examples
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Total Work done by a System: The total thermodynamic work done by a system on
its surrounding systems is the apparent theoretical amount of energy transferred by
the system to all other systems in the surroundings.
(1.6)
Macroscopic Translational Kinetic Energy : Kinetic energy associated with the translational
motion of the center of mass of a system.
Internal Kinetic Energy : Kinetic energy associated with the motion of the components of a
system, relative to the center of mass of the system.
Macroscopic Potential Energy : Potential energy associated with the position of the center
of mass of a system.
Internal Potential Energy : Potential energy associated with the relative positions of the
components within a system.
Total Macroscopic Energy of a System :
(1.7)
Total Internal Energy of the System :
(1.8)
Total Energy of the System :
(1.9)
Heat (Thermodynamics): The heat transferred from system 1 to system 2 is the net
amount of energy transferred from system 1 to system 2 after discounting the theoretical
amount of energy transferred from system 1 to system 2 in the form of thermodynamic
work.
(1.10)
where is the amount of energy transferred from system 1 to system 2.
As it can be seen, Thermodynamic Work is determined by interaction forces between the
systems. In general, interacting forces can be classified depending on:
Understanding Work, Heat, and the First
Law of Thermodynamics 2: Examples
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The time- or position-dependence of the force:
o Variable forces: This is the most general case. The interaction forces between
the elements of the systems change over time as the relative positions between
the elements change.
o Invariant forces: The interaction forces acting between the different elements of
the systems remain constant (or relatively constant) for the time interval
considered. Thus, they are practically independent from time and/or position. In
this case, the energy transfer considering only an invariant force is:
(1.11)
where and represent the number of elements in systems 1 and 2,
respectively; the term represents the -th element present in system 1;
and the term represents the -th element present in system 2.
The temporal scale of the interaction (relative to the scale of observation):
o Long-range interaction forces: This is the most general case. There is a significant
net change in the interaction potential energy of the interaction during the time
interval considered ().
o Short-range interaction forces: For these forces, the interaction potential energy
changes during the time interval considered, but the net change over the whole
period is negligible. These forces will be denoted by the subscript . In this case,
only kinetic energy changes need to be considered:
(1.12)
The spatial effect of the interaction:
o Local forces: This is the most general case. Typically, only a fraction of elements
in both systems are mutually interacting, corresponding to those close to the
boundary between the systems. Thus, most force terms are zero.
o Field forces: The interaction force per unit mass
1
exerted by an element of a
system is approximately the same for all elements in the other system. These
forces will be denoted by the subscript . In this case,
(1.13)
1
Equivalent to the translational acceleration of the system when it is the only force exerted on it.
Understanding Work, Heat, and the First
Law of Thermodynamics 2: Examples
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The origin of the force:
o Passive forces: Passive forces emerge spontaneously due to the relative position
between two systems, and no action is needed. Most forces in Nature are
passive forces.
o Active forces: Active forces do not emerge due to the relative position of the
systems, but because of an action of the system. Active forces are not
spontaneous, and thus, they require a continuous supply of energy by the
system to sustain the force. As an example, consider the force exerted by
muscles, motors, engines or other devices. Active forces will be denoted by the
subscript . In this case, a consumption of internal energy
in the
system originating the force is observed:
(1.14)
where represents the power consumed by the system per unit of
active force exerted on (having units of velocity). Notice that when an active
force is exerted on a system, an opposite, passive reaction force will
spontaneously emerge.
The First Law of Thermodynamics represents the principle of conservation of energy [2]
considering the interaction of systems. A very general mathematical representation of the first
law for any arbitrary system, derived in the first part of this report series [1], is the following:
(1.15)
where represents the change in total energy of an arbitrary system of mass ,
interacting with different surrounding systems () after a time interval ; is the
energy transferred from surrounding system in the form of heat, due to the unobserved
translation of the system caused by local, short-range forces;
is the energy transferred
from surrounding system in the form of heat, due to internal motion of the elements of the
system caused by local, short-range interaction forces; is the energy transferred from
surrounding system with mass to system in the form of heat, due to the unobserved
Understanding Work, Heat, and the First
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translation of the system caused by long-range field forces; is the energy transferred
from system to surrounding system in the form of heat, due to the unobserved translation
of the surrounding system caused by long-range field forces;
is the amount of internal
energy consumed by the system to sustain an active force on surrounding system ,
is the amount of energy transferred by surrounding system to system in the form of
[apparent macroscopic] work, caused by local, short-range forces; is the amount of
energy transferred by surrounding system to system in the form of [apparent
macroscopic] work, caused by long-range field forces; is the amount of energy
transferred by system to surrounding system the in the form of [apparent macroscopic]
work, caused by long-range field forces; is the specific energy of the mass
transferred from surrounding system to system ; and is the specific energy of the
mass transferred from system to surrounding system .
The first sum in Eq. (1.15) represents the net amount of energy transferred to system from the
surroundings in the form of heat; the second sum is the net amount of energy transferred to
system from the surroundings in the form of work done on the system; and the third sum is
the total amount net amount of energy transferred to system from the surroundings
associated to mass transfer.
Eq. (1.15) can also be expressed in terms of the work done by the system as follows:
(1.16)
where all the heat and work terms included in these expressions are not reflexive. That is, the
sum of each of heat and work terms considered in both directions is not zero.
The first law can also be expressed in terms of the internal energy change of the system as
follows (neglecting the effect of mass transfer on the macroscopic energy of the system):
(1.17)
Understanding Work, Heat, and the First
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or in terms of macroscopic work:
(1.18)
Particularly, the term
can be expressed as an approximate function of macroscopic
variables as follows:
(1.19)
where is a heat transfer coefficient between system and surrounding system ;
is the area of the boundary between system and surrounding system ; and
are the temperatures of system and surrounding system , respectively; is the
average mass of the elements of surrounding system at the boundary with system ;
is the translational velocity of surrounding system ; and is the Boltzmann constant.
Now, if the following conditions are met:
No field forces are considered .
Forces are invariant .
No active forces are present
.
The velocity difference between the systems can be considered negligible
.
The system is closed .
Then, the first law applied to the total energy of the system simplifies into:
(1.20)
Alternatively, if the following conditions are met:
No short-range work is done on the system of interest .
Forces are invariant .
No active forces are present
.
Understanding Work, Heat, and the First
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The velocity difference between the systems can be considered negligible
.
The system is closed .
Then, the first law applied to the internal energy of the system becomes:
(1.21)
While Eq. (1.20) and (1.21) resemble the expressions of the first law commonly found in the
literature, they implicitly require fulfilling the following additional condition (reflexive work) to
satisfy the principle of conservation of energy:
(1.22)
2. Macroscopic Systems without Volume Change
2.1. Free Fall of an Object in a Vacuum
As a first example, let us consider one of the simplest situations: The free fall of certain rigid
object (having constant shape and volume) under the effect of gravity with no resistance to its
fall (such as a fall in perfect vacuum). The situation considered is depicted in Figure 1.
Only two systems are considered: The free-falling body (System 1) and the Earth (System 2).
Notice that for this example, Earth is a massive system, and therefore and
.
The force exerted by Earth on the falling body is:
(2.1)
where represents the gravitational acceleration vector, represents the magnitude of the
gravitational acceleration, and is a unit direction vector heading upwards with respect to the
surface of Earth.
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Figure 1. Free fall of a rigid body (system 1) under the effect of Earth’s gravity (system 2) in vacuum. Red
arrows: Interaction forces. Blue arrow: Displacement of the center of mass of system 1.
From Newton’s third law, we may conclude that the falling body also exerts a force on Earth, in
this case having the same magnitude but opposite sign, as follows:
(2.2)
This pair of interaction forces represents a macroscopic attraction between the systems and will
be assumed to remain approximately constant during the fall. Thus, gravitational forces can be
considered as constant field forces. For this reason, the subscript is used in these forces.
The interaction forces can be summarized as a matrix of vectors (represented by tensor ) as
follows:
(2.3)
Notice that the matrix of interaction force vectors is antisymmetric (skew-symmetric) [3]. The
diagonal of zeros indicate that self-interaction forces do not exist
2
.
Considering only the vertical direction, we obtain the following vertical interaction force matrix:
2
Internal forces within a system may exist, but they will be pairs of interaction forces between different
subsystems. The net effect of these internal interaction forces is zero. For this reason, no macroscopic
self-interaction of the systems is considered to exist.
Understanding Work, Heat, and the First
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(2.4)
The interaction force matrix in horizontal directions is a null or zero matrix.
The vertical displacement of the falling body can be expressed in terms of the vertical
direction unit vector as follows:
(2.5)
Assuming a negligible initial velocity of the bodies, the vertical displacement of system 2 would
be approximately given by:
(2.6)
With the previous information, we may now proceed to calculate thermodynamic work,
thermodynamic heat, and changes in total energy, internal energy and macroscopic energy for
each system.
The total energy change for each system can be determined using Eq. (1.15), as follows:
(2.7)
(2.8)
Since no mass transfer takes place between the systems in this example, then the mass change
terms are zero. This can be represented in matrix form as follows:
(2.9)
Since no short-range forces are considered, all terms related to short-range forces are also zero:
(2.10)
Understanding Work, Heat, and the First
Law of Thermodynamics 2: Examples
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(2.11)
(2.12)
Since no active forces are considered, the corresponding heat terms are also zero:
(2.13)
Also, since the interaction forces are assumed invariant, then:
(2.14)
Notice that all terms involved in the internal energy change of the systems (Eq. 1.17) are zero for
this example, as can be seen in the following matrix notation:
(2.15)
where the superscript indicate the transposed matrix, the operator extracts the diagonal
of a squared matrix as a column vector, the term represents a column vector of ’s with
rows, where is the number of systems considered, and the matrices and are in this
case:
(2.16)
(2.17)
Replacing Eq. (2.9) to (2.14) in Eq. (2.15) it can be observed that a vector column of zeros is
obtained. Thus, it can be concluded that the internal energy of the systems remains constant
during the free fall of an object:
(2.18)
Understanding Work, Heat, and the First
Law of Thermodynamics 2: Examples
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In addition, the total energy changes of the systems (Eq. 2.7 and 2.8) can be expressed in matrix
form as follows:
(2.19)
Considering Eq. (2.9) to (2.14), Eq. (2.19) simplifies into:
(2.20)
Where the thermodynamic work done between the systems can be determined from Eq.
(1.4):
(2.21)
and therefore,
(2.22)
In principle, no significant energy is transferred between the systems during the free fall of a
body.
These results and additional calculations are summarized in Table 1. The additional calculations
include the determination of the total work done on each system (Eq. 1.5), also representing the
change in macroscopic kinetic energy of the system; the macroscopic potential energy change
(only for field forces); the macroscopic energy change (macroscopic kinetic plus macroscopic
potential, Eq. 1.7); the internal energy change (Eq. 2.15); and the total energy change (Eq. 2.19).
We can see that the free fall of a body in the absence of any environmental resistance would
result in no significant exchange of energy between the body and Earth. The increase in
macroscopic kinetic energy of the body during fall is exactly compensated by a decrease in the
corresponding fraction of the interaction potential energy associated with the falling body.
However, since , the change in macroscopic energy of Earth can be considered
negligible.
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Table 1. Results summary for the free fall of an object in vacuum.
Observed in / Relative to / Exerted
on
Total
1 (Object)
2 (Earth)
System Mass
Vertical force exerted
by System
1 (Object)
2 (Earth)
Net Force
Vertical Displacement
Net Work done on System
Macroscopic Kinetic Energy Change
Macroscopic Potential Energy Change
Macroscopic Energy Change
Internal Energy Change
Total Energy Change
In conclusion, the body uses its own potential energy to fall. This result is perfectly consistent
with basic concepts introduced by physics textbooks. Now, if air resistance is considered, a third
system (atmospheric air) must be included in the analysis. This will be done in the next example.
2.2. Free Fall of an Object with Air Resistance
The situation considered now involves three systems as illustrated in Figure 2. The additional
system (system 3) is atmospheric air. Air, being a gas, exerts a short-range force on every
surface in contact with it. For instance, it exerts a downwards vertical force on the
ground, where is the pressure of atmospheric air, and is the total surface of the
Earth in contact with atmospheric air. This force indirectly includes the effect of the
gravitational force exerted by Earth on the mass of the atmosphere [4]. Thus, it will be the only
interaction force considered between both systems. The overall force exerted by air is a short-
range, local (non-field) force since it is not exerted on all elementary components of the system,
but only on the surface of the system. In addition, air exerts forces on all sides of the falling
object. The forces applied on non-vertical directions will exactly compensate resulting in a zero
net force between air and the object. However, the force exerted by the molecules of air below
the body will be greater than the force exerted by the molecules of air above the body, due to a
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difference in relative molecular fluxes of air [5,6]. Thus, the net force exerted by air on the body
will be considered as a local friction force. A force with the same magnitude and opposite
direction is then exerted by the falling body on the mass of air. Air is also a massive system with
negligible change in its center of mass during the process . However, air is not rigid,
as local flows of air molecules may emerge during the fall of the body, as the air molecules
below the body move upwards trying to occupy the empty space left behind by the body. In
addition, we may assume in this case that .
Figure 2. Free fall of a rigid body (system 1) under the effect of Earth’s gravity (system 2) with air (system
3) resistance. Thick arrows: Interaction forces. Narrow arrow: Displacement of the center of mass of
system 1.
For this situation, the vertical interaction force matrix becomes:
(2.23)
with vertical displacement of the systems:
(2.24)
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Notice that and are considered negligible in this case.
In addition, all other matrices required for the analysis of energy change of the systems are
presented next, considering:
No mass transfer between the systems .
Earth’s interaction forces are invariant .
The falling body is rigid, but air is not .
No active forces are considered .
Heat transfer between air and Earth is neglected.
(2.25)
(2.26)
(2.27)
(2.28)
(2.29)
(2.30)
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(2.31)
(2.32)
(2.33)
The internal energy changes of the systems become (Eq. 2.15):
(2.34)
while the total energy changes of the systems are (Eq. 2.19):
(2.35)
Now, considering the conservation of energy, the following equation is obtained:
(2.36)
which simplifies into:
(2.37)
where is the vertical speed of the object.
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The term represents the amount of macroscopic energy “lost” by the falling object
due to air resistance. Such energy is not “lost” but rather transformed into two types of motion,
represented by the two terms at the right-hand side of Eq. (2.37). The term represents
the energy associated to the local (mesoscopic) flow of air from the bottom to the top of the
falling object, as depicted in Figure 3. The term
represents the
increase in molecular kinetic energy (temperature) resulting from the collision between the
falling body and surrounding air molecules.
Figure 3. Local air currents developed during the free fall of a rigid object
As the speed of the object increases, the increase in molecular kinetic energy overcomes the
increase in mesoscopic kinetic energy,
, and thus, the
air resistance force can be expressed as:
(2.38)
indicating that the magnitude of the upwards resistance force exerted by air on the falling body
is approximately proportional to the contact surface and to the downwards velocity of the
falling body.
On the other hand, for low falling velocities, most of the macroscopic energy “lost” by the
falling body transforms into mesoscopic air currents with no significant increase in temperature.
Additional calculations assuming a differential change in position are summarized in
Table 2. This assumption implies that air resistance can be considered approximately constant
during the elapsed time .
According to these results, the increase in macroscopic kinetic energy of the falling body is less
than for the free fall in vacuum, due to air resistance. Part of this missing macroscopic energy
stays in the body in the form of heat (increased temperature), while the rest goes to the
surrounding air in the form of mesoscopic air currents and temperature rise.
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Table 2. Results summary for the free fall of an object in atmospheric air (differential change)
Observed in / Relative to / Exerted on
Total
1 (Object)
2 (Earth)
3 (Air)
System Mass
Vertical
force
exerted by
System
1 (Object)
2 (Earth)
3 (Air)
Net Force
Vertical Displacement
Net Work done on
System
Macroscopic Kinetic
Energy Change
Macroscopic Potential
Energy Change
Macroscopic Energy
Change
Internal Energy Change
Total Energy Change
2.3. Rising of Helium Balloon
This example is very similar to the previous one. The main difference now is that the force
exerted by air molecules on the helium balloon is greater than the gravitational force on the
balloon, resulting in a positive (instead of negative) vertical displacement. Figure 4 illustrates
the current situation.
Compared to the previous example, only the work terms and total energy changes are different:
(2.39)
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Figure 4. Rising of a light body (system 1) under the effect of Earth’s gravity (system 2) and air (system 3)
lifting. Thick arrows: Interaction forces. Narrow arrow: Displacement of the center of mass of system 1.
(2.40)
(2.41)
All remaining calculations, considering a differential change in position , are
summarized in Table 3.
The net effect of air on the object is an increase in the macroscopic energy of the body, whereas
for the previous example, the net effect was a decrease in macroscopic energy. Let us recall
that gravity does not change the energy of small systems
3
. In this case, the macroscopic energy
of the rising body, as well as the increase in temperature due to air resistance, is provided by the
surrounding air itself as can be seen from the conservation of energy expression:
3
Systems with small masses compared to the mass of Earth.
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(2.42)
The term is negative and its magnitude is greater than
,
indicating that air at the bottom transfers to the body more energy than it receives back at the
top. The ultimate source of energy used by air is gravity, which increases the speed of the
molecules (increasing the temperature of air) [4], and is then transferred to the balloon (with a
subsequent decrease in temperature of air at the bottom of the body).
Table 3. Results summary for the rising of a light object in atmospheric air
Observed in / Relative to / Exerted on
Total
1 (Object)
2 (Earth)
3 (Air)
System Mass
Vertical
force
exerted by
System
1 (Object)
2 (Earth)
3 (Air)
Net Force
Vertical Displacement
Net Work done on
System
Macroscopic Kinetic
Energy Change
Macroscopic Potential
Energy Change
Macroscopic Energy
Change
Internal Energy Change
Total Energy Change
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2.4. Lifting of a Box
Increasing the level of complexity, let us now consider the lifting or elevation of a rigid box,
having constant volume and mass, against the force of gravity. Particularly, let us consider the
situation depicted in Figure 5. Here, a lift truck at rest is lifting a box of mass a certain
height difference . At least four different systems can be identified here: 1) The box of mass
, 2) the lift truck, 3) the ground, and 4) the atmosphere. Since the lifting process usually
takes place at low speeds, the friction effect of the atmosphere can be neglected, and since the
only system changing its center of mass position is the box, interactions with all other systems
can also be neglected. Thus, the atmosphere will not be included as a system in the present
analysis.
Figure 5. Lift truck elevating a box. Adapted from an image by Vectorportal.com CC BY
The forces acting on each system (only forces with vertical components are considered) are
presented in Figure 6. The force exerted by the truck on the box
is an active force,
because such force does not naturally occur unless an internal action is performed by the truck.
However, the reacting force is passive and short ranged . On the other hand, two types
of interaction forces between the ground (Earth) and the truck are considered: A field force
(gravity) and a short-range force also known as normal force. All systems are
again assumed to be in thermal equilibrium, that is, all system temperatures are the same.
The corresponding vertical interaction force matrix for this situation is:
(2.43)
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Figure 6. Forces acting on the different systems during lifting. Green: Forces acting between the lift truck
and the box. Blue: Field forces acting between the box and the ground. Purple: Field forces acting
between the truck and the ground. Red: Short-range forces acting between the truck and the ground.
Adapted from an image by Vectorportal.com CC BY
And the vertical displacement of the systems is:
(2.44)
Vertical position changes of the truck and the ground are negligible.
Considering a net zero force acting on the truck, we have:
(2.45)
Thus, the normal force exerted by the ground on the truck compensates both the
weight of the truck and the active force exerted on the box. Also notice that the normal force
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acting on the truck is actually distributed among the four wheels in such a way that a net zero
torque is obtained
4
.
The vertical interaction force matrix then becomes:
(2.46)
Now, considering:
No mass transfer between the systems .
Field forces are invariant .
Earth’s mass is larger than the masses of the truck and the box .
Contact, passive forces are invariant ()
No additional forces are considered.
The matrices required for the analysis of energy change of the systems become:
(2.47)
(2.48)
(2.49)
(2.50)
(2.51)
4
When the normal force exerted on the wheels of the truck results in a non-zero torque, then the truck
overturns.
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(2.52)
(2.53)
(2.54)
(2.55)
The internal energy changes of the systems are (Eq. 2.15):
(2.56)
and the total energy changes of the systems are (Eq. 2.19):
(2.57)
Thus, from the conservation of energy we may conclude that:
(2.58)
At the end of the lifting process, the macroscopic kinetic energy of the box returns to zero as its
vertical motion stops. Clearly, the active force exerted by the truck is not constant during this
process. At first, it is greater than the weight of the box, and at the end, it is equal to the weight
of the box. As a limiting case we may assume that the initial force is equal to the weight of the
box, but then the time required to lift the box would be infinite (the system remains at
mechanical equilibrium without any vertical displacement). Taking all this into consideration, the
results summary for this example is shown in Table 4.
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Table 4. Results summary for the lifting of a box
Observed in / Relative to / Exerted on
Total
1 (Box)
2 (Truck)
3 (Ground)
System Mass
Vertical
force
exerted by
System
1 (Box)
2 (Truck)
3 (Ground)
Net Force
Vertical Displacement
Net Work done on System
Macroscopic Kinetic
Energy Change
Macroscopic Potential
Energy Change
Macroscopic Energy
Change
Internal Energy Change
Total Energy Change
From this example we may observe that the energy of the box increases when lifted
(assuming at rest at the end of the process), and this energy is provided by the truck while
sustaining the active force. The internal energy used for this lifting process is
. However, the total chemical energy employed (fuel consumption) will be greater as
part of the internal chemical energy increases the temperature of the truck (which is another
type of internal energy). The amount of chemical energy transformed into thermal energy
increases with the work done by truck on the box ()
5
. The excess work (greater than
) initially becomes macroscopic kinetic energy of the box, but then it dissipates as
thermal energy of the truck due to friction and/or braking. Air resistance may also consume part
of the excess work, but this effect was not considered in this example.
We may thus conclude that the minimum work required to lift a mass is but this is not
the work done by the truck when lifting a mass. The work done by the truck when lifting a mass
is
.
At any arbitrary time , the net force acting on the box is
, and therefore, the box
will move upwards with the following acceleration:
5
The term represents the mean active force exerted by system 2 on system 1 during the time
interval considered.
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(2.59)
2.5. Pushing a Mass on a Horizontal Surface
In the following example, the displacement of a mass takes place on a perfectly horizontal
surface, as illustrated in Figure 7.
Figure 7. A person pushing a mass on a horizontal surface and the respective forces involved. Green:
Horizontal interaction forces between the person and the mass. Purple: Horizontal interaction forces
between the person and the floor. Red: Horizontal interaction forces between the mass and the floor.
Three different systems can be clearly identified: 1) The mass being pushed, 2) the person
pushing the mass, and 3) the horizontal surface (floor). Only horizontal forces are shown in
Figure 7. The upper image illustrates the pushing force exerted by the person on the mass
(
), as well as on the floor (
), and the friction force exerted by the floor opposing
the motion of the mass (). All these forces are local, short-range forces, and particularly,
the forces exerted by the person on the mass and the floor are active forces, since they will not
occur unless the person stimulates the body muscles using electric (neural) signals. In the lower
image we observe the corresponding counterparts of these forces (all passive and short
ranged), emerging according to Newton’s third law.
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The object moves a distance along the horizontal surface when the force applied by the
person overcomes the maximum static friction force. Such motion is assumed monotonic, that
is, the object is not coming back and forth during this process. It will also be assumed that the
person is not moving from its position as the distance considered is short. For longer distances,
the person also must move with the mass taking steps, moving forward the back leg while
supported on the other leg, but this additional motion will not be considered here.
It would be possible to include a fourth system in the analysis, corresponding to the atmosphere
around both the person and the object. However, we will assume again that the velocity of the
displacement is relatively low, and thus, that the air provides only a negligible resistance to
motion (friction).
The horizontal interaction force matrix for this scenario is the following:
(2.60)
The horizontal displacement vector for this example is:
(2.61)
Since the floor is massive compared to the pushed object, the displacement of the floor can be
considered negligible. Now, since the person is not moving, a balance of forces acting on the
person yields:
(2.62)
Thus,
(2.63)
This means that the force exerted by the legs of the person on the floor must have the same
magnitude as the force exerted by the arms of the person on the object, but in opposite
direction. With this consideration, the interaction force matrix becomes:
(2.64)
Now, considering that:
No mass transfer takes place between the systems .
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The floor is massive .
Monotonic motion and rigid bodies are considered (no mesoscopic motion) ().
No horizontal field forces are present .
The heat and work matrices become:
(2.65)
(2.66)
(2.67)
(2.68)
(2.69)
(2.70)
(2.71)
(2.72)
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(2.73)
The internal energy changes of the systems become (Eq. 2.15):
(2.74)
with the corresponding total energy changes (Eq. 2.19):
(2.75)
Let us recall that the terms are associated to changes in the molecular motion of the
systems during the interaction, and in this example, it is mostly associated to heat generated by
friction. On the other hand, the terms are associated to internal energy consumption to
sustain the active forces. In this example, it is energy consumed by the person to exert forces.
Even if no work is done (e.g. ) the terms are not necessarily zero, resulting
in physical tiredness or muscular fatigue of the person.
The energy conservation principle then implies that:
(2.76)
indicating that the energy associated to the net work done on the object is provided by the
internal energy sustaining active forces, after discounting energy losses due to friction.
If no friction between the object and the floor is considered
6
(), heat derived from
friction is neglected and all systems are assumed at thermal equilibrium (, then the
energy changes of the systems simply become:
(2.77)
6
Friction between the person and the floor is mandatory because
. If there is
no such friction, then it is impossible to exert an active force on the object.
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(2.78)
implying that all internal energy consumed by the person while exerting active forces become
work done on the object, increasing its macroscopic kinetic energy. Since no object-floor friction
is considered, the velocity gained by the object would be conserved after removing the active
forces. In the presence of object-floor friction, such macroscopic motion would be dissipated in
the form of molecular motion (temperature increase) of both the object and the floor, until the
object stops again.
Table 5. Results summary for the scenario where a mass is pushed horizontally without friction between
the object and the floor with an invariant force
Observed in / Relative to / Exerted on
Total
1 (Object)
2 (Person)
3 (Floor)
System Mass
Horizontal
force
exerted by
System
1 (Object)
2 (Person)
3 (Floor)
Net Force
Horizontal Displacement
Net Work done on System
Macroscopic Kinetic
Energy Change
Macroscopic Potential
Energy Change
Macroscopic Energy
Change
Internal Energy Change
Total Energy Change
A precise determination of the terms would require knowledge about friction factors and
contact areas between the systems. In the case of the terms, it would be possible to
assume that the nature of the active forces is similar and therefore, . Since the forces
have the same magnitude, we may then assume:
.
Additional calculations neglecting object-floor friction and assuming an invariant pushing force
are summarized in Table 5.
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2.6. Pushing a Mass on an Inclined Surface
An additional degree of complexity can be introduced to the previous example, by considering
an inclined surface instead of a completely horizontal surface. Figure 8 illustrates this situation.
Only the most representative forces are shown. For simplicity, let us consider only the forces
exerted in the direction parallel to the inclined surface. Forces perpendicular to the inclined
surface are perfectly counteracted, and no system is displaced in such direction.
Figure 8. A person pushing an object on an inclined surface and the respective forces involved. Green:
Interaction force between the person and the object. Purple: Interaction force between the person and
the floor. Red: Local interaction forces between the mass and the floor. Orange: Field interaction force
between the mass and the floor (Earth). Cyan: Field interaction force between the person and the floor
(Earth).
The matrix of interaction forces which are parallel to the inclined surface is the following:
(2.79)
The force has two components, one perpendicular to the surface corresponding to the
normal response of the surface to the weight of the object, and the other parallel to the
surface, representing the net friction force observed between the object and
the floor. The angle , which is practically independent of the angle of the surface , and is
mainly determined by the magnitude of the friction force, relative to the normal force of the
surface. If the net friction between the object and the floor is negligible or zero, then
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and . A similar passive force exists between the person and the floor , which
was not shown in Figure 8. The term represents a passive friction force emerging
as a natural response of the floor to the weight of the person only. It does not include the
passive force emerging as the reaction to the active force exerted by the legs of the
person. The angle , which is independent of and , also depends on the magnitude of the
friction force, relative to the normal force of the surface.
Assuming that the person is not moving, then:
(2.80)
Indicating that the active force exerted by the legs of the person must compensate the active
force exerted by the arms and the weight of the person, discounting the contribution of friction
on the support of the weight of the person. If the passive friction force is enough to support the
weight of the person, then , as in the previous example. At some point, when
the friction force reaches its limit, the legs of the person will slip. Assuming that it is not the
case, and neglecting the friction between the object and the floor for illustrative purposes, the
interaction force matrix can be simplified into:
(2.81)
Now, considering that:
No mass transfer takes place between the systems .
The floor is massive .
Monotonic motion and rigid bodies are considered (no mesoscopic motion) ().
Field forces are invariant .
Heat derived from friction is neglected and all systems are assumed at thermal
equilibrium (.
Then the heat and work matrices become:
(2.82)
(2.83)
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(2.84)
(2.85)
(2.86)
(2.87)
(2.88)
(2.89)
(2.90)
The internal energy changes of the systems become (Eq. 2.15):
(2.91)
with the corresponding total energy changes (Eq. 2.19):
(2.92)
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Additional calculations for this example are summarized in Table 6.
Table 6. Results summary for the scenario where a mass is pushed over an inclined surface without
friction between the object and the floor with a constant force
Observed in / Relative to / Exerted on
Total
1 (Object)
2 (Person)
3 (Floor)
System Mass
Parallel
force
exerted by
System
1 (Object)
2 (Person)
3 (Floor)
Net Force
Parallel Displacement
Net Work done on System
Macroscopic Kinetic
Energy Change
Macroscopic Potential
Energy Change
Macroscopic Energy
Change
Internal Energy Change
Total Energy Change
These are the same results that were obtained in the previous example (neglecting the object-
floor friction), with the only difference that the active force that must be exerted by the person
on the object must be , and the net acceleration of the body will be:
(2.93)
For we simply obtain the behavior described in the previous example. For we
obtain the behavior described in example 2.4, for the vertical lifting of an object.
3. Volume Changes in Macroscopic Gas Systems
Example 2.2 and example 2.3 already considered atmospheric air as a macroscopic system.
However, no volume change of air was involved. In this section, different examples involving
the change in volume (either expansion or compression) of macroscopic gas systems are
discussed.
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3.1. Horizontal Adiabatic Expansion of Gas Systems
Let us first consider the most representative example of classical thermodynamics: Gas
expansion in a piston-cylinder system. The system is illustrated in Figure 9. The piston is
assumed horizontal to remove gravitational considerations from this analysis. Both the piston
and the cylinder are assumed to be thermally insulated, so no energy is transferred to the air or
the ground by thermal conduction. As it can be seen, at least different systems are involved in
this relatively simple example: 1) The gas contained in the piston-cylinder system, 2) the moving
piston, 3) the air surrounding the piston-cylinder system, 4) the fixed cylinder, and 5) the
ground.
Figure 9. Horizontal expansion of a gas in a piston-cylinder system.
Gas expansion takes place only when the pressure of the gas is greater than the surrounding air
pressure. The typical assumption done in conventional thermodynamics is that both pressures
are practically identical, with only an infinitesimally small difference between them (reversible
process). In practice, this implies that the piston displaces a distance after an almost infinite
time, or that no expansion takes place at all.
Here, we are not considering how the gas pressure became greater than the external air
pressure. There are different ways to achieve this under adiabatic conditions. For example, the
system is previously pushed by the piston and then released, or the external air pressure is
suddenly decreased. Regardless of the system history, we will only focus on the subsequent
results.
Before proceeding to the mathematical analysis of this system, let us discuss additional
considerations:
The gas is a macroscopic system composed of molecules moving at random,
independent velocities in all three directions.
The velocity of the gas molecules in a single direction is normally distributed, initially
with a mean value of zero, and a variance proportional to the system temperature.
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The temperature of all systems considered initially is the same, so that they are all at
thermal equilibrium.
The collision of gas molecules with cylinder walls results in complete reversion of the
molecular velocities in the direction perpendicular to the wall. These collisions are
assumed perfectly elastic. This assumption relies on i) neglecting thermal energy
changes of the cylinder during collisions (neglecting internal energy changes of the
cylinder), and ii) considering that the cylinder is fixed to the ground in any way (using
bolts, glue, or due to strong friction), and is impossible for it to move in any direction
(neglecting macroscopic energy changes of the cylinder).
The collisions of gas molecules with the piston are not completely elastic, as a significant
portion of the kinetic energy of the gas molecule is transferred to the piston in the form
of macroscopic kinetic energy (work). Thermal energy changes between the gas and the
piston are neglected.
The loss of kinetic energy by the gas molecules when hitting the piston is reflected in a
decrease in the thermal energy of the gas.
The pressure of the gas is the result of molecular collisions between gas molecules and
container walls, and it depends on the thermal energy and density of the gas.
As the piston moves, the volume available for the gas system increases (expansion). This
implies a net macroscopic displacement of the gas, and a decrease in gas density.
The decrease in gas temperature and gas density results in a decrease in the gas
pressure, eventually equating the external air pressure and reaching mechanical
equilibrium. At this point, the displacement of the piston stops.
Friction forces between the piston and the cylinder are not negligible. In the absence of
friction forces, mechanical equilibrium is never achieved as the piston would perpetually
oscillate. As the magnitude of the friction force increases, the motion of the piston is
dampened, and at some point, oscillations disappear (overdamped conditions). This is
the situation assumed in the present example (no oscillation of the piston).
The pressure of external air remains constant during the whole process.
All interaction forces are passive, local, short-range forces. There are no field forces, and
there are no active forces present. The interaction forces are considered in general, non-
invariant.
There are no direct interactions between the gas and external air, between the gas and
the ground, and between the piston and the ground. The horizontal interaction
between air and the ground is also negligible.
There are no gas leaks, so no mass transfer is observed.
As it can be seen, this relatively simple example is not simple at all. There are many systems and
assumptions that must be considered. Taking all of this into account, let us obtain the matrix of
interaction forces in the direction of the piston axis :
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(3.1)
where represents the cross-section area of