This document provides an overview of key concepts in thermodynamics. It begins with contact information for the instructor, Dr. Sabar D. Hutagalung, and lists the main topics to be covered, including the four laws of thermodynamics. It then provides more detailed explanations of these topics, such as definitions of the zeroth, first, and second laws. It also explains concepts like heat, work, internal energy, and processes involving gases like isobaric, isothermal, and adiabatic. In addition, it discusses mechanisms of heat transfer including conduction, convection, and radiation, and defines important related terms.

1. Instructor:
Dr. Sabar D. Hutagalung
Physics Department, Faculty of Science, Jazan University, Jazan, Saudi Arabia
Email: sdhutagalung@gmail.com

3.  Raymond A. Serway & John W. Jewett, Jr., Physics
for Scientists and Engineers with Modern Physics,
9th Edition, Brooks/Cole, 2014.

4.  Zeroth (0th) law of thermodynamics
 Thermometers and Celcius temperature
scale
 Work and heat in thermodynamic processes
 The first law of thermodynamics
 Energy transfer mechanism in thermal
processes
 Heat engines and the second law of
thermodynamics

5.  The 0th law:
◦ If two systems are each in thermal equilibrium with a
third, they are also in thermal equilibrium with each
other.
 The 1st law:
◦ The change in internal energy of a system (ΔU) is due to
heat gain or loss (Q) and work done (W).
 The 2nd law:
◦ Heat flows spontaneously from a substance at a higher
temperature to a substance at a lower temperature and
does not flow spontaneously in the other direction.
 The 3rd law:
◦ It is not possible to lower the temperature of any system
to absolute zero in a finite number of steps.

6.  “If objects A and B are separately in thermal
equilibrium with a third object C, then A and
B are in thermal equilibrium with each other.”
Heat will always flow in a
direction from hot objects to
colder ones, but never the other
way around.
Thermal equilibrium
(same temperature)

7.  Thermometers are devices
used to measure the
temperature of a system.
 All thermometers are based
on the principle that some
physical property of a
system changes as the
system’s temperature
changes (volume,
dimensions, pressure,
electric resistance, color).
A common thermometer
consists of a mass of liquid –
usually mercury or alcohol -
that expands into a glass
capillary tube when heated.

8.  Any temperature change in the range of the thermometer can
be defined as being proportional to the change in length of
the liquid column.
 The thermometer calibrated with a natural system that
remains at constant temperature.
 A mixture of water and ice in thermal equilibrium at
atmospheric pressure is defined to have a temperature of
zero degrees Celsius, 0oC; called the ice point of water.
 A mixture of water and steam; is defined as 100oC, which is
the steam point of water.
 The length of the liquid column between the two
points is divided into 100 equal segments to
create the Celsius scale.

9.  Mercury and alcohol have different thermal
expansion properties: they may indicate a slightly
different value for a temperature.
 Mercury and alcohol have limited range of
temperatures over which it can be used.
 A mercury thermometer cannot be used below -39oC
(the freezing point of mercury).
 An alcohol thermometer is not useful for above 85oC
(the boiling point of alcohol).
 Therefore, need a universal thermometer
whose readings are independent of the
substance used in it (the gas thermometer).

10.  The height h, the difference between the mercury
levels in B and A, indicates the pressure in the flask
at 0oC by equation:
 The flask is then immersed in water at the steam
point for the gas pressure at 100oC.
 These two pressure and temperature values are
then plotted.
 One version of a gas thermometer is the constant-volume apparatus.
 The physical change exploited is the variation of pressure of gas with
temperature.
 The flask is immersed in an ice-water bath, and mercury reservoir B is
raised or lowered until the top of the mercury in column A is at the zero
point on the scale.

11. Conversion formulas:
Example 19.1.
On a day when the temperature reaches
50oF, what is the temperature in
degrees Celsius and in Kelvins?
Answer:

12.  Internal energy is all the energy of a system
that is associated with its microscopic
components—atoms and molecules—when
viewed from a reference frame at rest with
respect to the center of mass of the system.
 Usually the internal energy consists of the sum of
the potential and kinetic energies of the working
gas molecules.

13.  Heat is defined as a process of
transferring energy across the
boundary of a system because of a
temperature difference between the
system and its surroundings.
 It is also the amount of energy Q
transferred by this process.

14.  State variables in thermodynamics: pressure
(P), volume (V), temperature (T), and internal
energy (Eint).
 A second category is transfer variables (heat
(Q) and work (W) as the transfer variables).
 The heat is positive or negative, depending
on whether energy is entering or leaving the
system.
 The work is positive if system compressed
and negative if system expands.

15.  As the piston is pushed downward by an external
force through a displacement of
the work done on the gas is:
 Because Ady is the change in volume of the gas dV,
the work done on the gas:
If gas is compressed, dV is negative and dW work
done on the gas is positive. If gas expands, dV
positive and dW work done on the gas negative.

16.  The total work done on the gas as its
volume changes from Vi to Vf is
The curve on a PV diagram is
called the path taken between the
initial and final states.

17. The volume is first
reduced from Vi to Vf at
constant pressure Pi and the
pressure of the gas then
increases from Pi to Pf by
heating at constant volume Vf .
The pressure is increased from
Pi to Pf at constant volume Vi
and then the volume of the gas
is reduced from Vi to Vf at
constant pressure Pf .
Both P and V change
continuously, the work done
on the gas has some value
between the values obtained
in the first two processes.

18.  The gas in the cylinder is at a pressure equal to
1.01 x 105 Pa and the piston has an area of 0.1
m2. As energy is slowly added to the gas by
heat, the piston is pushed up a distance of 4 cm.
Calculate the work done by the expanding gas
on the surroundings, Wenv, assuming the
pressure remains constant.

19.  Find the change in volume of the gas, dV,
 The work done:
W = -P (Vf - Vi) = -P dV = P dV
(gas expands: dV is positive)

20.  The 1st Law of Thermodynamics
(Conservation) states that “energy is always
conserved, it cannot be created or
destroyed.”
 Energy can be changed from one form to
another, but it cannot be created or
destroyed.
 The 1st law:
◦ The change in internal energy of a system (ΔU) is
due to heat gain or loss (Q) and work done (W).

21.  The sum of the heat transferred into the
system and the work done on the system
equals the change in the internal energy of
the system.
DEint - change internal energy;
Q - energy or heat transfers; and
W - work.

22. Isolated system:
 No energy transfer by heat and the work done
on the system is zero; hence, the internal
energy remains constant.
Isolated system

23. Cyclic process:
 A process that starts and ends at the same
state.
 The change in the internal energy is zero
therefore, the energy Q added to the system
must equal the negative of the work W done.

24.  Adiabatic process is a process which no
energy enters or leaves the system by heat;
that is, Q = 0.
◦ An adiabatic process can be achieved either by
thermally insulating the walls of the system or by
performing the process rapidly so that there is
negligible time for energy to transfer by heat.

25.  Free expansions. A process in which no heat
transfer occurs between the system and its
environment and no work is done on or by the
system. Thus, Q =W = 0.
 Isobaric process is a process that occurs at
constant pressure.
◦ The values of the heat and the work are both
usually nonzero.

26.  Isovolumetric process is a process that takes
place at constant volume.
◦ Because the volume does not change in such a process, the
work done is zero (W = 0).
 Isothermal process is a process that occurs at
constant temperature.
◦ Because the temperature does not change in a
process involving an ideal gas, DEint = 0.
Q = -W

27. The first law of Thermodynamics:
Special cases
The General Law: DEint = Q + W
Process Restriction Consequence
Isolated system Q = W = 0 DEint = 0
Cyclic DEint = 0 Q = -W
Adiabatic Q = 0 DEint = W
Free expansions Q = W = 0 DEint = 0
Isobaric W = -P(Vf - Vi) DEint = Q + W
Isovolumetric W = 0 DEint = Q
Isothermal DEint = 0 Q = -W

29. 0 + +
+ 0 +
0 0 0

30.  Suppose an ideal gas is allowed to expand quasi-statically at
constant temperature;
PV = nRT = constant
 Work:
 Because T is constant:
Figure 20.9 PV diagram
for an isothermal
expansion of an ideal.
W = - nRT ln (Vf-Vi)
= nRT ln (Vi-Vf)

31.  Characterize the paths in Figure 20.10 as isobaric,
isovolumetric, isothermal, or adiabatic. For path B,
Q = 0. The blue curves are isotherms.
Figure 20.10

32.  Characterize the paths in Figure 20.10 as isobaric,
isovolumetric, isothermal, or adiabatic. For path B,
Q = 0. The blue curves are isotherms.
Answer:
A: isovolumetric
B: Adiabatic
C: Isothermal
D: Isobaric
Figure 20.10

33.  A 1.0-mol sample of an ideal gas is kept at
0.0°C during an expansion from 3.0 L to 10.0
L.
◦ (A) How much work is done on the gas during the
expansion?
◦ (B) How much energy transfer by heat occurs
between the gas and its surroundings in this
process?
◦ (C) If the gas is returned to the original volume by
means of an isobaric process, how much work is
done on the gas?

34. (A) How much work is done on the gas during the
expansion? This is isothermal process:
T = 0 oC = 273 K (constant)
n = 1; R = 8.31 J/mol
Vi = 3 L = 3x10-3 m3
Vf = 10 L = 10x10-3 m3
Work negative (gas expands).

35. (B) How much energy transfer by heat occurs between
the gas and its surroundings in this process?
In isothermal process, DEint = 0

36. (C) If the gas is returned to the original volume by
means of an isobaric process, how much work is
done on the gas?
Returned to original volume by isobaric process:
Vi = 10 L = 10x10-3 m3
Vf = 3 L = 3x10-3 m3
Work positive (gas compressed).
= 1.6 x 103 J

37. Hotter Colder

38.  Conduction is the process energy transfer by heat
(Q) through the material of the body.
 Conduction occurs only if there is a
difference in temperature between two
parts of the conducting medium.
 The rate of energy transfer by heat is:
where P is power (watt, W), Q is heat (Joule, J),
k is the thermal conductivity (W/m °C), and
|dT/dx| is the temperature gradient.
Figure 20.11

39.  Uniform rod of length L is thermally insulated
so that energy cannot escape from its surface
except at the ends.
 The temperature gradient:
The rate of energy transfer by conduction
through the rod is:

40.  A compound slab containing several materials of
thicknesses L1, L2, . . . and thermal conductivities
k1, k2, . . .
 The rate of energy transfer through
the slab at steady state is
 where Th and Tc are the temperatures
of the outer surfaces.

41.  You have two rods of the same length and
diameter, but they are formed from different
materials. The rods are used to connect two
regions at different temperatures so that energy
transfers through the rods by heat. They can be
connected in series as in Figure 20.13a or in
parallel as in Figure 20.13b.
 In which case is the rate of energy transfer by heat
larger?
(a) The rate is larger when the rods are in series.
(b) The rate is larger when the rods are in parallel.
(c) The rate is the same in both cases.
Answer: (b)
Because larger contact area, A
when connected in parallel.

42.  Two slabs of thickness L1 and L2 and thermal conductivities k1 and k2
are in thermal contact with each other as shown in Figure 20.14. The
temperatures of their outer surfaces are Tc and Th, respectively, and
Th > Tc. Determine the temperature at the interface and the rate of
energy transfer by conduction through an area A of the slabs in the
steady-state condition.
Figure 20.14

43.  Express the rate at which energy is transferred through
an area A of slab 1:
 Express the rate at which energy is transferred through the
same area A of slab 2:
 Set these two rates equal to represent the steady-state situation:
 Solve for T:
 Substitute Equation (3) into either Equation (1) or Equation (2):
Figure 20.14

44.  Convection is energy transfer by
the movement of molecules
within substances/fluids such as gases and
liquids.
 Natural convection: Convection resulting from
differences in density.
◦ Example: Airflow at a beach.
 Forced convection: The heated substance is
forced to move by a fan or pump.
◦ Example: Hot-water heating systems

46.  Radiation is energy transfer by
electromagnetic waves.
 The rate of energy transfer by radiation is
proportional to the fourth power of the absolute
temperature of the surface. This known as Stefan’s
law, is expressed as:
where P is the power, s is a constant = 5.6696 x 10-8 W/m2 K4,
A is the surface area, e is the emissivity, and T is the surface
temperature in Kelvins. The value of e can vary between zero
and unity.

47.  The emissivity, e is equivalent to the absorptivity,
which is the fraction of the incoming radiation that
the surface absorbs.
 A mirror has very low absorptivity & emissivity
because it reflects almost all incident light.
 A black surface has high absorptivity & emissivity.
 An ideal absorber is defined as an object that absorbs
all the energy incident on it, e =1, is often referred as
a black-body.
 So, the emissivity, e is defined as the ratio of the
energy radiated from a material's surface to that
radiated from a black-body at the same temperature
and wavelength.
 Every second, ~1370 J of electromagnetic radiation
from the Sun passes perpendicularly through each 1
m2 at the top of the Earth’s atmosphere.

48.  If an object is at a temperature T and its
surroundings are at an average temperature
T0, the net rate of energy gained or lost by
the object as a result of radiation is:

49.  The Dewar flask is a container designed to minimize energy
transfers by conduction, convection, and radiation.
 A Dewar flask is used to store cold or hot liquids for long
periods of time.
 Dewar flash is commonly used to store liquid nitrogen
(boiling point 77 K) and liquid oxygen (bp 90 K).
 An insulated bottle or Thermos, is equivalent of a Dewar
flask.
 For liquid helium (bp 4.2 K), necessary to use
a double Dewar system in which the Dewar
flask containing the liquid is surrounded by
a second Dewar flask.

50. Energy is transferred by
direct contact.
Energy is transferred by
the mass motion of
molecules.
Energy is transferred by
electromagnetic radiation.

51.  A poker is a stiff, nonflammable rod used to
push burning logs around in a fireplace. For
safety and comfort of use, should the poker
be made from a material with (a) high thermal
conductivity, (b) low thermal conductivity, (c)
average thermal conductivity?

53.  A heat engine is a device that takes in energy
by heat and, operating in a cyclic process,
expels a fraction of that energy by means of
work.
 A heat engine carries some working
substance through a cyclic process:
(1) the working substance absorbs energy by heat
from a high-temp energy reservoir,
(2) work is done by the engine, and
(3) energy is expelled by heat to a lower-temp
reservoir.

54.  Example of heat engine, a steam engine,
which uses water as the working substance.
1. The water in a boiler absorbs energy from
burning fuel,
2. water evaporates to steam,
3. does work by expanding against a piston.
4. After the steam cools and condenses,
the liquid water produced returns
to the boiler and the cycle repeats.

55.  The engine absorbs a quantity of energy |Qh| from
the hot reservoir. We use absolute values.
 The engine does work Weng (so that W = -Weng) and
then gives up a quantity of energy |Qc| to the cold
reservoir.
 Cyclic process: DEint = Q+W = Q -Weng = 0
Weng = Qnet ; Qnet = |Qh| - |Qc|
 Therefore,
 The thermal efficiency, e of a heat engine:

56.  Absorbs heat, Qh
 Performs work, Weng
 Rejects heat, Qc
A heat engine is any device
which through a cyclic process.
Cold Res. Tc
Engine
Hot Res. Th
Qh
WengQc

57.  Kelvin–Planck form of the second law
of thermodynamics states the
following:
“It is impossible to construct a heat
engine that, operating in a cycle,
produces no effect other than the
input of energy by heat from a
reservoir and the performance of an
equal amount of work.”

58. Cold Res. Tc
Engine
Hot Res. Th
400 J
300 J
100 J
• A possible engine. • An IMPOSSIBLE
engine.
Cold Res. Tc
Engine
Hot Res. Th
400 J
400 J

59.  The energy input to an engine is 4.00 times greater
than the work it performs.
 (i) What is its thermal efficiency? (a) 4.00, (b) 1.00,
(c) 0.250, (d) impossible to determine.
 (ii) What fraction of the energy input is expelled to
the cold reservoir? (a) 0.250, (b) 0.750, (c) 1.00, (d)
impossible to determine.

60.  The energy input to an engine is 4.00 times greater
than the work it performs.
 (i) What is its thermal efficiency? (a) 4.00, (b) 1.00,
(c) 0.250, (d) impossible to determine.
 (ii) What fraction of the energy input is expelled to
the cold reservoir? (a) 0.250, (b) 0.750, (c) 1.00, (d)
impossible to determine.
Qh = 4 Weng
e = Weng/Qh = Weng/4Weng
= 0.25
e = 0.25
Qc /Qh = 1 – 0.25 = 0.75

61.  An engine transfers 2.00 x 103 J of energy from a
hot reservoir during a cycle and transfers 1.50 x
103 J as exhaust to a cold reservoir.
(A) Find the efficiency of the engine.
(B) How much work does this engine do in one
cycle?
What If ? Suppose you were asked for the power
output of this engine. Do you have sufficient
information to answer this question?

62.  (A)
 (B)
What If ?
Answer:
 No, we do not have enough information. The power of an
engine is the rate at which work is done by the engine. We
know how much work is done per cycle, but you have no
information about the time interval associated with one cycle.