Tuesday, June 25, 2019

Unizor - Physics4Teens - Energy - Heat Transfer - Radiation




Notes to a video lecture on http://www.unizor.com

Heat Transfer - Radiation

Heat radiation IS NOT the same as radioactivity. Though, under certain circumstances (like an explosion of an atomic bomb) the heat radiation and radioactivity are both present. When we discuss the heat radiation we talk about a process that occurs in any object with a temperature greater than absolute zero, while radioactivity occurs in extreme cases of very high energy output.

Heat transfer through radiation is totally different from conduction and convection. The most important property of heat transfer by radiation is that heat transfer occurs without any visible material conduit that carries the heat, like molecular movement in two other cases.

Let's start from an example.
The brightest example is our Sun, as a source of heat energy. Between Sun and Earth there is no visible material conduit, yet the heat comes to Earth and is a source of life on our planet.

The fundamental concept that lies in the foundation of a process of heat radiation is a concept of a field.
The field is a region of space, where certain forces act on certain objects without visible material medium.
As an example of the field, consider gravity. The Sun keeps planets on their orbits, the Earth keeps the Moon circling around, people are walking on the ground without flying away to stars etc. We did not know much about WHY the gravitational field exist, yet we did study its behavior, the forces involved and the laws of motion in this field.

There are other fields.
Magnetic field around our planet, acting on a compass, forces the arrow to point North.
Electric field exists around electrically charged objects, so other electrically charged objects are attracted to or repulsed from it.

In Physics we successfully study these fields, but complete understanding of WHY they have the properties that we observed is not completely clear. So, we will concentrate on properties, answering the question HOW?, not on a more fundamental question WHY?.

Let's start with a particular field called electro-magnetic. Very simplified description of this field is as follows.

Any electron creates an electric field around itself. Moving electrons, which we call electric current or electric field that changes in time, also create a magnetic field around them. So, changing in time electric field creates magnetic field that changes in space. It's an experimental fact, and we have the whole theory about properties of these fields.

Consider an experiment, when you move a metal rod or any other electrical conductor in a magnetic field or change a magnetic field around any electrical conductor, thus creating a magnetic field that changes in time. You will observe that there is an electric current in the conductor, thus creating an electric field that changes in space. It's an experimental fact, and we also have the whole theory about properties of this process.

So, electric field creates magnetic field, which, in turn, creates electric field etc. This is a loop of energy conversion that propagates extremely fast, with a speed of light, about 3·108m/sec.

The combination of electric and magnetic forces form electro-magnetic field that propagates much faster than its physical medium - electrons. So, the propagation of the electro-magnetic field seems to be a self-sufficient process, occurring without the medium. This is a very brief and unsatisfactory explanation of the nature of the electro-magnetic field. We will not go much further in this explanation, but rather concentrate on the properties of the electro-magnetic field.

Assuming that we accept the existence of the electro-magnetic field and, however uncomfortably we feel about it, but accept that there is no need for a medium to propagate this field, we can talk about frequency of electro-magnetic transformations, that can be considered similar to oscillation of molecules in a solid. Inasmuch as the oscillations of molecules in a metal are propagated, thus transferring heat energy from hot area to cold one, oscillations of the electric and magnetic components of the electro-magnetic field transfers energy.
This energy transfer by electro-magnetic field is called radiation.

As in a case of oscillating molecules in a solid, carrying more heat energy when oscillation is more intense (higher frequency), the electro-magnetic field oscillation carries energy with higher frequencies being more "energetic" than lower.

Interestingly, receptors in our skin feel the temperature of a solid object, that is, we feel the intensity of oscillation of its molecules. Similarly, we feel the warm rays of Sun on our skin, that is, we feel the intensity of electro-magnetic oscillation of the electro-magnetic field.
What's more remarkable, we see the light. Apparently, electro-magnetic oscillations in certain frequency range act upon censors in our eyes, thus we see the light. Moreover, in this visible range of frequencies different frequencies of electro-magnetic oscillation produce effect of different colors in our eyes.

As you see, the light and heat of radiation have the same source - the oscillation of electro-magnetic field, the only difference is the frequency. In other words, the light and heat radiation are manifestations of the same process of transferring energy by the oscillations of the components of the electro-magnetic field.

An object does not have to have a temperature of the Sun to emit heat radiation. All objects that have temperature higher than absolute zero emit thermal radiation of some frequencies. Usually, the whole spectrum of frequencies of electro-magnetic oscillations is emitted by objects. Lower frequencies (usually called infrared) are felt by skin receptors, higher frequencies are visible by an eye. Frequencies higher than those visible by a human eye are called ultraviolet. Even higher frequencies are called X-rays, which can be produced by special equipment and, depending on intensity and time of exposure, can represent a health hazard. Even higher intensity and high frequencies are called gamma rays, and they are produced in extreme cases like nuclear explosion or nuclear reactor meltdown, and they are extremely dangerous and are usually meant, when the term radioactivity is used. All frequencies can be observed using some scientific instruments.

Any object, placed in the outer space will emit its heat energy through radiation until its temperature will reach absolute zero. Our Sun emits huge amounts of energy in all spectrum of frequencies in all directions and, eventually, run out of heat energy and go dark.

The intensity of radiation, that is amount of heat radiated per unit of time per unit of area of an object depends, as in other cases of heat transfer, on the temperature of an object and temperature of surrounding environment.
In the complete vacuum with no other source of energy the radiation intensity of an object is proportional to the fourth degree of its absolute temperature in °K:
q = σ·T4 where
σ = 5.67·10−8 W/(m2·°K4)
is the Stefan-Boltzmann constant.
This is the Law of Stefan-Boltzmann. Its derivation is complex and is outside of the scope of this course.

Radiation is not only emitted by objects with temperatures above absolute zero, but also can be absorbed by them and even reflected. While ability to absorb the heat is common for other heat transfer types (conduction and convection), reflection is a specific property of heat radiation. More precisely, it's a specific property of oscillations of the electro-magnetic field.
Obvious application of this property is the usage of mirrors that reflect the oscillations of the electro-magnetic field in a very broad spectrum of frequencies, including the visible light.
An example of absorbed radiation is a slice of bread toasted in the electric toaster. It absorbs the thermal radiation emitted by electric coils, that changes the bread's structure.

Friday, June 21, 2019

Unizor - Physics4Teens - Energy - Heat Transfer - Convection





Notes to a video lecture on http://www.unizor.com

Heat Transfer - Convection

As in a case of conduction, we start with a statement: heat is a form of internal energy that is related to molecular movement.
However, while heat transfer during the process of conduction occurs between molecules oscillating around their relatively fixed positions and transferring their internal energy by "shaking" the neighboring molecules, convection occurs when molecules are free to travel in different directions and carry their internal energy with them.

In other words, conduction is a pure transfer of energy on a micro level from one oscillating molecule in a relatively fixed position to another such molecule, while convectionoccurs when molecules freely fly away from their positions, carrying their internal energy with themselves, thus transferring energy on a macro level.

It should be noted that, when dealing with solid objects, conduction is a prevailing way of heat transfer, while in liquids and gases the main way of heat transfer is convection. It does not mean that conduction does not occur in liquids or gases, it does, but it does not constitute the major way of heat transfer. Much more heat is transferred through the mechanism of convection

Here are a few examples of heat transfer through convection:
(a) heating up water in a pot; heat is carried from hot bottom of a pot up by hot (fast moving with high kinetic energy) molecules;
(b) circulation of air in the atmosphere from hot places to cold;
(c) circulation of water in oceans from hot places to cold.

Describing convectionmathematically is not a simple task.
While in case of conduction we can use a relatively simple Fourier's Law of Thermal Conduction
q(x) = −k·dT(x)/dx
that describes the heat flow as a function of how fast the temperature between the layers of conducting material changes (dT(x)/dx) and properties of the material itself (conductivity coefficient k), the process of convection is significantly more complex, described by convection-diffusion differential equations that are beyond the scope of this course.

However, for practical purposes we can use a similar formula that puts the amount of heat transferred by convectionprocess in a liquid or gas during a unit of time through a unit of area as proportional to a difference of temperatures between the layers of liquid or gas and a convective heat transfer coefficient h that depends on the physical properties of this liquid or gas:
q = −h·(T2−T1)

This formula puts amount of heat q going through a layer of a unit area of liquid or gas during a unit of time as proportional to a difference of temperatures between bounding surfaces of this layer T2−T1 and some physical properties of liquid or gas expressed in convective heat transfer coefficient h that, in turn, depend on such properties as viscositydensity, the type of flow (turbulent or laminar) etc.

Consider an example.
A round steam pipe of temperature 100°C goes through a room with air temperature 25°C. We have to calculate the amount of heat from the pipe to select an air conditioner required to neutralize the heat from a pipe and keep the room temperature at that level.
Assume that the pipe's length is 4m, diameter 0.2m and the convective heat transfer coefficient of air is 40J/(sec·m²·°C). As we know, J/sec is a unit called "watt", so we will use W instead of J/sec.

The heat transfer per unit of time through a unit of area of a pipe is, therefore,
q = 40·(100−25) = 3000(W/m²)
The pipe's area is
A = π·0.2·4 = 2.512(m²)
Therefore, the pipe is producing the following amount of heat:
Q = 3000·2.512 = 7536(W)

So, we need an air conditioner that can extract 7536W of heat from the room to maintain stable temperature of 25°C.
Usually, the power of air conditioners is measured in BTU/hr (1 watt = 3.41 BTU/hr). So we need an air conditioner of approximately 2200 BTU/hr - a relatively small one.

Another example.
Outside temperature is 40°C, inside a room we want temperature 25°C. The glass wall between a room and outside air has an area of 20m². What kind of air conditioner is needed to maintain the room temperature at 25°C, assuming the convective heat transfer coefficient of air is 40W/(m²·°C)?

Q = 40·(40−25)·20 = 12000(W)
This is equivalent to about 3500 BTU/hr.

Thursday, June 20, 2019

Unizor - Physics4Teens - Energy - Heat Transfer - Conduction





Notes to a video lecture on http://www.unizor.com

Heat Transfer - Conduction

As we know, heat is a form of internal energy that is related to molecular movement.

For solids the molecular movement is usually restricted to molecules' oscillation around some neutral positions.

For liquids the freedom of molecular motion is greater, but still restricted by external forces, like gravity, and surface tension. The average distance between molecules of liquids is relatively constant.

Gas molecules are usually taking all the space available for them. Such forces as gravity also restrict their movement (otherwise, the air molecules would fly away from our planet), but still allow substantial freedom. The average distance between molecules of gas mostly depends on a reservoir the gas is in, the larger the reservoir - the larger average distance between molecules.

Transfer of heat is transfer of molecular movement from one object or part of an object to another object or part of an object, from an object or part of an object with more intense molecular movement (relatively warmer) to an object or part of an object with less intense movement (relatively cooler).

There are three major ways to transfer heat from a hot object to a cold one:
Conduction,
Convection,
Radiation.
This lecture explains a concept of conduction.

Conduction

Conduction of heat energy is a transfer of molecular movement mostly applicable to solid objects - between two solid objects that touch each other, having an area of a contact, or within one object, one part of it having different temperature than another.
The conductivity is present in heat transfer in liquids and gases, but there it's usually combined with another form of heat transfer - convection, while in solids it's not the case, and we can study conductivity by itself.

An example is a building wall, one side of which towards the outside having temperature of the air outside, while inner surface of the wall having room temperature. The heat energy constantly flows from a warmer surface of the wall to the opposite cooler one with some rate of flow that depends on the thermal conductivity of the wall material. The wall material with higher level of thermal conductivity will transfer more heat energy to a cooler side within unit of time per unit of area, which is usually not a desirable property of the building walls.

The mechanism of heat transfer through conductivity can be explained as follows.
Imagine two objects (or two parts of the same object), a hot one with higher intensity of molecular movement and a cold one with lower intensity level of molecular movement, that touch each other along some surface, while completely insulated from heat around them. For example, you put a cold metal spoon into a styrofoam cup with hot tea.

Molecules of a hot object are hitting the molecules of a cold one, thus forcing the molecules of a cold object to move faster. These faster molecules of a cold object, in turn, hit their neighbors, forcing them to move faster. This process of transferring heat energy through contacting surfaces continues until the intensity of molecular movement gradually equalizes on average. A hot object will lose some energy of molecular movement, while a cold one will gain it. As a result, the temperatures of both will equalize.

Since the heat energy is a kinetic energy of molecular movement, that is a sort of mechanical energy, we expect that the total amount of energy for an isolated system will remain constant, whatever a hot object loses in its kinetic energy of molecular movement will be gained by a cold object. The total amount of heat energy will remain the same.

If you put a silver spoon into a cup of hot tea, it will heat up faster than a spoon made of steel, which, in turn, will heat up much faster then a spoon made of wood. The reason for this is that the thermal conductivity of different materials is different.

We can experimentally measure the thermal conductivity of different solids by having a standard rod of any solid material at certain starting temperature and heating its one end by bringing it to contact with some hot object. Measuring the temperature on the other end after different time intervals will give us a picture of growing temperature.

Some materials with higher thermal conductivity will have the temperature at the opposite end of a rod growing faster than in case of other materials.
Metals have much higher heat conductivity then plastic or wood, for example. That's why the handle of a tea kettle is usually made of plastic or wood. Diamonds have one of the highest thermal conductivity, even higher than silver.

More precise definition of thermal conductivity is related to a concept of heat flux(sometimes, called heat flow density or thermal flux, or thermal flow density). Heat flux is an amount of heat energy flowing through a unit of area during a unit of time.

Let's examine how heat flows through a building wall made of some uniform material from a warm room to cold air outside the building.
Assume, the room temperature is Troom and the cold air outside the building has temperature Tair . If the thickness of a wall is L, the temperature inside the wall T(x), as a function of the distance x from the surface facing outside, gradually changes from T(0)=Tair to T(L)=Troom.

It is intuitively understandable and experimentally confirmed that amount of heat energy flowing through a unit of area of such a wall during a unit of time (heat conductivity of a wall) is proportional to a difference between temperatures Troomand Tair and inversely proportional to a thickness of a wall L:
q = −k·(Troom − Tair / L
(negative sign is used because the flow of heat is opposite to a direction of temperature growth).

The situation with heat fluxmight be compared with a water flow down a river between two points A and B. The difference in levels above the sea level of these points is similar to a difference in temperature between the inside and outside walls of a building. The distance between points A and B is similar to a thickness of a wall. It's reasonable to assume that amount of water flowing through a unit of area in a unit of time will be proportional to a difference between the levels of points A and B above the sea level and inversely proportional to a distance between these two points.

To make this definition of the heat flux more precise and independent of the way how the heat flows inside the wall, let's consider a thin slice of wall parallel to both sides from a point at distance x from the outdoor cold side to a point at distance x+Δx.
The heat flow through this thin slice of a wall, as a function of distance x, can be expressed similarly to the above:
q(x)=−k·[T(x+Δx)−T(x)] /Δx

Next step is, obviously, to reduce the thickness of the slice by making Δx infinitesimal, that is Δx→0, which leads to the following definition of the heat flux:
q(x) = −k·dT(x)/dx
This definition was formulated by Fourier in 1822 and is called Fourier's law of thermal conduction.
The coefficient k is called thermal conductivity.

To find the amount of heat Q(x,A) going through an area Aat distance x from the outside wall during a unit of time, we have to multiply the heat flux by an area:
Q(x,A) = −k·A·dT(x)/dx
As in a case of water flow along the river, if the temperature is linearly dependent on the distance from the outside wall, the derivative is constant and the flow of heat is constant. But, if the wall material is uneven, like in case of a river bed not being a straight line down, the heat flow rate will change.

Wednesday, June 12, 2019

Unizor - Physics4Teens - Energy - Measuring Heat - Problems





Notes to a video lecture on http://www.unizor.com

Measuring Heat - Problems

Problem 1

How much heat energy is required to raise the temperature of 1 kg of water from 20°C to a boiling point of 100°C?
Assume the specific heat capacity of water is
Cw = 4184 J/(kg·°C).

Answer
Q = C·m·(Tend−Tbeg) =
= 4184·1·(100−20) =
= 334,720 J


Problem 2

A piece of unknown metal of mass Mm and temperature Tm was put into an isolated reservoir filled with Mw mass of water at temperature Tw. After the system of water and metal came to thermal equilibrium, its temperature became T.
Assume that the metal is not too hot (so, water will not vaporize) and not too cold (so, the water will not freeze).
Assuming that the water's specific heat capacity is known and equals to Cw, what is the specific heat capacity Cm of the unknown metal?

Answer

Cw·Mw·(T−Tw) =
= Cm·Mm·(Tm−T)

from which Cm equals to
Cw·Mw·(T−Tw)/[Mm·(Tm−T)]

Problem 3

A burger has about 300 kcal of energy in it.
1 kcal = 4184 J.
A person, who ate it, wants to spend this energy by climbing up the stairs. A person's mass is 75 kg, the height between the floor is 3 m.
Assume that only 25% of energy in the food can be used for climbing, while the other 75% is needed to maintain our body's internal functions.
Counting from the ground floor as floor #0, to what floor can a person climb using that energy from a burger?

Answer
Ebur = 300kcal · 4184J/kcal =
= 1,255,200
J

Eclimb = 0.25·Ebur =
= 313,800
J

Efloor = 75kg · 9.8m/sec² · 3m =
= 2205
J

N = Eclimb / Efloor ≅ 142 floors

Problem 4

An ice of mass 0.1kg has temperature −10°C.
What's the minimum amount of water M at temperature 20°C needed to melt it?
Assume, specific heat capacity of water is 4183J/(kg·°C) and that of ice is 2090J/(kg·°C). Assume also that the amount needed to melt ice at 0°C is 333,000J/kg.

Answer
Ewarm = 2090·0.1·10 = 2090J
Emelt = 333000·0.1 = 33300J
Eneed = 2090 + 33300 = 35390J
Ewater = 4183·M·20 = 83660·M
Ewater = Eneed
35390 = 83660·M
M = 0.423kg

Monday, June 10, 2019

Unizor - Physics4Teens - Energy - Heat - Heat & Temperature





Notes to a video lecture on http://www.unizor.com

Heat and Temperature

Specific Heat Capacity

Let's discuss the relationship between heat and temperature.
From the unscientific standpoint these two concepts are almost identical. Heating an object results in increase of its temperature. Increasing a temperature of an object constitutes its heating.
Yet, from the strictly scientific viewpoint these two concepts are different.

Heat is energy of some specific type, that can be transferred from one object to another, while temperature is an average kinetic energy of molecules of an object.
The same amount of heat, transferred into different objects, will result in different growth of their temperatures, depending on many factors, like mass, chemical composition, state etc. of these objects. Analogy of this is that same amount of fuel in different cars results in achieving very different speeds and, therefore, different kinetic energies in different cars, even if the gas pedal is pushed all the way down for all cars. It's just because cars are different and their internal structure converts fuel into movement differently.

It has been experimentally observed that to increase a temperature of an isolated object of a unit mass by a unit of temperature is independent (within reasonable level of precision) of the initial temperature of an object, but depends only on the type of object's material, composition, state etc. In other words, amount of heat needed to increase a temperature of 1 kg of water from 20°C to 21°C is the same as from 50°C to 51°C. If, instead of water, we take copper, the amount of heat, needed to increase its temperature from 20°C to 21°C, will be the same as to increase it from 50°C to 51°C, but different than that for water.

The above experimental fact allowed to establish a concept of specific heat capacity for each material as an amount of heat required to increase the temperature of a unit of mass of this material (1 kg in SI) by a unit of temperature (1°C or 1°K in SI).

Thus, specific heat capacity of water is, as we know, one kilocalorie per kilogram per degree - 1 kcal/(kg·°K), that is about 4184 joules per kilogram per degree - 4183 J/(kg·°K).
For copper the specific heat capacity is 385 J/(kg·°K).
Gold has the specific heat capacity of 129 J/(kg·°K).
Uranium's specific heat capacity is 116 J/(kg·°K).
Cotton's specific heat capacity is 1400 J/(kg·°K).
Hydrogen's specific heat capacity is 14304 J/(kg·°K).
Generally speaking, but not always, more dense, more solid materials have less specific heat capacity than less dense or liquids, which have, in turn, less specific heat capacity than gases.

Knowing specific heat capacityC of material of an object and its mass m, we can easily determine amount of energy ΔQneeded to heat it up by ΔTdegrees:
ΔQ = C·m·ΔT
Inversely, knowing the amount of heat supplied, we can determine an increase in temperature:
ΔT = ΔQ/(C·m)
Notice, that increment of temperature ΔT and increment of heat energy ΔQ can be both positive or both negative, which means that an object, that has increased its temperature, has increased (gained, consumed) energy, and the object that decreased its temperature, has decreased (lost, released) energy.

Change of State

Consider specific heat capacity of ice and water:
Ice: 2090 J/(kg·°K)
Water: 4183 J/(kg·°K)
Both these substances exist at temperature about 0°C=273°K. That means that we can heat 1 kg of ice up to 0°C spending 2090 joules per each degree of temperature, but to increase the temperaature of the water around 0°C we have to spend 4183 joules per each degree. But, while ice is melting into water, which takes some time, both states, solid and liquid, exist side by side and the temperature of the water will not rise while the ice is not completely melted, we have to spend heat energy just on melting without actually changing the temperature of a substance, that remains around 0°C during the melting process.

This experimental observation leads us to believe that, while graduate change of temperature for any specific state of matter linearly depends on the amount of heat supplied, change of state (like melting or freezing, or evaporating etc.) brings an element of non-linearity to this dependency.
More precisely, the temperature, as a function of amount of heat supply, in case of an object going through transformation of state from solid to liquid, looks like this (for better view, right click on the picture and open it in another tab of a browser):

As seen on this graph, supplying heat to ice will increase its temperature proportionally to amount of heat supplied.
Then, when the temperature reaches 0°C, ice will start melting and new heat will not change the temperature of ice/water mix, but will be used to change the state of matter from solid (ice) to liquid (water).

This process of melting consumes heat without increasing the temperature until all ice is melted. Such a transformation of state that requires supply of heat energy is called endothermic.
Then, when the process of melting is complete, and all ice is transformed into water, the temperature of water will start increasing, as the new heat is supplied, but with a different coefficient of linearity relatively to the heat than in case of ice, since water's specific heat capacity is different than that of ice.
The graph above is characteristic to any heating process, where the transformation of the state of matter is involved. It is relatively the same for transforming liquid to gas (evaporation) or solid to gas (sublimation).

In case of a reversed transformation from liquid to solid (freezing) or gas to liquid (condensation) or gas to solid (deposition) the heat energy must be taken away from a substance. It is the same amount in absolute value as was needed to supply to solid to melt it into liquid or to liquid to vaporise it into gas, or to solid to vaporize it into gas. So, in case of freezing, condensation or deposition we deal with the process of decreasing heat energy of a substance. Such a process is called exothermic.

Amount of heat energy needed to transform a substance from one state to another is also experimentally determined and, obviously, depends on a substance, its mass and a kind of transformation it undergoes.
For example, melting of 1 kg of ice requires 333,000 joules of heat energy to be supplied. The same amount of heat energy should be extracted from the water at 0°C to freeze it into ice.

Unizor - Physics4Teens - Energy - Heat - Measuring





Notes to a video lecture on http://www.unizor.com

Measuring Heat - Calorie

Heat and temperature were known and researched by scientists long before the molecular movement was determined as their essence. As a result, attempts of measuring the amount of heat were unrelated to kinetic energy of the molecules.

The unit of amount of heat was defined as amount of heat needed to increase the temperature of one gram of water by one degree Celsius (or Kelvin), and this unit of amount of heat was called a calorie (cal).

Obviously, this definition has its flaws. For example, the amount of heat needed to heat the same amount of water by the same temperature depends where exactly on Earth we are. The higher we are above the sea level - the less heat is needed. It also depends on the chemical purity of water. Also it's not obvious that the amount of heat needed to heat the water from 1°C to 2°C is the same as amount of heat needed to heat it from 88°C to 89°C, though within reasonable level of precision we do assume that this is true.

With the development of molecular theory of heat and establishing relationship between heat and kinetic energy of molecules there was a need to put into correspondence existing units of heat (calories) and units of mechanical energy (joules). A simple experiment allowed to do just that.

Imagine a standing on the ground large reservoir with known amount of water M of depth H from the surface to the bottom, kept at certain known temperature T°, and a relatively small stone of known mass m, having the same temperature T°, kept on the level of the surface of the water in a reservoir. This stone has certain known amount of potential energy E=m·g·Hrelatively to the bottom level of a reservoir.
Now we let the stone go down the reservoir to its bottom. Its potential energy relatively to the bottom of a reservoir decreases to zero. Where did the potential energy go? It's used to stir the water, thereby increasing the kinetic energy of its molecules.

Because of this more intense movement of molecules, the water will increase its temperature. Potential energy of a stone will turn into kinetic energy of the molecular movement of water. If the temperature of the water has risen by ΔT°, potential energy of a stone m·g·H (in joules) equals to M·ΔT (in calories).

Precise experiments like the above allowed to determine the correspondence between historical measure of the amount of heat in calories under different conditions and contemporary one in units of energy in SI - joules. This correspondence had been established approximately as
1 calorie = 4.184 joules
but under different conditions (initial temperature, air pressure etc.) it might be equal to a slightly different value.

Besides calorie, which is a relatively small amount of heat, the unit kilocalorie (kcal) had been introduced.
As is obvious from its name, 1 kilocalorie = 1000 calories.

The amount of energy contained in food and in some other practical cases very often is measured in kilocalories, which sometimes are called large calories, while a calorie is sometimes called small calorie. Unfortunately, the word "large" in many cases is omitted, which might cause misunderstanding.
From the experimental viewpoint, 1 kcal is amount of heat needed to heat 1 kg of water by one degree Celsius (or Kelvin).

To avoid problems with the definition of calorie as amount of heat needed to warm up one gram of water by one degree Celsius, the contemporary scientific definition of thermocalorie is
1 thermocalorie = 4.1833 joules

Monday, June 3, 2019

Unizor - Physics4Teens - Energy - Ideal Gas Kinetics - Problems

Notes to a video lecture on http://www.unizor.com

Problem 1

How much kinetic energy have all the molecules in the room?
How fast should an average size car move to have this amount of kinetic energy?

Assume the following:
(a) the room dimensions are 4x4x3 meters (that is, V=48m³);
(b) normal atmospheric pressure is 100,000 Pascals (that is, p=100,000N/m²);
(c) a mass of an average size car is 2,000 kg (that is, M=2,000kg).

Solution

From the lecture on Kinetics of Ideal Gas we know the relationship between the pressure on the walls of a reservoir, volume of a reservoir and total kinetic energy of gas inside this reservoir:
p = (2/3)Etot /V

From this we derive a formula for total kinetic energy:
Etot = (3/2)·p·V

Substituting the values for pressure and volume, we obtain
Etot = (3/2)·100,000·48 =
= 7,200,000
(joules)


The kinetic energy of a car is
E = M·v²/2
Therefore, given the kinetic energy and mass, we can determine the car's speed:
v = √2·E/M
Substituting calculated above Etot=7,200,000(J) for E and the value for mass M=2,000(kg), we obtain
v = √2·7,200,000/2,000
≅ 85
(m/sec)
≅ 306
(km/hour)
≅ 190
(miles/hour)



Problem 2

Given the temperature, pressure and volume of the air in a room, determine the number of gas molecules in it.

Assume the following:
(a) the room dimensions are 4x4x3 meters (that is, V=48m³);
(b) normal atmospheric pressure is 100,000 Pascals (that is, p=100,000=105N/m²);
(c) temperature is 20°C (that is, T=20+273=293°K).

Solution
Recall the combined law of ideal gas
p·V/T = kB·N = const
where
kB = 1.381·10−23 (J/°K) is Boltzmann's constant and
N is the number of gas molecules in a reservoir.
From this we derive the number of molecules
N = p·V/(kB·T)
Substituting the values,
N = 105·48/(1.381·10−23·293) = 0.12·1028
It's a lot!

Monday, May 6, 2019

Unizor - Physics4Teens - Energy - Heat - Temperature, Pressure, Volume o...





Notes to a video lecture on http://www.unizor.com

Temperature, Pressure
and Volume of Ideal Gas


Temperature is an observable macro property of an object. It's related to a particular instrument we use to measure this property.
Let's examine the mechanism of this measurement using a classic mercury-based or alcohol-based thermometer.

Our first step in measurement is to make a physical contact between a thermometer and an object of measurement (for example, a human body or air in a room). When accomplished, we expect that the measurement of a thermometer would correspond to a state of average kinetic energy of molecules of an object. The reason it happens (and it takes some time to happen) is that on a micro level molecules at the surface of an object are colliding with molecules at the surface of a thermometer, and exchange the kinetic energy, eventually equalizing it. The molecules close to a surface, in turn, collide with surface molecules and also eventually equalize their average kinetic energies. This process continues until the average level of kinetic energy in all parts of an object and in a thermometer equalize.

What is important in this case is that the total amount of kinetic energy of all molecules of an object and a thermometer remains the same. So, if an object has more intense movement of molecules and a thermometer's molecules are moving slower, the kinetic energy is transferred from an object to a thermometer. If the molecules of a thermometer are, on average, faster, then the exchange of kinetic energy will be from a thermometer to an object.
 In any case, the average kinetic energy of molecules of both an object and a thermometer equalize.

An important consideration is that the contact between an object and a thermometer changes the average level of kinetic energy in both. The process of measuring, therefore, is not completely neutral towards an object. However, what happens in most cases is that the number of molecules inside an object we measure is usually significantly greater that the number of molecules in a thermometer. As a result, equalizing the average kinetic energy of all molecules does not significantly change the level of average kinetic energy of molecules of an object, and the level of average kinetic energy of the molecules of a thermometer is a good representation of this characteristic of an object.

As explained in the Heat and Energy lecture of this course, the temperature in mercury or alcohol thermometers is an observable expansion of the volume of liquid inside a thermometer. We also indicated in that lecture that this thermo-expansion is proportional to an average of squares of velocities of molecules, that is proportional to average kinetic energy of the molecules of a thermometer, which, in turn, is equalized with average kinetic energy of molecules of an object.

Thus, by observing the expansion of liquid in a thermometer we measure the average kinetic energy of molecules of an object, which allows us to write the following equation:
T ≅ AVE(Ekin) = Eave
where T is an observable level of liquid inside a thermometer in some units (that is, temperature) and
AVE(Ekin) = Eave is average kinetic energy of molecules of an object.
This looks more natural in a form
AVE(Ekin) = Eave ≅ T
since average kinetic energy of molecules Eave (micro characteristic) cannot be easily observed, while the temperature T (macro characteristic) can.

So, we have established that the temperature (in some units, starting from absolute zero) and average kinetic energy of molecules are proportional. The coefficient of proportionality between the temperature and average kinetic energy of molecules remains unknown and is different for different substances.

Obviously, the measure of liquid in a thermometer should be calibrated and, for this equation to be true, we have to assign the zero temperature to a state of an object when all its molecules are at rest, which happens when there is no source of energy around, like in open space far from stars.
The convenient scale is the Kelvin scale with zero temperature on this scale corresponding to this state of molecules at complete rest and a unit of measurement of the temperature is a degree with the distance between the temperature of melting ice and boiling water assigned as 100 degrees on this scale.

From the previous lecture about kinetics of ideal gas we know the relationship between the pressure of ideal gas on the walls of a reservoir, the volumeof a reservoir and average kinetic energy of the molecules
p = (2/3)N·Eave /V
where
p is the gas pressure against the walls of a reservoir,
N is the number of gas molecules in a reservoir,
V volume of a reservoir,
Eave is average kinetic energy of molecules,

Alternatively, it can be written as
Eave = (3/2)·p·V/N

Comparing this with a derived above relationship between temperature (counting from the absolute zero) and average kinetic energy of molecules, we have established a relationship
Eave = (3/2)·p·V/N ≅ T
and
p·V/N = k·T
where k is an unknown coefficient of proportionality.

The only thing that prevents us from determining average kinetic energy of molecules by its temperature is an unknown coefficient of proportionality k.

Now we will concentrate attention on gases, as an object of measuring temperature and average kinetic energy of molecules.
Different gases have different molecules and different molecular mass. Using certain theory and using chemical and physical experiments, we can compare the masses of different molecules and even measure this mass in certain "units of mass" called atomic mass units.
For such a unit of molecular mass scientists used 1/12 of a mass of a single atom of carbon. So, in this system of units hydrogen molecule of 2 hydrogen atoms H2 has approximate mass of 2, oxygen molecule of two oxygen atoms O2 - approximately 32, carbon dioxide molecule CO2 had a measure of, approximately, 44 atomic mass units, etc.

Using this measurement, we can always establish experiments with the same number of molecules of different gases. For example, if the mass of certain amount of oxygen (atomic mass of molecules O2 is 32) is 16 times greater than the mass of certain amount of hydrogen (atomic mass of molecules H2 is 2), we can assume that the number of molecules in both cases is the same.

It has been experimentally established that, if the same number of different gas molecules are placed in reservoirs of the same volume and hold them at the same temperature, the pressure on the walls in both cases will be the same. Alternatively, if the pressure is the same, the temperature will be the same too.
In other words, the coefficient kin formula
p·V/N = k·T
does not depend on the type of gas we deal with, it's a universal constant called the Boltzman's constant, which is equal to
kB = 1.381·10−23 (J/°K)
This was the reason to introduce a concept of ideal gas. All gases are, approximately, ideal to a certain degree of precision. This is related to the fact that molecules of the gas are flying with high speeds and on large distances from each other, much larger than their geometric sizes.

Now we can write the equation between the temperature T (in degrees °K from absolute zero), average kinetic energy of molecules (in units of SI joules J), pressure p (in units of SI newton/m²), volume V (in ) and number of molecules in a reservoir for ideal gas:
Eave = (3/2)·p·V/N = (3/2)·kB·T

Consequently, if we are dealing with certain fixed amount of gas (N molecules) then
p·V/T = kB·N = const
That means that changing the pressure, volume and temperature of the same amount of gas preserves the expression p·V/T
which is called the Combined Ideal Gas Law.

For example, if the absolute temperature remains the same, but volume taken by certain amount of gas increases (decreases) by some factor, the pressure will decrease (increase) by the same factor, that is pressure and volume are inversely proportional to each other (Boyle-Mariotte's Law).

If the pressure remains the same, but the volume taken by certain amount of gas increases (decreases) by some factor, the absolute temperature will increase (decrease) by the same factor, that is volume and absolute temperature are proportional to each other (Charles' Law).

If the volume taken by certain amount of gas remains the same, but the absolute temperature increases (decreases) by some factor, the pressure will increase (decrease) by the same factor, that is absolute temperature and pressure are proportional to each other (Gay-Lussac's Law).

Out of curiosity, let's use the formula
Eave = (3/2)·kB·T
to calculate how fast the molecules of oxygen are flying in the room at some normal temperature.
Assume the pressure at the ground level is about 100,000N/m² and the temperature in the room is about 20°C=293°K. Then the average kinetic energy of a molecule of oxygen is
Eave = (3/2)·1.381·10−23·293 =
= 6.06·10−21 J

Mass of a molecule of oxygen O2 is m=5.31·10−26 kg
From the formula for kinetic energy E=m·v²/2 we derive the average of squares of velocities of oxygen molecules as
AVE() = 2·Eave /m =
= 2·6.06·10−21/5.31·10−26 =
= 2.28·105

Therefore, the average speed of oxygen molecule will be equal to a square root of this number:
AVE(v) = 478 m/sec
Pretty fast moving! Take into consideration, however, that real oxygen molecules, as molecules of any real gas, are chaotically colliding with other and change the direction all the time.

Wednesday, May 1, 2019

Unizor - Physics4Teens - Energy - Heat - Kinetics of Ideal Gas





Notes to a video lecture on http://www.unizor.com

Kinetics of Ideal Gas

In this lecture we will discuss a concept of pressure of gas, enclosed in some reservoir, against the walls of this reservoir.

Usually, speaking about pressure, we have in mind an object of certain weight on a horizontal table, in which case the constant force of its weight exerts a pressure on a table equal to this weight divided by an area of the table occupied by an object or, as it's sometimes presented, the weight per unit of area.

As an introduction to kinetics of gases, let's consider a reservoir shaped as a cube with sides aligned parallel to coordinate planes and with the length of each edge L.
Assume that there is a single molecule of gas of mass mflying between the opposite walls perpendicularly to them, elastically reflecting off these walls.
Keep in mind that, according to the Third Newton's Law, the force exerted by the wall towards the molecule, which causes the reflection of a molecule from the wall, is equal to the force exerted by a molecule towards the wall, which causes pressure.

The area of each of these opposite walls equals to , which will be used to calculate the pressure.

In this case of a single gas molecule, flying between the opposite walls of a reservoir, the force is variable. It exists during the time when a molecule hits the wall and then the force becomes zero until the next time this molecule hits the same wall.
A proper definition of pressurein this case is based on a concept of the average force during certain amount of time.

Firstly, let's evaluate the average force during a single period of oscillation of a molecule between walls from a moment of time it's near one wall to a next moment it's at this position after flying to the opposite wall and returning back.

As we know, a change of a momentum of an object equals to an impulse of the force that caused this change
m·vend − m·vbeg = F·τ
where
m is a mass of an object
vbeg is the velocity of an object in the beginning of the action of the force
vend is the velocity of an object at the end of the action of the force
F is the force, acting on an object
τ is the time duration of the force, acting on an object.
Keep in mind that velocity and force are vectors in the above equation.

In our case we consider one period of oscillation of a molecule as the time period of average force acted on it.
So, τ is the time between two consecutive events when a molecule is at the wall opposite to the one it collides with:
τ = 2L/|v|.
where |v| is now an absolute speed of a molecule (scalar).

Since we are considering elastic reflection of a molecule in the opposite direction to its initial trajectory,
vend = −vbeg
and the equation above can be written in scalar form
2m·|v| = |F|·τ = |F|·2L/|v|
from which we determine the absolute average force
|F| = m·|v|²/L
Mass m of a molecule is, obviously, a constant during this process.
The average pressure of a molecule on a wall is the average force divided by the area of a wall:
p = m·|v|²/L³ = m·|v|²/V =
= 2E
kin /V

where V is the volume of a reservoir and Ekin is a kinetic energy of a molecule.

Our first step towards a comprehensive theory is to consider a case of many molecules flying parallel to each other in the same reservoir as above, but with, generally speaking, different speeds. Obviously, the pressure against the wall will be greater because each molecule contributes its own force against the wall during a collision.
If vi is a speed of the ithmolecule, the combined pressure will be
p = Σpi = Σm·vi²/V =
= 2E
tot /V

where Etot is a total kinetic energy of all molecules moving parallel to each other.

Considering the number of molecules N remains the same, it's more convenient to express this formula in terms of an average of squares of molecular speed and average kinetic energy of molecules:
ave = (1/N)Σvi²
Eave = (1/N)ΣEi
Using this average of squares of molecular speed, the pressure is
p = N·m·v²ave /V = 2N·Eave /V
which brings us to a principle of proportionality between pressure of the gas against the walls of a reservoir and an average of squares of the molecular speed or to an average of kinetic energy of the molecules.

Granted, we showed this only for a case of all molecules moving between two opposite walls parallel each other and perpendicularly to the walls they collide with.

Let's make another step towards comprehensive theory and consider the chaotic movement of all gas molecules.
To quantitatively approach molecular movement, we will use a model of an ideal gas.
This model assumes that molecules of ideal gas are point-mass objects completely chaotically moving in all directions with equal probabilities and elastically colliding at random times among themselves or with all the walls of a reservoir that contains this gas.

We further assume that no interacting forces (like gravity or electromagnetic forces) exist between them. So, only kinetic energy of these molecules plays the role in evaluation of the characteristics of the ideal gas.

Every vector of velocity of any molecule can be represented as a sum of three vectors along the XYZ-coordinates
v = vx+vy+vz

So, every movement of a molecule can be represented as simultaneous movement in three different directions. Depending on which wall of the reservoir is hit by a molecule, its pressureagainst the wall is determined only by one component - the one that is perpendicular to the surface of a wall.

If molecules inside the reservoir are moving completely chaotically, they are hitting all walls with approximately the same frequency. So, for each of the six walls of a cube we can use the same logic as above for one side, except for the walls parallel to YZ coordinate plane the pressures px will be proportional to the average of vx², for the walls parallel to XZ coordinate plane the pressures py will be proportional to the average of vy² and for the walls parallel to XY coordinate plane the pressures pz will be proportional to the average of vz².

Considering completely chaotic character of the molecular motion in the ideal gas, the three pressures pxpy and pzmust be the same. Similarly, averages of vx²vy² and vz² must be also the same.

Since v = vx+vy+vz, according to three-dimensional equivalent of Pythagorean Theorem,
v² = vx² + vz² + vz²
and, similarly, for averages
AVE() = v²ave =
=
AVE(vx²)+AVE(vy²)+AVE(vz²)


From the equality of averages for the ideal gas we come to the following equalities:
ave = 3·AVE(vx²)
AVE(vx²) = (1/3)·v²ave

Therefore,
px = N·m·AVE(vx²)/V =
= (1/3)N·m·v²
ave /V =
= (2/3)N·E
ave /V


Similarly,
py = N·m·AVE(vy²)/V =
= (1/3)N·m·v²
ave /V =
= (2/3)N·E
ave /V

and
pz = N·m·AVE(vz²)/V =
= (1/3)N·m·v²
ave /V =
= (2/3)N·E
ave /V


And, since pressures on all walls are the same in the ideal gas,
p = (1/3)N·m·v²ave /V =
= (2/3)N·E
ave /V = (2/3)Etot /V


The expression of a pressure in terms of average kinetic energy and volume is more general than in terms of mass, average of squares of molecular speed and volume, as it encompasses a case with molecules of different masses.

Wednesday, April 17, 2019

Unizor - Physics4Teens - Energy - Temperature and Kinetic Energy





Notes to a video lecture on http://www.unizor.com

Heat, Temperature &
Kinetic Energy of Molecules


The goal of this lecture is to demonstrate that temperature, as a measure of heat, is proportional to average of a square of molecular speed.

Temperature is what we see on a thermometer that measures the intensity of the molecular movement. Let's examine this process of measurement using the classical mercury-based thermometer.

The thermometer shows different values of temperature based on physical property of mercury to expand, as its molecules are moving faster. The reason for this is that, as the molecules move faster, they cover longer distance before they are stopped by collision with other molecules or the walls or by a surface tension forces. The faster molecules push against obstacles with greater forces, and that is the reason why the volume of hot mercury is larger than the volume of the same mass of cold mercury.

Let's quantify this increase in volume.
There is a force of resistance that acts against the movement of the molecule, as it pushes its way through, that stops this movement. In some way it's analogous to throwing a stone vertically up against the force of gravity. The force of gravity eventually stops the movement upward, and the stone falls back on the ground.

The height of the trajectory of a stone depends on the initial speed it has at the beginning of movement on the ground level. Assume, the initial speed is V. Then the kinetic energy in the beginning of motion is M·V²/2, where M is a mass of a stone. The potential energy relative to the ground in the beginning of motion is zero.
At the highest point at height Hthe kinetic energy is zero but potential is M·g·H, where g is a free fall acceleration.

As we know, the total mechanical energy is constant in this case, from which follows that kinetic energy in the beginning of motion (with potential energy zero) equals to potential energy at the highest point (with kinetic energy zero):
Ekin = M·V²/2 = M·g·H
From this follows
H = V²/(2g)
So, the height of the rise and the square of initial velocity are proportional to each other, independently of mass.

Obviously, one mercury molecule does not do much, but when the heat is relatively evenly distributed among all molecules, their combined efforts push the surface of the mercury in the tube sufficiently noticeably, and the height it rises is proportional to average of a square of molecular speed.

The thermometers are different, their design and construction differ, they all must be calibrated, but one thing remains - a principle of proportionality between geometrical expansion (that is, temperature, as we read it on a thermometer's scale) and an average of squares of the molecular speed.

It is important to point out that this proportionality should be understood in absolute terms. That is, if the speed is zero, the temperature is zero as well. So, we are talking about temperature scales with zero representing absolute zero on the Kelvin's scale, like in space far from any source of energy.
The coefficient of proportionality, of course, depends on the units of measurement.

So, if the average speed of molecules doubles, the temperature on the Kelvin's scale rises by a factor of 4.

Inversely, if an object is at temperature 0°K (absolute zero), the kinetic energy of its molecule is zero, that is no molecular movements.
If we observe that the temperature has risen by a factor of 4, we can safely assume that the average speed of molecules doubles.

Monday, April 15, 2019

Unizor - Physics4Teens - Energy - Heat - Molecular Movement



Notes to a video lecture on http://www.unizor.com

Molecular Movement

Molecules

In order to understand the nature of heat and temperature, we have to go inside the objects we experiment with.
If we divide a drop of water into two smaller drops, each of these smaller drops will still be water. Let's continue dividing a smaller drop into even smaller and smaller. There will be a point in this process of division, when further division is not possible without changing a nature of the object, it will no longer be water.

From what we know now, the water consists of hydrogen and oxygen - two gases, connected in some way. In our process of division of a drop of water, we will reach such a point that, if we divide this tiniest drop of water, the result will be certain amount of hydrogen and certain amount of oxygen, but no water.
That tiniest amount of water, that still preserves the quality of being water, is called a molecule.
Similarly, tiniest amount of any substance, that retains the qualities of this substance, is called a molecule.
Incidentally, if we divide a molecule of any substance, the result will be certain atoms, that possess completely different qualities.

In our discussion about heat we will not go deeper than the molecules because our purpose at this stage is to study the nature of heat as it relates to different objects and substances, so the preservation of the qualities of these objects and substances is important. That's why in this part of a course we will not cross the border between molecular and more elementary atomic level.

Any object or substance we are dealing with consists of certain number of molecules - the smallest particles that retains the qualities of this object or substance. This number of molecules, by the way, for regular objects we see and use in practical life, is extremely large because the size of molecules is extremely small. We cannot see individual molecules with a naked eye. Only special equipment, different in different cases, can help us to see individual molecules. And, being so tiny, molecules of different substances are different in size among themselves.
And not only in size. Since the molecules contain different, more elementary particles called atoms, the configuration of these atoms that form a molecule is different for different molecules. Thus, a molecule of water contains two atoms of hydrogen and one atom of oxygen that connects hydrogen atoms into some three-dimensional construction. A molecule of protein consists of many different types of atoms and its structure and size are quite different from the molecule of water.

States of Matter

The next topic we would like to address is the states of matter.
When the word matter is used in physics, it means any object or substance that occupies certain space and has certain mass, thus consisting of certain molecules interacting among themselves.

There are three major states of mattersolidliquid and gas.

When an object is solid or is in solid state, it means that it retains its shape and form regardless of surrounding environment, not intended to change its form. The molecules of this object are strongly connected to each other. Their movement relative to each other is rather restricted. This movement can be oscillating around some point, maintaining an orderly three-dimensional structure, for crystal (or crystalline type of) solids or just slow movement, changing their relative position, but not changing the overall form for amorphous (non-crystalline) type.
Examples of solid objects are ice (crystal), steel (crystal), plastic cup (amorphous).

When an object or substance is liquid or is in liquid state, it takes the shape of a vessel or reservoir it's in. The connection between the molecules in case of liquid is strong enough to hold the molecules together, but not strong enough to preserve the overall shape.
Examples of liquid substances are water, mercury ("quick silver"), oil.

When an object or substance is gas or is in gaseous state, it takes as much space as it is available. Connections between the molecules are weak and they fly in all directions in completely chaotic fashion. Examples of gases are air, helium, oxygen.

Some examples above represent objects or substances that contain only one type of molecules, like ice or mercury, or helium. Some other examples are objects or substances that contain more than one type of molecules mixed together, like steel, oil or air.

Nature of Heat

Now we are in position to talk about heat.
Heat is the energy of molecular motion inside any object or substance.
As we mentioned, molecules are in constant motion inside any object. The more intense this motion is - the more heat this object possesses. This implies that heat is mechanical energyof molecules inside the object or internal energy of the object or substance.

As we know, mechanical energy can be transferred from one object to another, like during the collision of two billiard balls. Similarly, mechanical energy of one molecule can be transferred to its neighbors, from them - to their neighbors etc. This is a process of dissipation of heat. All what's necessary for this is the relative proximity of the molecules. This is exactly the way how heat is transferred from one body to another, from flame to pot, from pot to water, from water to vegetables in it, making soup.

Since heat or internal energy of an object is related to motion of its molecules, and increased heat means faster movement of the molecules, and, as we see, different states of matter are related to the strength of connection between the molecules, we can expect that the state of an object (solid, liquid, gas) might change with increasing or decreasing its internal energy by supplying or taking away the source of energy.
Indeed, it's true. Heat the ice - it will transform into water. Heat the water - it will transform into vapor. Heat the steel - it will melt. Freeze the helium - it will transform into liquid helium. Freeze the mercury - it will solidify.

As we see, the same molecules can form objects in different states. It only depends on the amount of internal energy, that is amount of heat, the object possesses.

Even without transformation from one state to another, heat causes certain changes in the object visible without any special instruments. We all know that mercury thermometer is working based on the property of mercury to expand as the temperature is rising.
This is a general property of most of the objects - to change physical dimensions with increase or decrease of amount of heat (internal energy) carried by their molecules during their constant motion.
This property is the principle, on which measuring of the intensity of molecular movement is based.

Heat and Temperature

Now let's address the issue of measuring the heat, that is amount of internal energy inside any object.
The term temperature is related to average intensity of the molecular movement inside an object or a substance. So, when we say that the temperature of an object has increased or decreased, we mean that average intensity of the molecular movement in it has increased or decreased correspondingly.

Our obvious task now is to quantitatively evaluate the temperature, thus measuring the intensity of the molecular movement inside an object.

It would be great, if we knew kinetic energy of each molecule at each moment of time and average it up to get the temperature in the units of energy. Alas, it's impossible. We have to find some easier method, not necessarily 100% accurate, but sufficient for day-to-day practical purposes.

Convenient instrument for this is a classic thermometer, whose indications are directly related to a change in physical size of objects with change of intensity of the molecular motion inside them.

A simple thermometer consists of a small reservoir with mercury and thin tube coming from it, so the mercury level in the tube will go up with increase of intensity of the molecular motion of the mercury or down, when the intensity decreases.

If we want to measure the temperature of any object, we bring it in contact with our thermometer and, when the temperatures equalize, which might take some time, the level of mercury in a tube of a thermometer will correspond to intensity of the molecular movement inside the object.

All, which is left to establish is the scale and units of measure.
There are three major systems of measurement of temperature: Celsius, Fahrenheit and Kelvin. Celsius system is used everywhere, except United States and its territories. Fahrenheit system is used in United States and its territories. Kelvin system is used everywhere in scientific research and equations of Theoretical Physics.

The unit of measurement in each system is called a degreeand the temperature is written with an indication of the system as follows:
0°C, 20°C, -40°C for temperatures in Celsius system;
32°F, 68°F, -40°F for temperatures in Fahrenheit system;
273.15°K, 293.15°K, 233.15°K for temperatures in Kelvin system.
Above are the examples of three different temperatures in three different measurement systems, correspondingly.

The conversion formulas are:
X°F = 5(X−32)/9°C
X°C = (X+273.15)°K

In the Celsius system the temperature 0°C corresponds to the temperature of melting ice at the sea level on Earth. Temperature of 100°C is the temperature of boiling water at the sea level on Earth. This range is divided into 100 degrees making up a scale.
Degrees in the Fahrenheit system are also connected to some natural processes. 0°F is the temperature of freezing of some chemical solution, while 100°F is approximately the temperature of a human body.
Finally, in Kelvin system 0°K is so-called absolute zerotemperature - the temperature of outer space far from any source of energy. The unit of one degree in Kelvin system equals to that of one degree in Celsius.

Tuesday, April 2, 2019

Unizor - Physics4Teens - Energy - Potential Energy - Gravity



Notes to a video lecture on http://www.unizor.com

Energy of Gravity

An object of mass m is raised on certain height h above the surface of the Earth.
Analyze its potential energy, as a function of height h.
Do not assume that the force of gravity is constant, but rather use the Newton's Law of Universal Gravitation.
Assume that the radius of Earth is R and its mass is M.

Solution:

If x is the distance from the center of the Earth to an object, the force of gravity equals to
F(x) = G·M·m/
(where G is a gravitational constant approximately equaled to 6.674·10-11N·m2·kg−2).

Potential energy of an object at height h above the ground is the work performed by the force of gravity as an object falls down to the surface of the Earth from the distance R+h from the center to the distance R.

Since the force of gravity varies, we should use the calculus to calculate this work.
An infinitesimal increment of work dW(x), assuming the force F(x) acts at a distance dx, equals to
dW(x) = F(x)·dx

Integrating this by x from R+hto R, we get the full work and the potential energy of an object at height h above the ground:
Epot = W(h) = RR+hF(x)·dx =
= G·M·m·
[1/R − 1/(R+h)]

Obviously, when an object is at the ground level (h=0), its potential energy equals to zero and, as it moves higher (h is increasing), its potential energy grows to its maximum value of G·M·m/R, as the object moves farther and farther from the ground.

When the height of the object his significantly smaller than the radius of the Earth (h << R), our formula for potential energy can be approximated with a simpler expression:
Epot = W(h) =
= G·M·m·
[1/R − 1/(R+h)] =
= G·M·m·h/
[R·(R+h)] ≅
≅ G·M·m·h/R² = m·g·h
,
where g=G·M/ is the free fall acceleration at the ground level, approximately equaled to
g ≅ 9.81 m/sec²
and P = m·g is the weight of an object at the ground level, assumed constant for small heights h.
In this case the potential energy is simply a product of weight (force of gravity) and distance - a classical expression for work performed by a constant force:
Epot = W(h) ≅ P·h

Let's also analyze the kinetic energy of the object.
Firstly, we assume that our object starts its free falling from the height H above the ground with zero initial velocity.

Its potential energy at any height h we know from the above calculation:
Epot = G·M·m·[1/R − 1/(R+h)]
Its kinetic energy depends on its speed V:
Ekin = m·V²/2
The problem is, we don't know how the speed depends on the height h.

What we do know is the dependency of the force of gravity F on the height h:
F = G·M·m/(R+h)²
We also know that speed V is the first derivative of height hby time t and the acceleration is the second derivative:
V(t) = h'(t)
a(t) = h"(t)
Therefore, according to the Newton's Second Law,
F = m·a = G·M·m/(R+h)²
Now we have an expression for acceleration a:
a = G·M/(R+h)²
Let's underscore that acceleration is the second derivative of distance h, as a function of time t:
h"(t) = G·M/[R+h(t)]²

We will not attempt to solve this differential equation to get a function h(t), take its derivative to get the speed and substitute into a formula for kinetic energy.
Instead, we will apply a clever trick.
Let's multiply the last expression by 2h'(t):
2h'(t)h"(t) =
= 2G·M·h'(t)/
[R+h(t)]²
The reason for this operation is the following.
The left side of this equation is the derivative of the square of the speed:
d/dt{[h'(t)]²} = 2h'(t)·h"(t)
and on the right side of this equation we find another derivative:d/dt{2G·M/[R+h(t)]} =
= 2G·M·h'(t)/
[R+h(t)]²
Therefore, we have the equality of the derivatives:
d/dt{[h'(t)]²} =
d/dt{2G·M/[R+h(t)]}

If derivatives are equal, the functions differ by a constant:
[h'(t) = 2G·M/[R+h(t)+ C
The constant C can be determined, using the conditions at the beginning of motion at time t=0:
h(0) = H and h'(0) = 0
Therefore,
[h'(0) = 2G·M/[R+h(0)+ C
[0 = 2G·M/[R+H+ C
C = − 2G·M/[R+H]
Now we can write
[h'(t) =
= 2G·M/
[R+h(t)]−2G·M/[R+H]
From this we can express the kinetic energy
Ekin = m·V²(t)/2 =
= m·
[h'(t)]²(t)/2 =
= G·M·
{1/[R+h(t)]−1/[R+H]}

Since our task was to express the energy in terms of height above the ground, we can simply write it as
Ekin = G·M·[1/(R+h)−1/(R+H)]

Finally, let's find the full mechanical energy of an object falling from the sky.
Epot = G·M·m·[1/R−1/(R+h)]
Ekin = G·M·[1/(R+h)−1/(R+H)]
Efull = Epot + Ekin =
= G·M·
[1/R−1/(R+H)]
As we see, the full mechanical energy is constant, it does not depend on the height. As an object falls from the sky, its potential energy is decreasing, but kinetic energy increasing, and the sum of both types of energy remains constant.

Monday, February 25, 2019

Unizor - Physics4Teens - Energy - Potential Energy of a Spring



Notes to a video lecture on http://www.unizor.com

Energy of Spring

The energy discussed in the previous introductory lecture was an energy related to constant force acting on an object. In this lecture we will address a more difficult case of a variable force of a spring acting horizontally on an object according to the Hooke's Law.
Let's remind the Hooke's Law. It states that the force of a spring, fixed on one end, onto a point-object attached to its free end is proportional to the length the spring is extended or compressed from its neutral position.
The corresponding formula for the force F of a spring onto an object is
F = −k·L
where L is a displacement of the spring's free end and k is a coefficient of elasticity - a specific characteristic of "stiffness" of each spring.
The minus sign signifies that the direction of the force is opposite to a displacement. The frame of reference is associated with the neutral position of a spring and the X-axis is directed towards spring's stretching. Then, for positive displacement L(stretching) the force is directed towards neutral point (that is, negative), while for negative displacement (compression) the force also directed towards a neutral position, (that is, positive).

At this point we recommend to refresh the material in the "Work and Elasticity" lecture of the "Mechanical Work" chapter of the "Mechanics" part of this course.
As stated in that lecture, the total amount of work W to compress (or stretch) a spring with elasticity k by length Lequals
W = k·L²/2
regardless of how exactly we compress or stretch a spring.

Assume now that we have compressed a spring by the length L, performing work equaled to W=k·L²/2. At a spring's end we have attached an object of mass m. Then we let the spring go by itself with the object attached to its end.

Intuitively, we think that, at first, the spring should return back to its neutral position. During this time the object's speed will increase because the force of elasticity of a spring will push it.

But then, when a spring reaches its neutral position, the object, having certain speed, will continue its move by inertia, stretching the spring. During this time the force of elasticity of a spring will slow down the movement of the object until it stops at the end of the stretching process.

At that time the force of elasticity will pull the object back to a neutral position, increasing its speed.

At the neutral position object continues moving by inertia, compressing a string, until the force of elasticity stops its movement when the spring will be in the compressed position, as in the beginning of our process.

At this point the process repeats itself and can continue like this indefinitely.

Well, our intuition is correct, a spring with an object attached to it will oscillate indefinitely. Let's analyze quantitatively these oscillations.

Our first goal is to find the dependency of the object's position on time. This can be done using the Newton's Second Law, taking into account the initial position of the object at the end of a compressed spring with no initial speed.
Let S(t) be a function describing the position of the object at the end of a spring relatively to its neutral position with stretching being a positive direction. Then S(0)=−L and S'(0)=0. The acceleration is the second derivative of S(t), that is a=S"(t). The force of elasticity is proportional to S(t) as F=−k·S(t). Now the Newton's Second Law F=m·a looks like this differential equation:
−k·S(t) = m·S"(d)
with two initial conditions:
S(0) = −L
S'(0) = 0
This differential equation is fully solvable and its general solution is
x(t) = C1·cos(t·√k/m) +
+ C2·sin(t·√k/m)


Constants C1 and C2 can be determined using the initial conditions:
C1 = −L
C2 = 0
Therefore,
S(t) = −L·cos(t·√k/m)

This function describes simple sinusoidal oscillations with amplitude L and period
T=2π√m/k.

We can also calculate the speed of an object as the first derivative of distance:
V(t) = S'(t) = dS(t)/dt =
= L·√k/m·sin(t·√k/m)


During the first quarter of this time an object accelerates under the influence of the elasticity of a spring, moving from the extremely compressed position on a spring to its neutral position. During the next quarter of the period the spring's elasticity slows it down, while it moves to extremely stretched position of a spring. During the next quarter of the period elasticity of a spring accelerates it again to a string's neutral position. Finally, the elasticity slows the object down, as it moves to the initial most compressed position.

Let's calculate the work that spring performs during the first quarter of its period from complete compression to a neutral point by accelerating the object from initial speed V(0)=0to its maximum at a spring's neutral position V(T/4).
It can be done by two methods.

1. By integrating by displacement of a spring's end from value −L (full compression) to value 0 (neutral position)
W[−L,0] =
0−LF(S)·
dS =
0−L(−k·S)·
dS =
= −k·S²/2|0−L = k·L²/2


2. By integrating by time from t=0 to t=T/4=(π/2)·√m/k(calculations are more complex but lead to the same result)
W[0,T/4] =
0T/4F(S(t))·
dS(t) =
0T/4F(S(t))·(
dS(t)/dt)·dt =
0T/4F(S(t))·V(t)·
dt =
0T/4k·L·cos(t·√k/m
·L·√k/m·sin(t·√k/m
dt =
0T/4k·L²·cos(t·√k/m
·sin(t·√k/m
d(t·√k/m) =
0T/4k·L²·sin(t·√k/m
·
d(sin(t·√k/m)) =
= k·L²sin²((1/2)π√m/k·√k/m)/2

After cancellation of factors and taking into account that sin(π/2)=1 we obtain the same expression for work:
W[0,T/4] = k·L²/2

What we have proven now is that the same amount of work done to compress a spring (W=k·L²/2) is performed by a spring when it returns to its neutral position.

That's why we say that the spring accumulates potential energy, when it's compressed, and releases it, when it returns back to neutral position.

Now, let's see where this energy, released by a spring during its return to a neutral position, goes. Our assumption is that it is transformed into kinetic energy of the object attached to its free end. Let's check it out.
Ekin = m·V²(T/4)/2 =
= m·L²·(k/m)·sin²(π/2)/2 =
= k·L²/2

So, kinetic energy of the object, when a spring is in the neutral position, exactly equals to a spring's potential energy when it's fully compressed.
Compressing a spring results in transferring some external energy to its potential energy. Then, as the spring returns to its neutral position, it releases this potential energy into a kinetic energy of the object attached to its end. At the neutral position the potential energy of a spring is zero, but kinetic energy of the object equals to potential energy of a spring in a compressed position.

Our last step is to calculate full energy of the system (potential energy of a spring and kinetic energy of the object attached to it) at any moment of time t.
Epot(t) = k·S²(t)/2 =
= k·L²·cos²(t·√k/m)/2

Ekin = m·V²(t)/2 =
= m·L²·k/m·sin²(t·√k/m)/2

Efull = Epot + Ekin = k·L²/2
(since sin²(φ)+cos²(φ)=1)

As we see, the full energyremains constant, it does not depend on the attached object's mass, only on the properties of the spring - coefficient of elasticity k and its initial displacement from the neutral position.