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Heat Transfer
Thermal energy of a system can change in many ways. The engine of an idling car becomes warm because much of the energy released
when gasoline burns in the engine is converted into thermal energy. If you rub your hands together vigorously, they become warm because work
done by the muscles in your arms is converted into thermal energy caused by friction as your hands rub together.
Thermal energy can also be changed by a process called heat transfer. Heat transfer occurs when two objects at different temperatures
are brought into contact with each other. Thermal energy in the hotter object is transferred to the cooler one.
The terms heat and thermal energy are often used interchangeably. They are however, different quantities. The thermal energy in a system
is the amount of random kinetic energy of the atoms and molecules of objects in the system. Heat, on the other hand, is energy transferred between
a system and its environment because of a difference in their temperatures. If the system gains energy, Q (heat) is positive; if the system loses
energy, Q negative.
Typically, when scientists discuss energy transformations that involve heat, the energy-conservation idea is expressed in the following form:
W + Q = ΔEsystem
= ΔKE + ΔPEg + ΔEth + ΔEchem + ...
Work W and heat Q represent the energy added to or removed from a system during a certain time period or during some process of
interest. This energy added to or removed the system equals the net charge in all the different forms of energy within the system. (Ref.#1)
 Heat Transfer. The transfer of energy from one body or system to another as a result of a difference in temperature.
Effects of Heat Transfer
1.
Change in the temperature of a substance
2.
Change in the state of a substance
3.
Change in the volume
* Atoms and molecules spread farther apart when the substance is hot than when cold.
Methods (modes or mechanisms) of Heat Transfer
1.
Conduction. The transmission of heat through a substance from a region of high temperature to a region of lower temperature. In gases
and most liquids, the energy is transmitted by collisions between atoms and molecules with those possessing lower kinetic energy. In solid
and liquid metals, heat conduction is predominantly by migration of fast-moving electrons, followed by collisions between these electrons
and ions. In solid insulators, the absence of free electrons restricts heat transfer due to the vibrations of atoms and molecules within crystal
lattices.
Heat transfer by conduction is likewise defined as the transmission of heat by molecular collisions. Or, the transfer of heat
through a medium without any obvious motion of the medium.
Heat moves from the hot end of the rod to the cold end of the rod by conduction.
Conductor. A substance that has a high thermal conductivity. Metals are good conductors on account of the high concentration of free
electrons they contain (or, because the valence electrons in the metal move about freely and in so doing, carry kinetic energy
along with them. Good electrical conductors are good conductors of heat.
Valence Electron. An electron in one of the outer shells of an atom that takes part in forming chemical bonds.
Thermal Conductivity. Ability or a measure of the ability of a substance to conduct heat. It is usually denoted by the symbol lambda () a proportional constant. However, in some books, K is used.
Non-metals (such as glass, mica, porcelain, and wood) are poor conductors of heat.
The heat-transfer rate (or Heat Current) H is the amount of heat ΔQ that passes from one region of an object to another region
divided by the time Δt needed for that heat transfer:
𝐻𝑒𝑎𝑡 − 𝑇𝑟𝑎𝑛𝑠𝑓𝑒𝑟 𝑅𝑎𝑡𝑒 = 𝐻 =
∆𝑄
∆𝑡
H is expressed in terms of watts, which is the ratio of a unit heat, the joule, and a unit of time, the second: That is, 1 W = 1 J/s.
The heat-conduction rate across an object, such as shown in the preceding figure, can be computed using the equation,
∆𝑄
( )
∆𝑡 𝑐𝑜𝑛𝑑𝑢𝑐𝑡𝑖𝑜𝑛
= 𝐻𝑐𝑑 =
𝐾𝐴(𝑇2 −𝑇1 )
𝐿
where Hcd is the conductive heat-transfer rate from region 1 of an object at temperature T1 to another part of the object, region 2, at
temperature T2; L is the distance the heat must travel from region 1 to region 2; A is the cross-sectional area across which the heat travels;
and K is the thermal conductivity, which indicates the ease or difficulty of heat transfer by conduction through different types of materials
(refer to the table below). Metals, which are good conductors of heat, have large values of K; poor heat conductors such as air, wood,
and Styrofoam have small values of K. The latter materials are often called insulators and are used to prevent the escape of heat from
ice chests, thermos bottles, and houses. In the SI metric system of units, K is in terms of W/mC; A is in m2; T1 and T2 ar in C; and
consequently, Hcd is in W.
* considerations:
∆𝑄
∆𝑡
∆𝑄
∆𝑡
is positive when heat is entering region 1 - that is, T1 is less than T2.
is negative when heat is leaving region 1 - that is, T1 is greater than T2.
Thermal Conductivity K of Different Materials
------------------------------------------------------------------------------------Material
K (W/mC)
------------------------------------------------------------------------------------Metals
Aluminum
205.0
Brass
109.0
Copper
385.0
Gold
314.0
Iron
79.5
Lead
34.7
Silver
406.0
Steel
50.2
Non-Metals
Asbestos
0.25
Brick (insulating)
0.15
Dry Soil
0.20
Dry Wall
0.16
Red Brick
0.60
Concrete
0.80
Cork
0.04
Felt
0.04
Fiber Glass
0.04
Glass
0.80
Ice
1.60
Polyurethane
0.024
Rock Wool
0.04
Rubber
0.20
Sheathing
0.55
Styrofoam Insulation
0.01
Water
0.60
Wood
0.02-0.04
Fat
0.021
Muscle
0.042
Bone
0.042
Gases (at 20)
Air
0.024
Argon
0.016
Helium
0.140
Hydrogen
0.140
Nitrogen
0.0234
Oxygen
0.0238
------------------------------------------------------------------------------ Temperature Gradient. The temperature difference per unit length,
(𝑇𝐻 − 𝑇𝐶 )
𝐿
Example: One end of a 1.25-m cylindrical brass rod with a 2.4 cm diameter is in boiling water at 100C. The other end is on top of an
ice cube at exactly 0C. (a) Compute the conductive heat-transfer rate to the ice. (b) How much time is required to
melt 17 grams of ice?
Solution
The available information are:
Kbrass = 109 W/mC (refer to the table)
L = 1.25 m
𝑑𝑐𝑖𝑟𝑐𝑙𝑒 = 2.4 𝑐𝑚 (
1𝑚
100 𝑐𝑚
) = 0.024 𝑚
𝑟𝑐𝑖𝑟𝑐𝑙𝑒 = ½ 𝑑𝑐𝑖𝑟𝑐𝑙𝑒 = ½(0.024 𝑚) = 0.012 𝑚
𝐴𝑐𝑖𝑟𝑐𝑙𝑒 = 𝜋𝑟 2 = 𝜋(0.012 𝑚)2 = 4.5239 𝑥 10−4 𝑚2
(a)
The ice (cold) side of the rod is chosen to be the region 1 (implicated by the requirement (a) as the
reference). It follows then that the other (hot) side of the rod is the region 2. Thus,
𝑇1 = 0𝐶 𝑎𝑛𝑑 𝑇2 = 100𝐶
It is expected that Hcd is positive because heat is entering region 1 (the cold region)
Substituting the given values:
𝐻𝑐𝑑 =
𝐾𝐴(𝑇2 −𝑇1 )
𝐿
=
(109
𝑊
𝑚𝐶°
)(4.5239 𝑥 10−4 𝑚2 )(100−0)𝐶°
1.25 𝑚
= 3.945 𝑊
* cancelled units are C˚ and m2.
(b)
We know that the heat-transfer rate is, Hcd = _Q/_t = 3.945 W and wish to determine the time needed
to transfer enough heat to melt 17 grams of ice. To solve for _t, we have to manipulate the equation into
the following form:
∆𝑄
∆𝑡 =
𝐻𝑐𝑑
To determine the heat needed _Q to melt 17 g of ice, we have to recall our lesson on changes of state. To
melt ice, Q = m L f. To melt 17 grams (or 0.017 kg) of ice,
𝑄 = ∆𝑄 = 𝑚𝐿𝑓 = 0.017 𝑘𝑔 (335,000
𝐽
) = 5695 𝐽
𝑘𝑔
* Lf of water is equal to 335,000 J/kg
The time required to melt 17 grams of ice is,
∆𝑡 =
5695 𝐽
3.945 𝑊
= 1443.6
𝐽
(
1𝑊
𝑊 1 𝐽/𝑠
) = 1443.6 𝑠 (
1 𝑚𝑖𝑛
60 𝑠
) = 24 𝑚𝑖𝑛
Example: (a) Calculate the heat-transfer rate from the inside of a 47-m² mobile house, that has a 0.035 m thick walls with a
thermal conductivity of 0.15 W/mC, to the outside, if the inside temperature is 27C and the outside
temperature is 1C. (b) Compare the conductive heat-tranfer rate of the said mobile house to that of the
roof of a 47-m² room in an earth house covered with a half meter of dry soil.
Solution:
(a) In this problem the heat travels a short distance L = 0.035 m across a large area A = 47 m². The small value of
L and large value of A contribute to a high heat transfer rate by conduction. The implied region 1 is the inside
house (notice the phrase "from the inside of the house to the outside"). Thus T1 = 27C and T2 = 1C. Substituting
the available values in the equation,
𝐻𝑐𝑑 =
𝐾𝐴(𝑇2 −𝑇1 )
𝐿
=
(0.15
𝑊
𝑚𝐶°
)(47 𝑚2 )(1−27)𝐶°
0.035 𝑚
= −5237.142857143 𝑊
The negative sign indicates that the heat flows away from region 1, the region inside the home. The home is losing
energy to the outside by heat conduction at that rate.
(b) Similarly, Hcd for the roof covered with 0.5m thick of dry soil could be solved with K of dry soil based on the table
equals to 0.2 W/mC and L equals to 0.50m.
𝐻𝑐𝑑 =
𝐾𝐴(𝑇2 −𝑇1 )
𝐿
=
(0.2
𝑊
𝑚𝐶°
)(47 𝑚2 )(1−27)𝐶°
0.5 𝑚
= −488.8 𝑊
The extra thickness of the soil provides a barrier that greatly reduces the rate of heat loss from the house.
2.
Convection. The transfer of heat energy from one place to another by mass that moves between places and carries the energy with it.
Or, simply the heat transferred by the movement of heated substance.
 Natural Convection. When the movement results from the difference in density, as it does in air around a fire.
 Forced Convection. When the movement is caused by the force from a fan or pump, as in some hot-air and hot-water heating systems.
The calculation of the rate of convective heat transfer is a complex subject and depends on many factors. We can approximate the
convective heat-transfer rate using the following equation:
∆𝑄
( )
∆𝑡 𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛
= 𝐻𝑐𝑣 = ℎ𝐴(𝑇2 − 𝑇1 ) = ℎ𝐴(𝑇𝑓𝑙𝑢𝑖𝑑 − 𝑇𝑠𝑢𝑟𝑓𝑎𝑐𝑒 )
where h is the convection coefficient and accounts for many of the other variables on which the convective heat-transfer rate depends, such
the type of fluid that crosses the surface, the speed of the fluid, and the shape and texture of the surface across which the fluid
moves. It must be determined for each unique situation. Tables and empirical equations are often used to estimate its value.
It is important when using the above equation to consider the sign of the convective heat-transfer rate. If the fluid is warmer than
the surface across which it moves, the T2 - T1 is greater than zero, and Hcv is positive. Heat is transferred from the warm, moving fluid to the
cooler surface. Otherwise, Hcv is negative.
Wind-chill Temperature. The temperature of still, cold air that would cause the same heat-loss rate from our bodies as occurs when cool
air is moving at a particular speed.
-------------------------------------------------------------------------------------------------------------------------Wind Speed, m/s (mph)
Air Temp. 2 (4.5)
5 (11.2)
10 (22.4)
15 (33.6)
20 (44.7)
-------------------------------------------------------------------------------------------------------------------------2
1.0
-6.6
-12
-16
-18
0
-1.3
-8.4
-15
-18
-20
-5
-7.0
-15
-22
-26
-29
-10
-12
-21
-29
-34
-36
-20
-23
-34
-44
-50
-52
-------------------------------------------------------------------------------------------------------------------------3.
Radiation. A heat transfer method involving electromagnetic radiation, produced by vibrating and accelerating electric charges. When
electrons vibrate back and forth in the antenna of a radio station, radio waves, a form of electromagnetic radiation, leave the
antenna much as water waves move away from the beach ball pushed up and down in a pond. The radio waves travel at a
speed of 3.0 x 108 m/s (186,000 mi/s) in a vacuum and in air, and they travel slower in other media. A small fraction of the
waves leaving the antenna can be absorbed by the antenna of a car radio as electrons in the antenna are forced to moved up
and down on the passing wave. An energy transfer has thus occurred from the antenna of the radio station to that of the car.
This particular process is not called a heat transfer, but since energy moves from one place to another with the waves, it is an
important energy-transfer mechanism.
Another process, commonly termed heat transfer by radiation, is analogous to energy transfer by radio waves.
Consider a log burning in a fire. Because of their high temperature, many of the molecules in and near the fire vibrate
somewhat more violently than usual. These large-amplitude molecular vibrations produce a form of radiation called infrared
waves. The waves travel from the fire in all directions at the same speed as radio waves. Infrared waves falling on your skin
are absorbed and cause the molecules in your skin to vibrate with greater amplitude. Your skin feels warmer. This process is
called heat transfer by radiation.
Forms of Electromagnetic Radiation
1. Light. The most familiar - the visible radiation detected by our eyes.
2. Radio Waves
3. Micro Waves
4. Infrared and Ultraviolet Radiation
5. X-Rays
6. Gamma Rays
They all travel at the same speed and produced by electric charges that vibrate or experience some other type of accelerated
motion, and are absorbed by matter that lies in their path. The sun emits all of their forms, although most of the radiative energy leaving
the sun consists of light and infrared radiation. The earth is bathed approximately 1.7 x 1017 J of this radiative energy each second - more
than the total energy consumed by the U.S. each year.
The rate at which radiation is emitted from an object depends on its temperature T1, its surface area A, and the type of surface
it has. The surface is characterized by a quantity called emissivity, e. It is a unitless number ranging from 0 to 1. A good emitter, such as
dark or black surface, has an emissivity close to 1. For a poor emitter, such as a white or shiny surface (for example, a mirror), e is close to
zero. For the earth, the average value of e is 0.95. For the human body it is 0.98.
 The rate of emission from an object is,
∆𝑄
( )
∆𝑡 𝑒𝑚𝑖𝑠𝑠𝑖𝑜𝑛
where
= 𝑅𝑒 = 𝑒𝜎𝐴𝑇14
σ = Stefan-Boltzmann constant
= 5.6705119 x 10-8 W m-2 K-4
e = emissivity
T 1 = temperature of the surface in K
A = surface area
 The rate of absorption by the object is,
∆𝑄
( )
∆𝑡 𝑎𝑏𝑠𝑜𝑟𝑝𝑡𝑖𝑜𝑛
where
= 𝑅𝑎 = 𝑒𝜎𝐴𝑇24
T 1 = uniform temperature of the environment in K
 The Net Radiative Heat-Transfer Rate or the net energy gained or lost each second by the object as a result of radiation is
𝐻𝑟 = 𝑅𝑎 − 𝑅𝑒
𝐻𝑟 = 𝑒𝜎𝐴𝑇24 – 𝑒𝜎𝐴𝑇14
𝐻𝑟 = 𝑒𝜎𝐴(𝑇24 – 𝑇14 )
An object radiates at a rate given by the equation above. At the same time, the object also absorbs electromagnetic radiation.
If the latter process did not occur, the object would eventually radiate all of its energy and its temperature would reach absolute zero.
When an object is in equilibrium with its surroundings, it radiates and absorbs energy at the same rate, and so its temperature
remains constant. An ideal absorber is defined as an object that absorbs all of the energy incident on it; its emissivity is equal to unity (1.0).
Such as object is often referred to as a black body. An ideal absorber is also an ideal radiator of energy. In contrast, an object with a
zero emissivity absorbs none of the energy incident on it. Such an object reflects all the incident energy and so is a perfect reflector.
4.
Evaporation. The continual escape of energetic molecules from a liquid into the gas above.
Molecules in a liquid such as water continually move about and collide with each other. At any instant there is much variation in
their speed - some move fast and others slow. Slow-moving, less energetic molecules may reach the water surface, but they are usually
prevented from escaping because of their attraction to neighboring molecules at the surface. Nonetheless, energetic, fast-moving
molecules that reach the surface can escape (refer to the ensuing figure).
Evaporation tends to cool the liquid because only the energetic, "hot" molecules escape. Those that remain are less energetic and cooler.
Often, energy from some other source is added to the liquid to compensate for the decrease in thermal energy caused by evaporation.
When you perspire, a layer of moisture covers yours skin, and as moisture evaporates, it becomes cooler. However, thermal energy created by
metabolic processes inside your body is carried to the skin by your blood. The layer of moisture, then, is cooled by evaporation and warmed by
heat removed from inside your body. Sweating is one of the important mechanisms for maintaining body temperature when air temperatures are
high or when we exercise.
The rate of heat loss by evaporation depends on the rate of evaporation - that is, the ratio of the mass Δm of liquid evaporated and the
time Δt required for this evaporation.
𝐸𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑖𝑜𝑛 𝑅𝑎𝑡𝑒 =
Δ𝑚
Δ𝑡
Factors that affect evaporation rate:
1.
2.
Relative humidity of the air. If the humidity is high, the air contains many water molecules, which may hit the liquid surface and join the
liquid. This process, condensation, cancels the effects of evaporation. This is why it is more difficult to remain cool on a hot, humid day
than on a hot, dry day. On a humid day, water condenses on our skin almost as fast as it evaporates, and there is no net cooling
effect.
Whether the air near the body is moving.
* The average speed of air molecule is approximately 500 m/s in a zigzag path, its course changing abruptly at each collision, displacement is only
several cm/min and experiencing more than 1011 collisions with each collision causing its speed and direction of motion to change.
 Thermodynamics 
Thermodynamics is the study of relationship involving heat, mechanical work, and other aspects of energy and energy transfer. Or, the
field of physics that describes and correlates the physical properties of macroscopic systems of matter and energy.
An example of thermodynamic process is the liquefaction of gases. You may have wondered how gases are liquefied. One method is by
first compressing the gas to very high pressure while keeping the temperature constant, then insulating it and allowing it to expand. The gas cools so
much during the expansion that it liquefies.
Thermodynamic System. A system that can interact (and exchange energy) with its surroundings, or environment, in at least two ways, one of which
is heat transfer and the other is mechanical work. A familiar example is a quantity of popcorn kernels in a pot with lid. When the pot is
placed on a stove, energy is added to the popcorn by conduction of heat; as the popcorn pops and expands, it does work as it exerts an
upward force on the lid and moves it through a displacement. The state of popcorn changes in this process, since the volume, temperature,
and pressure of the popcorn all changes as it pops. The said process, in which there are changes in the state of thermodynamics system, is
called thermodynamics process.
Signs for Heat and Work in Thermodynamics
Energy relations in any thermodynamic process are described in terms of the quantity of heat Q added to the system and the work W
done by the system. Both Q and W may be positive, negative, or zero. A positive Q represents heat flow into the system, with a corresponding input
of energy to it; negative Q represents heat flow out of the system. A positive W represents work done by the system against its surrounding, such as
work done by an expanding gas, and hence corresponds to energy leaving the system. Negative W, such as work done during compression of a gas
in which work is done on the gas by its surrounding, represents energy entering the system.
Work Done During Volume Changes
A gas in a cylinder with a movable piston is a simple example of a thermodynamic system. Internal-combustion engines, steam engines,
and compressors in refrigerators and air conditioners all use some version of such a system.
When a gas expands, it pushes outward on it boundary surface as they move outward. Hence an expanding gas always does a positive
work. The same thing is true of any solid or fluid material that expands under pressure.
When one gas molecule collides with a stationary wall, it exerts a momentary force on the wall but does no work because the wall does
not move. But if the surface is moving, such as a piston in a gasoline engine, the molecule does work on the surface during the collision. If the piston
moves to the right (refer to the figure), so the volume of the gas increases, the molecules that strike the piston exert a force thru a distance and do
positive work on the piston. If the piston moves toward left (refer to the next figure), so the volume of the gas decreases, then positive work is done
on the molecule during the collision. Hence the gas molecules do negative work on the piston.
𝑑𝑊 = 𝐹 𝑑𝑥 = 𝑝𝐴 𝑑𝑥
But 𝐴 𝑑𝑥 = 𝑑𝑉,
where 𝑑𝑉 is the infinitesimal change of volume of the system. Thus we can express the work done by the system in this infinitesimal volume
changes as 𝑑𝑊 = 𝑝 𝑑𝑉
In a finite change of volume from 𝑉1 to 𝑉2 , 𝑊 = 𝛥𝑝 𝑑𝑉
First Law of Thermodynamics. A law that broadens the principle of conservation of energy to include internal energy, heat, and work. It states
that when heat Q is added to a system while it does work W, the internal energy U changes by an amount
𝑈2 − 𝑈1 = 𝛥𝑈 = 𝑄 − 𝑊 𝑜𝑟 𝑄 = 𝛥𝑈 + 𝑊
In an infinitesimal process,
𝑑𝑈 = 𝑑𝑄 − 𝑑𝑊
When we add a quantity of heat Q to a system and it does no work during the process, the internal energy increase by an amount equal
to Q; that is, ΔU = Q. When a system does work W by expanding against its surroundings and no heat is added during the process, energy leaves
the system and the internal energy decreases. That is, when W is positive, ΔU is negative, and conversely. So ΔU = -W. When both heat transfer
and work occur, the total change in internal energy is calculated with the formula of the First Law of Thermodynamics. (refer to the preceding
formula).
Internal Energy. The total of the kinetic energies of the atoms and molecules of which a system consists and the potential energies associated with
their mutual interactions. It does not include the kinetic and potential energies of the system as a whole nor their nuclear energies or other
intra-atomic energies. The value of the absolute internal energy of a system in any particular state cannot be measured; the significant
quantity is the change in internal energy, ΔU. For a closed system (one that is not being replenished from outside its boundaries) the
change in internal energy is equal to the heat absorbed by the system (Q) from its surroundings, less the work done (W) by the system on
its surroundings, i.e. 𝜟𝑼 = 𝑸 − 𝑾.
The internal energy of any thermodynamic system depends only on its state. The change in internal energy in any process
depends only on the initial and final states, not on the path. The internal energy of an isolated system is constant. While Q and W depend
on the path, 𝛥𝑈 = 𝑄 − 𝑊 is independent of path. The change in internal energy of a system during any thermodynamic process
depends only on the initial and final states, not on the path leading from on to the other.
Two Special Cases of the First Law of Thermodynamics Worth Mentioning
1. Cyclic Process. A process that eventually returns a system to its initial state. That is, the initial state is the same as final state, and so the
total internal energy change must be zero. 𝑼𝟐 – 𝑼𝟏 and 𝑸 = 𝑾
2. Another special case occurs in an isolated system, one that does no work on its surroundings and has no heat flow to or from its
surroundings. For any process taking place in an isolate system, 𝑾 = 𝑸 = 𝟎 and therefore 𝑼𝟐 – 𝑼𝟏 = 𝜟𝑼 = 𝟎
In a thermodynamic process, the internal energy of a system may increase, decrease, or stay the same.
a) If more heat is added to the system than the system does work, 𝜟𝑼 is positive and the internal energy increases.
b) If more heat flows out of the system, 𝜟𝑼 is negative and the internal energy decreases.
c)
If the heat added to the system equals the work done by the system, 𝜟𝑼 = 0 and the internal energy is unchanged.
Kinds of Thermodynamics Processes
 Adiabatic Process. No heat transfer into or out of a system; Q = 0. Heat flow can be prevented by either surrounding the system with
thermally insulating material or by carrying out the process so quickly that there is no enough time for appreciable heat flow.
𝑼𝟐 – 𝑼𝟏 = 𝜟𝑼 = − 𝑾
When a system expands adiabatically, W is positive (the system work on its surroundings), so ΔU is negative and the internal energy
decreases. When a system compressed adiabatically, W is negative (work is done on the system by its surroundings) and U increases. In
many (but not all) systems an increase of internal energy is accompanied by a rise in temperature.

Isochoric Process. A constant-volume process. When the volume of a thermodynamic system is constant, it does no work on its surroundings.
Then 𝑊 = 0, and 𝑼𝟐 – 𝑼𝟏 = 𝜟𝑼 = 𝑸
In an isochoric process, all the energy added as heat remains in the system as an increase in internal energy. Heating a gas in a closed
constant-volume container is an example of an isochoric process. (Note that there types of work that do not involve a volume change. For
example, we can do work on a fluid by stirring it. In some literature, "isochoric" is used to mean that no work of any is done.)

Isobaric Process. A constant-pressure process. In general, none of the three quantities ΔU, Q, and W is zero in an isobaric process, but
calculating W is easy nonetheless.
𝑊 = 𝑝(𝑽𝟐 – 𝑽𝟏 )

Isothermal Process. A constant-temperature process. For a process to be isothermal, any heat flow into or out of the system must occur
slowly enough that thermal equilibrium is maintained. In general, none of the quantities ΔU, Q, or W is zero in an isothermal process.
In some special cases the internal energy of a system depends only on its temperature, not on its pressure or volume. The most familiar
system having this special property is an ideal gas. For such systems, if the temperature is constant, the internal energy is also constant;
𝛥𝑈 = 0 and 𝑄 = 𝑊. That is, any energy entering the system as heat Q must leave it again as work W done by the system.
The 𝒑𝑽-diagram for each of thermodynamic processess
where:
 Adiabat the path followed in an adiabatic process.
 Isochor the vertical path or line (constant volume).
 Isobar the horizontal path or line (constant pressure).
 Isotherm the curve or path of constant temperature.
 The molar heat capacities Cv and Cp of an ideal gas are related by
𝐶𝑃 = 𝐶𝑣 + 𝑅
The ratio of heat capacities, 𝐶𝑃 /𝐶𝑣 is denoted by 𝜸:
𝐶𝑃
𝛾=
𝐶𝑣
 For an adiabatic process for an ideal gas the quantities TVγ-1 and pVγare constant. For an initial state (p1,V1,T1) and a final state (p2,V2,T2),
𝛾−1
𝛾−1
𝑇1 𝑉1
= 𝑇2 𝑉2
𝛾
𝛾
𝑝1 𝑉1 = 𝑝2 𝑉2
The work done by an ideal gas during an adiabatic expansion is
𝐶𝑣
1
(𝑝1 𝑉1 − 𝑝2 𝑉2 ) = −1 (𝑝1 𝑉1 − 𝑝2 𝑉2 )
𝑊 = 𝑛𝐶𝑣 (𝑇1 − 𝑇2 ) =
𝑅
𝛾
Cv
Zeroth Law of Thermodynamics:If two systems are in thermal equilibrium with a third system then they are in thermal equilibrium `with each other.