Physical constants. Basic laws and formulas, examples of problem solving What is angelic numerology
Each of us looks at the clock and often observes the coincidence of numbers on the dial. The meaning of such coincidences can be explained using numerology.
Thanks to numerology, it is possible to find out the main character traits of a person, his destiny and inclinations. Using a certain combination of numbers, you can even attract wealth, love and good luck. So what do these coincidences on the clock mean, and are they random?
Meaning of matching numbers
Repeating numbers often carry a message of warning and warning to a person. They can promise great luck, which should not be missed, or warn that you should carefully look at the little things and work thoughtfully to avoid mistakes and blunders. Particular attention should be paid to combinations occurring on Tuesday and Thursday. These days are considered the most truthful in relation to prophetic dreams coming true, random coincidences and other mystical manifestations.
Units. These numbers warn that a person is too fixated on his own opinion and does not want to pay attention to other interpretations of affairs or events, which prevents him from grasping the whole picture of what is happening.
Deuces. These coincidences force you to pay attention to personal relationships, try to understand and accept the current situation and make compromises in order to maintain harmony in the couple.
Threes. If these numbers on the clock catch a person’s eye, he should think about his life, his goals and, perhaps, rethink his path to achieving success.
Fours. The combination of numbers draws attention to health and possible problems with it. Also, these numbers signal that it is time to change something in life and reconsider your values.
Fives. Seeing these numbers means being warned that you will soon need to be more attentive and calm. Risky and rash actions should be postponed.
Sixes. The combination of these numbers calls for responsibility and honesty, not so much with others, but with yourself.
Sevens. Numbers denoting success often appear on the path of a person who has chosen the right goal and will soon realize everything planned. These numbers also indicate a favorable time for self-knowledge and identifying oneself with the world around us.
Eights. The numbers warn that in important matters an urgent decision must be made, otherwise success will pass by.
Nines. If the clock constantly shows you this combination, it means that you need to make efforts to eliminate the unpleasant situation before it provokes the appearance of a black streak in your life.
The meaning of identical combinations
00:00 - these numbers are responsible for desire. What you wish for will come true soon if you do not pursue selfish goals and do not act to the detriment of the people around you.
01:01 - ones in combination with zeros mean good news from a person of the opposite sex who knows you.
01:10 - the business or task you started is unsuccessful. It requires revision or abandonment, otherwise you will fail.
01:11 - this combination promises good prospects in the planned business. Its implementation will bring you only positive emotions and material stability. These numbers also mean success in teamwork.
02:02 - twos and zeros promise you entertainment and invitations to entertainment events, including going to a restaurant or cafe on a date.
02:20 - this combination warns that you should reconsider your attitude towards loved ones, compromise and be softer in your criticism and judgment.
02:22 - an interesting and fascinating investigation awaits you, a mystery that, thanks to your efforts, will become clear.
03:03 - triplets promise new relationships, romantic connections and adventures with a person of the opposite sex.
03:30 - this combination means disappointment in the man for whom you feel sympathy. Be careful and do not trust him with your secrets and plans.
04:04 — fours call for considering the problem from a different angle: its successful solution requires an extraordinary approach.
04:40 - this position of the numbers on the clock warns that you need to rely only on your own strength: luck is not on your side, be vigilant.
04:44 - Be careful when communicating with senior management. Your correct behavior and informed decisions will protect you from production errors and dissatisfaction with your boss.
05:05 - fives in this combination warn of ill-wishers who are waiting for your mistake.
05:50 - these meanings promise trouble and possible pain when handling fire. Be careful to avoid burns.
05:55 — you will meet with a person who will help solve your problem. Listen carefully to his rational opinion.
06:06 - sixes in this combination promise a wonderful day and good luck in love.
07:07 - Sevens warn of possible troubles with law enforcement agencies.
08:08 - this combination promises a quick promotion, occupation of the desired position and recognition of you as an excellent specialist.
09:09 - Monitor your finances closely. There is a high probability of losing a large sum of money.
10:01 - this meaning warns of a quick acquaintance with people in power. If you need their support, you should be more vigilant.
10:10 - tens mean changes in life. Whether they are good or not depends on you and your behavior strategy.
11:11 — units indicate a bad habit or addiction that needs to be gotten rid of before problems and complications begin.
12:12 - these numbers promise harmonious love relationships, rapid developments of events and pleasant surprises from your other half.
12:21 - a pleasant meeting with an old acquaintance awaits you.
20:02 - your emotional background is unstable and requires adjustment. Quarrels with loved ones and relatives are possible.
20:20 - these meanings warn of an impending scandal in the family. You need to take steps to avoid this incident.
21:12 - this meaning promises quick good news about the arrival of a new family member.
21:21 — the repeated number 21 indicates an imminent meeting with a person who will offer you a serious personal relationship.
22:22 — a pleasant meeting and relaxed communication with friends and like-minded people awaits you.
23:23 - this combination warns of envious people and ill-wishers who have invaded your life. Reconsider your attitude towards new acquaintances and do not talk about your plans.
Physical constants are constants included in equations that describe the laws of nature and the properties of matter. Physical constants appear in theoretical models of observed phenomena in the form of universal coefficients in corresponding mathematical expressions.Physical constants are divided into two main groups - dimensional and dimensionless constants. The numerical values of dimensional constants depend on the choice of units of measurement. The numerical values of dimensionless constants do not depend on the systems of units and must be determined purely mathematically within the framework of a unified theory. Among the dimensional physical constants, constants should be distinguished that do not form dimensionless combinations with each other; their maximum number is equal to the number of basic units of measurement - these are the fundamental physical constants themselves: the speed of light, Planck’s constant, etc.). All other dimensional physical constants are reduced to combinations of dimensionless constants and fundamental dimensional constants.
Physical constants
| Constant | Designation | Meaning | Error, % |
| Speed of light in vacuum | c | 299792458(1,2) m X s -1 | 0,004 |
| Fine structure constant | a a -1 | 0,0072973506(60) 137,03604(11) | 0,82 0,82 |
| Elementary charge | e | 1.6021892(46) x10 -19 TO | 2,9 |
| Planck's constant | h ћ=h/2 p | 6.626176(36) x10 -34 J X With 1.0545887(57) x10 -34 J X With | 5,4 5,4 |
| Avogadro's constant | N A | 6.022045(31) x10 23 mole -1 | 5,1 |
| Electron rest mass | m e | 0.9109534(47) x 10 -30 kg 5.4858026(21) x10 -4 A. eat. | 5,1 0,38 |
| Ratio of electron charge to its mass | e/m e | 1.7588047(49) x10 -11 k/kg -1 | 2,8 |
| Muon rest mass | m m | 1.883566(11) x10 -28 kg 0,11342920(26) A. eat. | 5,6 2,3 |
| Proton rest mass | m p | 1.6726485(86) x10 -27 kg 1,007276470(11) A. eat. | 5,1 0,011 |
| Neutron rest mass | m n | 1.6749543(86) x10 -27 kg 1,008665012(37) A. eat. | 5,1 0,037 |
| Faraday's constant | F=N A e | 9.648456(27) x10 4 k/mol | 2,8 |
| Magnetic flux quantum | F 0=h/ 2e | 2.0678506(54) x10 -15 wb | 2,6 |
| Rydberg constant | R ? | 1.097373177(83) x10 -7 m -1 | 0,075 |
| Bohr radius | a 0=a/4p R ? | 0.52917706(44) x10 -10 m | 0,82 |
| Compton wavelength of the electron | l c=a 2 /2 R ? l c/ 135 p =a a 0 | 2.4263089(40) x10 -12 m 3.8615905(64) x10 -13 m | 1,6 1,6 |
| Nuclear magneton | m N =eћ/2m p | 5.050824(20) x10 -27 J X Tl -1 | 3,9 |
| Bohr magneton | m B =eћ/2m e | 9.274078(36) x10 -24 J X Tl -1 | 3,9 |
| Magnetic moment of an electron in Bohr magnetons | m e/m b | 1,0011596567(35) | 0,0035 |
| Magnetic moment of a proton in nuclear magnetons | m p /m N | 2,7928456(11) | 0,38 |
| Electron magnetic moment | m e | 9.284832(36) x10 -24 J X Tl -1 | 3,9 |
| Proton magnetic moment | m p | 1.4106171(55) x10 -26 J X Tl -1 | 3,9 |
| Magnetic moment of a proton in Bohr magnetons | m p/ m N | 1.521032209(16) x10 -3 | 0,011 |
| Gyromagnetic ratio for proton | g p | 2.6751987(75) x10 8 s -1 X Tl -1 | 2,8 |
| Universal gas constant | R | 8,314441(26) J/( K x mole) | 31 |
| Boltzmann's constant | k=R/N A | 1.380662(44) x10 -23 J/ TO | 32 |
| Stefan–Boltzmann constant | s = (p 2 /60) k 4 / ћ 3 s 2 | 5.67032(71) x10 -8 W X m -2?K -4 | 125 |
| Gravitational constant | G | 6.6720(41) x10 -11 N X m 2 / kg 2 | 615 |
Example 18.
A small positively charged ball of mass m = 90 mg is suspended in the air on a silk thread. If an equal but negative charge is placed below the ball at a distance r = 1 cm from it, then the tension in the thread will increase threefold. Determine the charge of the ball. Solution. The suspended ball is initially acted upon by two forces: the force of gravity P, directed vertically downward, and the tension force of the thread T 1, directed upward along the thread. The ball is in equilibrium and, therefore,After a negative charge has been brought to the ball from below, in addition to the force of gravity P, it is acted upon by a force F k, directed downward and determined by Coulomb’s law (Fig. 4). In this case, the tension force Taking into account equality (1), we write
Expressing in (2) Fк according to Coulomb’s law to the force of gravity P through the body mass m and the acceleration of free fall g we obtain
Let's check the units of the right and left sides of the calculation formula (3):
Let's write down the numerical values in SI: m = 9·10 -5 kg; r = 10 -2 m; eε = 1; g = 9.81 m/s 2 ; ε 0 = 8.85·10 -12 F/m. Let's calculate the required charge:
Example 19.
Two positive charges Q = 5 nC and Q 2 = 3 nC are located at a distance of d = 20 cm from each other. Where should the third negative charge Q be placed for it to be in equilibrium? Solution.The charge Q 3 is acted upon by two forces: F 1, directed towards the charge Q 1 and F 2, directed towards the charge Q 2. The charge Q 3 will be in equilibrium if the resultant of these forces is zero:
that is, the forces F 1 and F 2 must be equal in magnitude and directed in opposite directions. The forces will be opposite in direction only if the charge Q 3 is located at a point on a straight line connecting the charges Q 1 and Q 2 (Fig. 5). For equality of forces, it is necessary that the charge Q 3 be closer to the smaller charge Q 2. Since the force vectors F 1 and F 2 are directed along the same straight line, vector equality (1) can be replaced by a scalar one, omitting the minus sign:
Having expressed the forces F 1 and F 2 according to Coulomb’s law, we write (2) in the form
Taking the square root of both sides of the equality, we find
Let us write down the numerical values of the magnitudes included in (3) in SI: Q 1 = 5·10 -9 C; Q 2 = 3·10 -9 C; d = 0.2 m. Calculations:
Of the two values of the root r 1 = 11.3 cm and r 2 = -11.3 cm, we take the first, since the second does not satisfy the conditions of the problem. So, in order for the charge Q 3 to be in equilibrium, it must be placed on a straight line, connecting charges Q 1 and Q 2 at a distance r = 11.3 cm from charge Q 1 (Fig. 5).
Example 20.
At the vertices of an equilateral triangle with side a = 20 cm there are charges Q 1 = Q 2 = -10 nC and Q 3 = 20 nC. Determine the force acting on a charge Q = 1 nC located in the center of the triangle. Solution.Three forces act on the charge Q located in the center of the triangle: (Fig. 6). Since the charges Q 1 and Q 2 are equal and located at the same distances from the charge Q, then
where F 1 is the force acting on the charge Q from the side of the charge Q 1 ; F 2 is the force acting on the charge Q from the side of the charge Q 2 . The resultant of these forces
In addition to this force, charge Q experiences the action of force F 3 from charge Q 3 . We find the required force F acting on the charge Q as the resultant of the forces F´ and F 3:
Since F´ and F 3 are directed along the same straight line and in the same direction, this vector equality can be replaced by a scalar one: or, taking into account (2),
Expressing F 1 and F 2 here according to Coulomb’s law, we get
From Fig. 6 it follows that
Taking this into account, formula (3) will take the form:
Let's check the calculation formula (4):
Let's write down the numerical values of magnitude in SI: Q 1 = Q 2 = -1·10 -8 C; Q 3 = 2·10 -8 C; ε = 1; ε 0 = 8.85·10 -12 F/m; a = 0.2 m. Let's calculate the required force:
Note. In formula (4), charge modules are substituted, since their signs are taken into account when deriving this formula.
Example 21.
An electric field is created in a vacuum by two point charges Q 1 = 2 nC Q 2 = -3 nC. The distance between the charges is d = 20 cm. Determine: 1) the intensity and 2) the potential of the electric field at a point located at a distance of r 1 = 15 cm from the first and r 2 = 10 cm from the second charge (Fig. 7). Solution.According to the principle of superposition of electric fields, each charge creates a field regardless of the presence of other charges in space. Therefore, the strength E of the resulting electric field at the desired point can be found as the geometric sum of the strengths E 1 and E 2 of the fields created by each charge separately: 
. The electric field strengths created in vacuum by the first and second charges are equal, respectively:
Vector E is directed along a straight line connecting charge Q 1 and point A, away from charge Q 1, since it is positive; vector E 2 is directed along a straight line connecting charge Q 2 and point A, towards charge Q 2 since the charge is negative. The modulus of vector E is found using the cosine theorem:
where α is the angle between vectors E 1 and E 2. From a triangle with sides d, r 1 and r 2 we find
Substituting the expression E 1 from (1), E 2 from (2) into (3), we obtain
Let's write down the numerical values of magnitude in SI: Q 1 = 2·10 -9 C; Q 2 = -3·10 -9 C; d = 0.2 m; r 1 = 0.15 m; r 2 = 0.1 m; ε = 1; ε 0 = 8.85·10 -12 F/m; Let's calculate the value of cosα using (4):
Let's calculate the required tension:
Note. Charge moduli are substituted into formula (5), since their signs were taken into account when deriving this formula.
2. The potential at point A of the field is equal to the algebraic sum of the potentials created at this point by charges Q 1 and Q 2:
Let's calculate the required potential:
Example 22.
What is the speed of revolution of an electron around a proton in a hydrogen atom if the orbit of the electrode is considered circular with a radius r = 0.53 10 -8 cm Solution.When an electron rotates in a circular orbit, the centripetal force is the force of electrical attraction between the electron and the proton, i.e. the equality is true
Centripetal force is determined by the formula
where m is the mass of an electron moving in a circle; u is the speed of electron circulation; r is the radius of the orbit. The force F to the interaction of charges according to Coulomb’s law will be expressed by the formula
where Q 1 and Q 1 are the absolute values of the charges; ε - relative dielectric constant; ε 0 - electrical constant. Substituting into (l) the expressions F cs from (2) and F to from (3), and also taking into account that the charge of the proton and electron, denoted by the letter e, is the same, we obtain
Let's write down the numerical values of magnitude in SI:
e = 1.6·10 -19 C;
ε 0 = 8.85·10 -12 F/m;
r = 0.53·10 -10 m;
m = 9.1·10 -31 kg.
Let's calculate the required speed:
Example 23.
The potential φ at a field point located at a distance r = 10 cm from a certain charge Q is equal to 300 V. Determine the charge and field strength at this point. Solution.The potential of a point in the field created by a point charge is determined by the formula
where ε 0 is the electrical constant; ε - dielectric constant. From formula (1) we express Q:
For any point in the field of a point charge, the equality is true
From this equality the field strength can be found. Let's write down the numerical values of greatness, expressing them in SI:
ε 0 = 8.85·10 -12 F/m.
Let's substitute the numerical values into (2) and (3):
Example 24.
An electron, whose initial velocity u 0 = 2 Mm/s, flew into a uniform electric field with a strength E = 10 kV/m so that the initial velocity vector is perpendicular to the strength lines. Determine the speed of the electron after time t = 1 ns. Solution.An electron in an electric field is acted upon by a force
where e is the electron charge. The direction of this force is opposite to the direction of the field lines. In this case, the force is directed perpendicular to the speed u 0. It imparts acceleration to the electron
where m is the electron mass.
where u 1 is the speed that the electron receives under the influence of field forces. We find the speed u 1 using the formula
Since the speeds u 0 and u 1 are mutually perpendicular, the resulting speed
Substituting the speed expression according to (3) into (4) and taking into account (1) and (2) we obtain:
Let us write down the numerical values of the quantities included in (5) in SI:
e = 1.6·10 -19 C;
m = 9.11·10 -31 kg;
t = 105·10 -9 s;
u 0 = 2·10 6 m/s;
E = 10·10 4 V/m.
Let's calculate the required speed:
Example 25.
At point M in the field of a point charge Q = 40 nC there is a charge Q 1 = 1 nC. Under the influence of field forces, the charge will move to point N, located twice as far from charge Q as NM. In this case, work A = 0.1 μJ is performed. How far will the charge Q 1 move? Solution.The work of field forces to move a charge is expressed by the formula
where Q 1 is a moving charge; Φ M – potential of point M of the field; Φ N – potential of point N of the field. Since the field is created by a point charge Q, the potentials of the starting and ending points of the path will be expressed by the formulas:
where r M and r N are the distance from charge Q to points M and N. Substituting expressions for φ M and φ N from (2) and (3) into (1), we obtain
According to the conditions of the problem, r N = 2r M. Taking this into account, we obtain r N - r M = r M. Then
Let's write down the numerical values of the quantities in SI:
Q 1 = 1·10 -9 C;
Q = 4·10 -8 C;
A = 1·10 -7 J;
ε 0 = 8.85·10 -12 F/m.
Let's calculate the required distance:
Example 26.
The electron passed through an accelerating potential difference U = 800 V. Determine the speed acquired by the electron. Solution.According to the law of conservation of energy, the kinetic energy T acquired by the charge (electron) is equal to the work A performed by the electric field when the electron moves:
Work done by electric field forces when moving an electron
where e is the electron charge. Electron kinetic energy
where m is the mass of the electron; u is its speed. Substituting expressions T and A from (2) and (3) into (1), we obtain 
, where
Let's write down the numerical values of the quantities included in (4), in SI: U=800 V; e = 1.6·10 -19 C; m = 9.11·10 -31 kg. Let's calculate the required speed:
Example 27.
A flat capacitor, the distance between the plates of which d 1 = 3 cm, is charged to a potential difference U 1 = 300 V and is disconnected from the source. What will be the voltage on the plates of the capacitor if its plates are moved apart to a distance d 2 = 6 cm? Solution.Before the plates move apart, the capacitance of a parallel-plate capacitor
where ε is the dielectric constant of the substance filling the space between the capacitor plates; ε 0 - electrical constant; S is the area of the capacitor plates. Capacitor plate voltage
where Q is the charge of the capacitor. Substituting into (2) the expression for the capacitance of the capacitor from (1), we find
Similarly, we obtain the voltage between the plates after they move apart:
In expressions (3) and (4), the charge Q is the same, since the capacitor is disconnected from the voltage source and no charge loss occurs. Dividing term by term (3) by (4) and making reductions, we get 
where
Let's write down the numerical values in SI: U 1 = 300 V; d 1 = 0.03 m; d 2 = 0.06 m. Let's calculate
Example 28.
A flat capacitor with an area of plates S = 50 cm 2 and a distance between them d = 2 mm is charged to a potential difference U = 100 V. The dielectric is porcelain. Determine the field energy and volumetric field energy density of the capacitor. Solution.The energy of a capacitor can be determined by the formula
According to the Stefan-Boltzmann law energetic luminosity(emissivity) absolutely black body is proportional T4:
Re T4,
On the other hand, this energy emitted per unit time per unit surface of a black body:
R e W S t.
Then the energy emitted during time t:
W Rе S t T4 S t . Let's do the calculations:
W 5.67 108 2.0736 1012 8 104 60 5643.5 5.64(kJ).
Answer: W 5.64 kJ.
In the radiation of a completely black body, the surface area of which is 25 cm2, the maximum energy occurs at a wavelength of 600 nm. How much energy is emitted from 1 cm2 of this body in 1 s?
m 600 nm | 600 10 9 m | The wavelength corresponding to the maximum energy |
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t 1 s | radiation, is inversely proportional to temperature |
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S 1cm2 | 10 4 m | re T (Wien displacement law): |
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R e = ? | ||||||
where b 2.9 103 m K is the first Wien constant, T is the absolute temperature.
Tb,
Energy emitted2 per unit time from unit surface –
energetic luminosity R e according to the law Stefan-Boltzmann:
Re T4,
where 5.67 10 8 W/(m2 K-4) is the Stefan-Boltzmann constant. Substituting (1) into (2) we obtain in the SI system (W/m2):
We need to go outside the system. Then we take into account that 1m = 100 cm, and 1m2 = 104 cm2, i.e. 1cm2 = 10-4 m2. We get energetic luminosity outside the system:
Let's substitute the numerical values: | ||||||||||||||||||||
R e 5.67 10 | 3094 (V |
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t/cm2 ).
Answer: R e = 3094 W/cm2.
Note: The surface area of 25 cm2 is given in order to confuse the student, in other words, to test the strength of the student’s knowledge of the theory.
Taking the thermal emissivity coefficient a t of coal at temperature
T 600 K equal to 0.8, determine:
1) energetic luminosity R e from coal;
2) energy W emitted from a coal surface with an area of 5 cm2 in a time of 10 minutes.
a T 0.8 | 1. According to the Stefan-Boltzmann law, the energy |
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T 600K | 5·10-4 m2 | tic luminosity (emissivity) greybody |
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S 5cm2 | proportional to T 4: |
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t 10 min | R ec a TR ea TT 4, |
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where 5.67 10 8 W/(m2 K4) is Stefan’s constant |
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1) R e s ? | ||||
2)W? | Boltzmann. |
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Let's do the calculations: |
R e s 0.8 5.67 108 1296 108 5879 5.88(kW/m2).
2. For equilibrium gray body radiation, radiation flux (power):
Фe Re with S,
where S is the surface area of the body. Energy emitted over time:
W e t. Then:
W R e s S t . Let's do the calculations:
W 5879 5 104 600 1764 1.76(kJ). Answer: 1. R e s 5.88 kW/m 2 ;
The muffle furnace consumes power P 1 kW. The temperature T of its inner surface with an open hole of area S 25 cm2 is equal to 1.2 kK. Assuming that the furnace opening radiates as a black body, determine how much of the power is dissipated by the walls.
Energetic luminosity (emissivity) R e black body - energy
The energy emitted per unit time per unit surface area of an absolutely black body is proportional to the fourth power of the absolute temperature of the body
T 4, expressed by the Stefan-Boltzmann law:
Re T4,
where 5.67 10 8 W/(m2 K4) is the Stefan-Boltzmann constant. From here:
P isS T 4 .
Part of the dissipated power is the difference between the power consumed by the furnace and the radiation power:
P S T 4, | |||||||||||
Pac | S T 4 | ||||||||||
8 1,24 1012 25 10 | |||||||||||
1 294 10 3 |
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We can conventionally assume that the Earth radiates as a gray body located at a temperature T 280K. Determine the thermal emissivity coefficient t
Earth, if the energy luminosity R e from its surface is 325
kJ/(m2 h). | |||||
T 280K | The earth radiates like a gray body. |
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R e s 325 kJ/(m2 h) | 90.278J/(m2 s) | Thermal coefficient | radiation |
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(degree of blackness) of the gray body is from- |
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and t - ? | |||||
carrying energy | luminosity |
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gray body to the energetic luminosity of black body, and is found by the formula:
a R e s.
T R e
Stefan-Boltzmann law for absolute black bodies, as if the Earth were a completely black body:
Re T4,
where 5.67 10 8 W/(m2 K4) is the Stefan-Boltzmann constant. Let's substitute into the thermal emissivity coefficient:
aT | Re s | |||||||||||
T 4 | 5,67 10 8 | |||||||||||
Answer: a T | 0,259 . | |||||||||||
Power | P radiation from a ball with radius R 10 cm at a certain constant |
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at a given temperature T is equal to 1 kW. Find this temperature, considering the ball gray |
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body with emissivity coefficient a T 0.25. |
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P 1 kW | Gray radiation power (flux) body is a product |
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R 10cm | energy luminosity of the ball per surface area S: |
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P Ф Rс S. | ||||||||||||
Surface area S of the ball: |
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4 R 2 . | ||||||||||||
Energetic luminosity (emissivity) R e from the gray body expresses |
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is governed by the Stefan-Boltzmann law: |
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Re s | aT T4, | |||||||||||
where 5.67 10 8 W/(m2 K4) is the Stefan-Boltzmann constant. Then the radiation power:
P aT T4 4 R2 .
Taking into account all formulas, body surface temperature:
4 aT R2 | ||||||||
4 0,25 5,67 10 8 | 3,14 10 2 | |||||||
The answer is T 866K.
The temperature of the tungsten filament in a twenty-five-watt electric lamp is 2450K, and its radiation is 30% of the radiation of a black body at the same surface temperature. Find the surface area S of the filament.
T 2450 K | The power consumed by the filament goes to radiation with a flat |
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P 25 W | S is considered to be a gray body, i.e. the radiation flux is determined by |
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a T 0.3 | ||
P = Fe = Re S. |
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Energetic luminosity(emissivity) gray te- |
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la according to the Stefan–Boltzmann law:
R e =a T σT4,
where 5.67 10 8 W/(m2 K4) is the Stefan-Boltzmann constant, T is the absolute temperature.
Then power consumption:
R a T 4 S. | ||||
Radiation area from here: | ||||
aT T4 | ||||
Let's substitute the numerical values: | ||||
0,41 10 4 | m2 = 0.41 cm2. |
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0,3 5,67 10 8 24504 |
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Answer: S = 0.41 cm2.
The maximum spectral density of energetic luminosity (r, T)max of the bright star Arcturus occurs at a wavelength of 580 nm. Assuming that the star emits as a black body, determine the temperature T of the star's surface.
m 580 nm | 580·10-9 m | The temperature of the radiating surface can |
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be determined from Wien's displacement law: |
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where b 2.9 10 3 m·K is the first Wien constant. Let us express the temperature T from here:
Tb.
Let's calculate the resulting value:
2,9 10 3 | 5000K 5(kK). | ||||||||||||
580 10 9 | |||||||||||||
Answer: T 5 kK. | |||||||||||||
Due to changes in black body temperature, the maximum spectral |
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emissivity densities (r, T)max | shifted from 1 to 2.4 µm | on 2 | 0.8 µm. |
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How and by how many times the energetic luminosity changed | R e body and maxi- |
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minimum spectral density of energy luminosity (r, T)max? | |||||||||||||
2.4µm | 2.4·10-9 m | Energetic luminosity | |||||||||||
0.8·10-9 m | activity) R e of a black body - energy emitted |
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0.8µm | |||||||||||||
per unit time unit surface area absolute |
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Re 2 | |||||||||||||
lute of the black body, proportional to the fourth |
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Re 1 | |||||||||||||
(r ,T ) max 2 | degree of absolute body temperature | T 4, you- |
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is expressed by the Stefan-Boltzmann law: | |||||||||||||
(r ,T ) max1 | |||||||||||||
Re T4, | |||||||||||||
where 5.67 10 8 W/(m2 K4) is the Stefan-Boltzmann constant.
The temperature of the radiating surface can be determined from Wien's displacement law:
m T b ,
where b 2.9 10 3 m·K is the first Wien constant. Expressing the temperature T from here:
and substituting it into formula (1), we get:
and b | are constants, then the energetic luminosity | R e depends |
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only from | Then the energetic luminosity will increase by: | ||||||||||||||||
Re 2 | 2.4nm | ||||||||||||||||
Re 1 | |||||||||||||||||
0.8nm | |||||||||||||||||
2) Maximum spectral density of energy luminosity is proportional to the fifth power of Kelvin temperature and is expressed by the formula Wien's 2nd law:
CT 5 | |||||
where coefficient C 1.3 10 5 W/(m3 ·K5) is the constant of Wien’s second law. Let us express the temperature T from Wien's displacement law:
Tb.
Substituting the resulting temperature expression into formula (3), we find:
(r ,T ) maxC | |||||
Since the spectral density is inversely proportional to the wavelength in
fifth degree | We find that change in density from the relation: |
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2.4nm | ||||||||||||||
(r ,T ) max1 | ||||||||||||||
0.8nm | ||||||||||||||
Answer: increased: by 81 times the energy luminosity R e and by 243 times the maximum spectral density of energy luminosity (r, T)max.
The radiation of the Sun is close in its spectral composition to the radiation of a completely black body, for which the maximum emissive speciality occurs at a wavelength of 0.48 μm. Find the mass lost by the Sun every second due to radiation.
m 0.48 µm | 0.48·10-6 m | Mass lost by the Sun at any time |
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t 1 s | we find from Einstein's law: W mc 2: | |||||||||||||||||||||||||
R C 6.95 108 m | ||||||||||||||||||||||||||
m c 2, | ||||||||||||||||||||||||||
where c is the speed of light. | ||||||||||||||||||||||||||
Energy emitted during time t (see derivation). | ||||||||||||||||||||||||||
task #2): | ||||||||||||||||||||||||||
W T 4 | S t , | |||||||||||||||||||||||||
where 5.67 10 8 W/(m2 K4) is the Stefan-Boltzmann constant. | ||||||||||||||||||||||||||
Taking into account the fact that the surface area of the Sun as a sphere | S 4 R2 | |||||||||||||||||||||||||
temperature T | according to Wien's displacement law formula (2) will take the form: |
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4 R C t, | ||||||||||||||||||||||||||
where b 2.9 10 3 m·K is the first Wien constant. | ||||||||||||||||||||||||||
Substituting (3) into (1) we get: | ||||||||||||||||||||||||||
4 R C t | ||||||||||||||||||||||||||
Mass lost by the Sun every second: | ||||||||||||||||||||||||||
4 R C | ||||||||||||||||||||||||||
Let's substitute the numerical values: | ||||||||||||||||||||||||||
2,9 10 3 | ||||||||||||||||||||||||||
10 8 | 4 6,95 108 | |||||||||||||||||||||||||
0,48 10 6 | ||||||||||||||||||||||||||
3 108 | ||||||||||||||||||||||||||
3441,62 108 | 6041,7 4 | 5.1 109 (kg/s). | ||||||||||||||||||||||||
9 1016 | 600 nm; 2) |
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energetic luminosity R e in the wavelength range from | 1,590 nm up to |
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2610 nm. Accept that the average spectral density of the body's energy luminosity in this interval is equal to the value found for the wavelength
T 2 kK | 1) . Spectral energy density |
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600·10-9 m | luminosity, according to Planck's formula: |
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590·10-9 m | 2 hс 2 | ||||||||||||||||||||
610 nm | |||||||||||||||||||||
1) (r, T)max? | where ħ = 1.05·10-34 | ||||||||||||||||||||
J s – Planck’s constant (from the drawing |
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2) R e? | |||||||||||||||||||||
that); c = 3·108 m/s – speed of light; k = 1.38·10-23 |
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J/K – Boltzmann constant. Let's substitute the numerical values: |
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6,63 10 34 3 108 | |||||||||||||||||||||
3,14 6,63 10 34 3 10 | 4.82 1015 e 12, |
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1,38 10 23 2 103 6 107 |
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6 10 7 5 | |||||||||||||||||||||
2.96 1010 W | 3 107 | ||||||||||||||||||||
m2 mm | |||||||||||||||||||||
m2 mm | |||||||||||||||||||||
2). Energy luminosity R e we find from the definition spectral |
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luminosity density r, T: | |||||||||||||||||||||
Re r, T d r, T d r, T (2 1 ) . | |||||||||||||||||||||
We took into account that the average spectral density of the body's energy luminosity r, T is a constant value and can be taken out of the sign of the integral. Let's substitute the numerical values:
P = ? | ||||
All supplied power will go towards the difference between the radiation of the tungsten filament and the absorption of heat (radiation) from the environment:
P = F e,isl – F e,abs.
We find the radiation (absorption) flux using the formula:
Fe = Re S,
where S =πd ·ℓ is the area of the lateral surface of the bottom
ti (cylinder). Then:
P = R e,islS – R e,absorb S = (R e,isl – R e,absorb) S ,
Energetic luminosity (emissivity) R e of the gray body - energy
energy emitted per unit time by a unit surface of a body is proportional to the fourth power of the absolute temperature of the body T4, expressed by the law Stefan-Boltzmann:
R e =a T ·σ ·T 4,
where σ is the Stefan-Boltzmann constant.
Let's substitute it and the area into the power input formula:
Р = (аТ σТ4 – аТ σТ4 okr) πdℓ= аТ σ(Т4 – Т4 okr) πdℓ , Let's substitute the numerical values:
P = 0.3 5.67 10-8 3.14 0.2 5 10-4 = 427.5 W. Answer: P = 427.5 W.
A black thin-walled metal cube with side a = 10 cm is filled with water at temperature T1 = 80°C. Determine the cooling time of the cube to a temperature T 2 = 30°C if it is placed inside a blackened vacuum chamber. The temperature of the chamber walls is maintained close to absolute zero.
Boltzmann's constant builds a bridge from the macrocosm to the microcosm, connecting temperature with the kinetic energy of molecules.
Ludwig Boltzmann is one of the creators of the molecular kinetic theory of gases, on which the modern picture of the relationship between the movement of atoms and molecules, on the one hand, and the macroscopic properties of matter, such as temperature and pressure, on the other, is based. In this picture, gas pressure is determined by the elastic impacts of gas molecules on the walls of the vessel, and temperature is determined by the speed of movement of the molecules (or rather, their kinetic energy). The faster the molecules move, the higher the temperature.
Boltzmann's constant makes it possible to directly relate the characteristics of the microworld with the characteristics of the macroworld - in particular, with thermometer readings. Here is the key formula that establishes this relationship:
1/2 mv 2 = kT
Where m And v— respectively, the mass and average speed of gas molecules, T is the gas temperature (on the absolute Kelvin scale), and k — Boltzmann's constant. This equation bridges the gap between the two worlds, linking the characteristics of the atomic level (on the left side) with volumetric properties(on the right side), which can be measured using human instruments, in this case thermometers. This connection is provided by the Boltzmann constant k, equal to 1.38 x 10 -23 J/K.
The branch of physics that studies the connections between the phenomena of the microworld and the macroworld is called statistical mechanics. There is hardly an equation or formula in this section that does not include Boltzmann's constant. One of these relationships was derived by the Austrian himself, and it is simply called Boltzmann equation:
S = k log p + b
Where S— entropy of the system ( cm. Second law of thermodynamics) p- so-called statistical weight(a very important element of the statistical approach), and b- another constant.
Throughout his life, Ludwig Boltzmann was literally ahead of his time, developing the foundations of the modern atomic theory of the structure of matter, entering into fierce disputes with the overwhelming conservative majority of the scientific community of his day, who considered atoms only a convention, convenient for calculations, but not objects of the real world. When his statistical approach did not meet with the slightest understanding even after the advent of the special theory of relativity, Boltzmann committed suicide in a moment of deep depression. Boltzmann's equation is carved on his tombstone.
Boltzmann, 1844-1906
Austrian physicist. Born in Vienna into the family of a civil servant. Studied at the University of Vienna on the same course with Josef Stefan ( cm. Stefan-Boltzmann law). Having defended his degree in 1866, he continued his scientific career, holding at different times professorships in the departments of physics and mathematics at the universities of Graz, Vienna, Munich and Leipzig. Being one of the main proponents of the reality of the existence of atoms, he made a number of outstanding theoretical discoveries that shed light on how phenomena at the atomic level affect the physical properties and behavior of matter.
