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## Area of Expertise - Quantum physics

Energy quantum is the name for the smallest amount of energy, e.g. the energy difference of an electron between two excited states. According to Max Planck's postulate, energy is $\mathrm{E.}$ of an energy quantum proportional to the frequency $\nu $ and it applies $\mathrm{E.}=H\nu $, whereby $H$ is Planck's quantum of action.

## Planck's quantum of action

That **Planck quantum of action** $ h $ is the ratio of energy ($ E $) and frequency ($ f $) of a photon or a particle, according to the formula $ E = h cdot f $.

It is the foundation of quantum theory and, since its discovery by Max Planck in 1899 & # 911 & # 93 and 1900 & # 912 & # 93 & # 913 & # 93, has linked properties that were ascribed either only to particles or only to waves in classical physics . This makes it the basis of the wave-particle dualism of modern physics.

Planck discovered that the effect of a physical process, i.e. & # 160h. the product of converted energy and time can only take on discrete values, namely integer multiples of $ h $, therefore *Effect quantum*.

In addition to the gravitational constant $ G $ and the speed of light $ c $, the quantum of action is one of the three fundamental natural constants of physics and the basis of the primordial (primordial = "of first order") natural system of units, the Planck units. & # 914 & # 93

The energy of a photon can be calculated as follows:

- E.
_{ph}= Energy of the photon - h = Planck's quantum of action (h = 6.626 070 040 10 −34 J s)
- f = frequency of the photon

## Table of contents

Quote: *The entropy S is a quantity for which non-physicists have a very good idea and with which they intuitively operate correctly.* *There is probably no other quantity where the physical term matches a slang term so well:* *Most slang statements in which the word warmth or Amount of heat occurs*

*remain physically correct if these words are replaced by the word “entropy”.*Source: F. HERRMANN SCRIPTS FOR THE EXPERIMENTAL PHYSICS DEPARTMENT OF DIDACTICS OF THE PHYSICS UNIVERSITY OF KARLSRUHE EDITION 2003

Understanding the **Entropy in physics** and chemistry is greatly facilitated by the work of Herrmann, Job and the Karlsruhe physics course. The following texts are available on the Internet and are highly recommended as an introduction:

Entropy in physics is explained very nicely in the following article: Handout Entropy Uni Koblenz

Above all, you should calculate the exercises listed there, then you have already understood a lot.

The ebook Georg Job, Regina Rüffler is also highly recommended: **Physical chemistry. Part 1: Fundamentals of material dynamics**. See http://www.job-stiftung.de/index.php?id=8,22,0,0,1,0

The theory of heat (thermodynamics) is explained very well in the following book:

Another attempt to explain entropy in physics clearly is provided by André Thess in his book **The entropy principle - thermodynamics for the dissatisfied, Oldenbourg-Verlag 2007, ISBN 978-3-486-58428-8** undertaken using the principle of adiabatic accessibility.

In physics, entropy is easy to understand if one looks at the transition of a substance into another physical state. Amazingly, there is only talk of latent heat, but rarely of entropy.

One considers z. B. a block of ice made of frozen water.

If you want to thaw this block of ice, you have to put energy into it at freezing point without the temperature increasing. One speaks of the heat of fusion. This energy only serves to dissolve the existing order of the solid water particles, to enable more disorder, the water becomes liquid, the disorder increases, the entropy increases.

The entropy at the phase transition solid = & gt liquid corresponds to the heat of fusion = the melting energy measured in joules, since the temperature does not change here.

This simple fact has also been used for the definition of the entropy unit:

The unit of entropy in physics is 1 joule / Kelvin.

**1 Joule / Kelvin = entropy with which one can melt 0.893 cm3 of water ice.** (Requirement: pressure p = normal pressure)

In the case of the phase transition, one can simply divide the energy by the temperature of the phase transition and obtain the entropy difference of the process in Joule / Kelvin.

**A very important diagram of the relationship between entropy and temperature** Unfortunately only in French.

A mixture of water from a warm and a cold part is considered.

For the colder or warmer part, the change in entropies dS (delta or difference S) is defined as follows:

The total change in entropy is obtained from the following equation:

Here, dS & gt 0, since T_{W.} & gt T_{K}. Note: The temperatures are in the denominator.

In this voluntary process, the change in entropy is dS> 0. If the temperature of the warm body reaches that of the cold body, an equilibrium is reached and the heat exchange comes to a standstill.

### Example 1: Mixture of the same substance, the same quantity, different temperature Edit

A styrofoam vessel is filled with 300 g of water at a temperature of 40 degrees Celsius. Then add 300 g of water at 10 degrees Celsius. As a good insulator, the styrofoam vessel absorbs only a minimal amount of heat. Then stir a little and measure the mixing temperature 24.8 degrees Celsius. As an estimate, a mixed temperature of 25 degrees should come out.

How can you calculate the mixing temperature?

The following rules apply:

- The energy output E
_{W.}the warmer liquid is equal to the energy consumption E_{K}the colder liquid.

- The mass of the warmer liquid is equal to the mass of the colder liquid.

- The specific heat capacity of water is:

The temperatures are designated as follows:

Since E_{W.} = E_{K} applies, one can simplify:

From this equation one abbreviates c_{water}. Divide by c on both sides_{water}.

Since in the above example m_{W.} = m_{K} applies, you can also shorten m. The following formula remains:

Now insert the temperatures and calculate the mixed temperature:

Now one calculates the equation according to t_{mix} the end. Add t on both sides_{mix}.

10 ° C. is added on both sides.

You divide both sides by 2.

What does all of this have to do with entropy?

You can now also calculate the entropy from the example above:

However, the temperature must be used in Kelvin:

The energy intake and output are the same:

From this one can calculate the change in entropy:

The interesting thing about the whole bill is the fact that

- the energy that was given off by the warmer water to the colder water is identical to the energy that the cold water has absorbed.
- but the entropy of the mixed water has become greater than the sum of the individual entropies before the mixture.

There are three basic quantities in thermodynamics:

The quantities find the following analogy in electricity theory:

thermodynamics | Electricity |
---|---|

temperature | electrical potential |

Temperature difference | tension |

entropy | electric charge |

energy | energy |

In contrast to the electrical charge (positive and negative values), entropy can only assume positive values.

With these 3 quantities one can formulate easily understandable sentences about the entropy: (source handout energy and entropy)

- Every body contains entropy when its temperature is above absolute zero.
- Entropy can only have positive values.
- The greater the mass of an object, the more entropy it contains, provided the temperature remains the same.
- The hotter a body is, the more entropy it contains. The entropy is therefore temperature dependent.
- Entropy flows by itself from a place of higher temperature to a place of lower temperature, if an exchange between the places is possible.
- A temperature difference Delta T is the reason for an entropy flow
- The greater the temperature difference, the greater the entropy flow, assuming that the materials remain unchanged.
- The entropy flow I is the entropy that has flowed per unit of time.

**CAUTION: Do not confuse large T (temperature) with small t (time).**

### Analogy Thermodynamics and Electricity for Derived Quantities Edit

- Every material offers resistance to the entropy flow flowing through it.
- This resistance is greater, the smaller the exchange area A of the line and the greater the length l of the line.
- This resistance also depends on the thermally conductive material. This material constant is called rho
_{S.}. - The following applies to the entropy resistance:

- The entropy is the reciprocal of the resistance. It is calculated as

This is called Fourier's law.

- The increase in entropy of a body when the temperature rises is called the entropy capacity C.
_{S.}. It is calculated as

- The entropy capacity depends primarily on the temperature, but also on the mass, the volume, the pressure and the amount of substance.
- During phase transitions such as melting and evaporation, the entropy capacity changes very strongly.

- Entropy can be regenerated by
- friction
- a chemical reaction
- by electrical currents in electrical resistances
- by entropy flows in thermal resistances.

If the time factor is omitted from the formula on both sides, the result is:

- The energy flow corresponds to an output (energy / time).

- If you try to extract more and more energy from a body with different types of heat extraction, you can find two facts:
- You come very close to the temperature zero Kelvin (-273.15 degrees Celsius), but you cannot go below or even reach it. At this temperature there is an absolute temperature zero.
- At this temperature one can no longer withdraw any entropy. Absolutely cold bodies no longer have any entropy.

Contrary to popular belief, entropy is relatively easy to measure in physics.

A simple example shows the measurement of the heat of fusion or the heat of vaporization of various substances. Since the temperature is constant here, the determined energy can be equated directly to the entropy.

### Edit Peltier element

Entropy flows through a surface can be measured with a calibrated Peltier element.

### Task 0 edit

You have 1 liter of water at 20 degrees. To do this, pour another liter of water at 20 degrees. Question a: What is the temperature of the water afterwards? Question b: How big is the entropy after that?

Answer: The temperature stays the same. The entropy doubles.

### Task 1 edit

1. You have 1 kg of ice at zero degrees Celsius and normal ambient pressure. Calculate the increase in entropy when the ice has completely melted into 0 degree water. To do this, use the specific heat of fusion of water.

1a. You have 5 kg of water at a normal boiling point of 100 degrees Celsius and normal ambient pressure. Calculate the increase in entropy when the water has completely evaporated to water vapor with the temperature of the boiling point. To do this, use the specific heat of evaporation (enthalpy of evaporation) of water. (See de.wikipedia.org/wiki/Verdampfungswärme#Verdampfungsenthalpie)

1b. Why does energy have to be used in the process and why does the evaporation not take place on its own? When does the evaporation seem to take place on its own?

1c. Does the transition from water to water vapor lead to an increase or decrease in entropy?

1d. How can you explain that the water evaporates even at normal air pressure even at temperatures well below the boiling point z. B. when drying an item of laundry on the clothesline?

### Task 2 edit

2. You heat 1 kg of water from 20 degrees Celsius to 70 degrees Celsius. How much energy do you have to use? How big is the entropy increase?

### Task 3 edit

3. You mix 1 kg of water at 20 degrees Celsius and 1 kg of water at 50 degrees Celsius. How much energy is transferred from the warmer water to the colder water? What is the mixing temperature? How big is the entropy increase?

### Task 4 edit

10 kg of nitrogen are compressed isobarically to 1/10 of the initial volume.

- How is the pressure changing?
- How does the temperature change?
- How does the entropy change?
- How much mechanical energy does it have to use?

### Task 5 edit

The specific heat capacity of water Cw is equal to 4.19 J / (g * K) at p = 1 atm. Explain this value for 1 g of water and a temperature increase of 1 degree Kelvin.

### Task 6 Edit

What amount of water has the same amount of heat (entropy) at 50 degrees as 2000 g of water at 20 degrees Celsius?

### Task 7 Edit

The heating wire of a 1200 W hair dryer has a temperature of 1000 K.

### Task 8 edit

Calculate the entropy that is necessary to melt 5 kg of ice at zero degrees Celsius and p = 1 atm.

How can you check the calculated result with an immersion heater?

How does the value of the entropy, which is necessary for this, differ from the calculated value of the energy?

### Task 9 edit

Do you calculate the entropy that is necessary to heat 5 kg of water from zero degrees Celsius and p = 1 atm to 10 degrees Celsius and constant pressure?

How can you check the calculated result with an immersion heater?

### Task 10 edit

How does the entropy S in a piece of metal with the mass m = 0.001 kg change when the temperature rises from 1 degree Kelvin to 11 degrees Kelvin?

Assume that the specific heat capacity c of the metal is a constant 500 J / kg / Kelvin.

Use the following formula:

T2 is the temperature 11 Kelvin T1 is the temperature 1 Kelvin ln = logarithm to base e (natural logarithm)

### Task 11 Mixing gases

A 10 liter container is divided into two equal halves. In one half there is nitrogen at a pressure of 1 bar, and in the other half there is hydrogen at 1 bar. The temperature is 20 degrees Celsius.

- How does the entropy S and the free enthalpy G of the entire system change if a mixture is allowed?
- How long does it take until the two gases are evenly mixed?
- How big are ΔS and ΔG if the hydrogen is replaced by nitrogen?

### Exercise 12 Mixing 2 ideal gases

We mix two gases adiabatically with the same volume, same temperature, same number of particles. For example, 1 mole of helium and 1 mole of neon.

Use the Boltzmann constant k and Avogadro's number to calculate the entropy of mixing.

N 1 = N 2 = N A = A v o g a d r o s c h e Z a h l < displaystyle mathrm

= N_ <2> = N_ = Avogadrosche , number>> - DeltaS = k * NA * ln ((NA + NA) / NA) + k * NA * ln ((NA + NA) / NA)
- DeltaS = 2 * k * NA * ln 2
- DeltaS = 2 * 1.380 * 10 ^ -23 J / K * 6.02214179 * 10 ^ 23 1 / mol * ln2
- DeltaS = 2 * 1.4 * 6.02 * 0.69 J / K / mol
- DeltaS = 11.68 J / K / mol

### Task 13 Edit

You consider a gas with the following specifications:

Calculate the total entropy of the gas.

- Which formula do you want to use?
- Where is the result?
- What is the unit of the result?
- How can you interpret the result?

Every body contains entropy.

Entropy can only be transferred from one body to another together with heat.

The following applies: transferred heat / transferred entropy = temperature

Entropy can be produced but never destroyed.

In a closed system, the total entropy cannot decrease. In the course of the development of this system over time, it either remains constant or increases.

The following applies to closed systems:

### References edit

For theoretical considerations, crystals should actually have 0 entropy at absolute zero. However, this is seldom the case. Usually real crystals, once they have solidified, no longer manage to adopt the ideally possible crystal order with 0 entropy. The measurable real entropy of the crystals different from 0 at the absolute zero point of the temperature is called zero point entropy.

Every closed thermodynamic system in which certain exchange processes are freely possible assumes a state of equilibrium after the processes have subsided. This is determined by the maximum of the entropy of this system.

see also definition of entropy using the principle of adiabatic accessibility.

There are many examples of a local decrease in entropy, which supposedly contradict the 2nd law, but which can be conclusively explained by taking an overall view. Ordered structures emerge from previously disordered ones. At the same time, however, the overall entropy is still greater

### Examples edit

- Star formation through gravity
- Crystallization of substances on cooling
- Origin of life on earth
- Evolution of life on earth.

The idea of the Big Bang is also based on a very low-entropy initial state. Somehow this low-entropy state must have been reached.

L. Boltzmann recognized that the entropy law is synonymous with the following probability statement: The kinetic energy distributed to the individual molecules of a body always changes from a less probable distribution state to a more probable one, but not vice versa. Are z. If, for example, all air molecules are initially in a corner of a room, they are distributed evenly in this room: the entropy increases. However, it is practically impossible that, conversely, the evenly distributed molecules all collect in one corner of the room.

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