In the series: Note 3.

Version: 0.2

By: Albert van der Sel

Doc. Number: Note 3.

For who: for beginners.

Remark: Please refresh the page to see any updates.

Status: Ready.

It describes Energy and heat transfer in relation with matter (solid, liquid, gas), and

touches upon some elementary subjects from Thermodynamics.

Contents:

1. Some basic concepts and theories.

1.1 The density, or Specific mass of materials.

1.2 The Specific heat of materials.

1.3 The heat Capacity of materials.

1.4 The "mole" in Chemistry and Physics.

1.5 The "idealized" Gas laws.

1.6 Conservation of Energy.

1.7 Heat-energy and phase transitions.

1.8 Heat transfer rate Q/s.

it's "specific mass".

Take for example Iron. It weighs about 7.9 grams/cm

Do you notice that we say "

That's exactly what a "specific mass" actually is.

Instead of "weight", physicist use the term "mass" which is a more universal concept.

Also take note of the used units when we see something like "7.9 grams/cm

used here is in "Mass per Unit of Volume", like grams/cm

So, what if you would have 5 cm

The answer would be 5 * 7.9 = 39.5 grams (looks like Volume x Specific Mass).

Taking along the "units" in that calculation, the answer is 5 cm

So, it's possible to define the "density" or "specific mass" ρ for some material as:

ρ | = | M -- V |
(equation 1) |

Where "M" is the mass (usually in g or Kg), and "V" is the volume (usually in cm

So, then it's also true that:

M = ρ * V

This was already demonstrated above: 5 cm

where we had 5 cm

As another example:

Suppose substance "X" has a density of 5.0 grams/cm

Now suppose we have 10 cm

How much mass do we have here?

M = ρ * V = 10 * 5.0 = 50 grams (or 5.0 grams/cm

temperatures, and contract somewhat at lower temperatures. That's why tables exists listing

materials and specifying the ρ's per temperature.

Note on units:

In the "SI" system, people agreed to use kg/m

But, in many countries people still use g/cm

-If we have x g/cm

-If we have y kg/m

Well: 1 kg = 1000 g, and 1 m

So, 1 kg/m

Thus if Iron is 7.850 g/cm

This so, since 7850 kg/m

So, you can "safely" use a conversion factor of:

- of 1000 going from g/cm

- use 1/1000 going from in kg/m

to that amount of substance, then it's temperature may rise.

Formulated in a slightly better way: if we transfer "heat-energy" (which is Energy) to that substance,

it's temperature will rise.

It just depends on the sort of material, how much the temperature will go upwards.

Some factors are, whether that substance is a liquid, or a solid, and how many "internal" degrees

of freedom is present etc..

For now, we can leave out all such dependencies.

The amount of "heat-energy" Q needed, to change the temperature from T

of the amount "M" (mass) of that substance is:

You may read the formula as:

Heat-energy added = specific heat * Mass * Change in Temperature.

The specific heat "c" is the amount of heat-energy, per unit mass, required to raise the temperature

by one degree Celsius (or one degree Kelvin).

Q is in Joules "J", Mass is in "kg", Temperature is in degree Celcius "C" (or Kelvin "K").

If you look at the upper formula in SI units, then we have: J = (J/kg .C) * kg * C = J

Equation 2, is a linear relation.

-Suppose we have a fixed ΔT change, then Q and M are linearly related.

-Suppose we have a fixed Mass, then Q and ΔT are linearly related.

So, when you start out with "M" of some substance, which does not change (as it indeed does not),

and ofcourse "c" is simply a constant (depending on the sort of material), then "Q" is directly

and linearly related to ΔT.

So, in simple words, if you would have 2xMass as before, you need 2x the heat-energy as before,

for the same change in temperature.

Equation 2 is very understandable: The needed heat-energy "Q", is dependent on the amount of mass M, and change in Temperature ΔT.

Next, the "specific heat" c determines how fast or how slow some substance reacts to the transferred heat-energy.

That's why we have Q=c*M*ΔT.

Equation works in many ΔT ranges. However, when a socalled "phase transition" occurs, like if some substance

changes from solid to liquid, the formula cannot be applied in situations of a phase transition.

Notes:

-Equation 2 is not "extremely" exact, in the sense that it "works" for all possible Temperature changes.

For example, you know that a solid may melt, and a liquid may turn into gasseous form.

In that case, additional considerations are needed.

-Also, there are some subtleties with variations in "pressure" etc..

-The specific heat also varies slightly over ranges of temperatures.

However, in many highschool examples it is considered as a constant over "not too wild" temperature changes.

Suppose copper has a specific heat of 0.385 J/g.C. That means that we need 0.385 J to heat

up 1 gram of copper to one additional degree (Celcius C or Kelvin K).

So, if we want to express that in J/kg.C, we know that we need 385 Joules to heat up

1 kg of copper with 1 degree Celcius.

However, you may use Celcius as well, since the "rate of change", is the same.

For example, a ΔT=30, expressed in K or C, means exactly the same value.

Question: how much heat-energy is needed to heat up 0.5 kg of copper from 20 to 50 degrees Celcius?

Answer:

Q = c*M*ΔT = 385 * 0.5 * 30 = 5775 Joules of energy.

can stretch and rotate, but "

Also, the change in kinetic- and potential energies in general, is different for different molecules

if we would supply the same amount of heat-energy.

This lead to the concept of "Heat Capacity" C, which in general is different, for different substances.

Heat capacity ( C ), of a substance, is a measure for the heat-energy needed to raise

the temperature with 1 Kelvin.

Indeed, that would be different, for different substances.

In formula:

You may wonder: where is the "mass"? That's already accounted for with "Q", which is the

total Heat-energy.

Note that we have:

Q = C * ΔT

and

Q = c*M*ΔT

Thus:

C * ΔT = c*M*ΔT => C = c*M

Note the capital letter "C", which is different from the lower case "c", which is the

"specific heat" of a specific substance.

A piece of lead, which has a Temperature of 200 C, cools down to 20 C.

Suppose Lead has a Heat-capacity of 25 J/K.

How much heat-energy is dissipated to the environment?

Q = C * ΔT = 25 * 180 = 4500 J.

You might argue about the sign of my result. If ΔT = T

then Δ = -180. That would lead to -4500 J.

Some folks then solve the "dillemma" by saying that the piece of lead

that energy to the environment. So, the "direction" of the heat flow, then says what

sign we should us.

There is no absolute consensus here. But one thing is for sure: the Heat-energy involved

in the process, is 4500 J.

As a variant, if you have Q and ΔT, you can calculate "C" for a certain substance.

Note that many textbooks, or sites, indeed differentiate between the "specific heat" (small "c"),

and the "heat capacity" (capital "C"). The formula's involved are:

Q = c*M*ΔT

Q = C*ΔT

is the "mole".

We may also encounter it in discussions on Heat and heat transfer between substances.

A "mole" represents a number, and it is a large number: about 6.022 * 10

The number is also called "Avogadro's number A".

From the context where the mole is used, the term "particles" should get clear, like for example

atoms or molecules (or other particles, but atoms or molecules are the most prominent).

The "former" definition of the mole, stated, that it is the number of atoms that would "fit"

It's remarkable, since the

there are 12 particles (protons/neutrons) in the nucleus of that atom.

They practically form the total mass of one

This immediately leads to a surprising useful "mapping" between number of particles in the nucleus

of some substance, and how many grams one mole of that substance is. Remarkable, indeed.

Keep in mind that one mole, always represents the same number of particles (6.022 * 10

-If we have

which corresponds to 6.022 * 10

-If we have H

which again corresponds to 6.022 * 10

You see that the mole in grams of some material, is related to the Massnumber, that is the sum

of the protons+neutrons?

the number of protons+neutrons in a molecule, you can immediately say how many grams is in

one mole of that substance.

As another example: What about one mole of CO

Let's count the protons+neutrons: C=12, O=16 (we must mutltiply that by 2), which delivers

us a total of 44 protons+neutron, meaning one mole is (about) 44 grams.

of neutrons, so the real accurate number must be a weighted average of the isotopes.

that's why accurate values are listed in tables, databases etc...

As a simple rule: each atom, or molecule, contains X number of protons+neutrons, we know

that one mole of that substance is (about) X grams.

So, suppose we have a pure element "Y", with a "molar mass" of 95 grams, then we know

that an atom (if it's an element), or a molecule of that substance, has 95 protons+neutrons

in it's atomic nucleus (element), or combined nuclei (molecule).

Indeed, since the mass of the proton, is practically the same as the mass of the neutron,

it follows that a mole of a certain substance, is the same as the "mass number" of the atom,

or the total mass number of a molecule.

For an atom, the mass number is the sum of the protons and neutrons in the nucleus.

For a molecule, you simply count all the protons and neutrons of the constituent atoms.

(Not entirely correct if we consider the isotopes too of elements. Many elements have isotopes

with the same number of protons, but one of a few additional neutrons).

If we have some chemical reaction, like for example:

N

Then we may also say "A x" (Avogadro's number) every term of the equation, thus:

1 mole N

Each term has a certain mass in grams. For example 1 mole N

So, you can write everything (such a chemical equation) in grams too.

Just take the mole equivalent for that substance.

Suppose someone asks you: we use 50 grams of N

is needed or produced? Then you only need to multiply each number by 50/28.

The understanding of the mole can be a bit spicy indeed. However, I think it was wise

to place some basic info in this text as well.

Note: since many "elements" have a number of "isotopes" (which nucleus may have one or more

additional neutrons), it may happen that when you look up a mole of such element on the Internet

or some table, then that it is a number which is a weighted average of all isotopes of that element.

Suppose we have the reaction:

2Mg + O

You can see that equation in the perspective of atoms/molecules.

You may also read it as:

2 mole Mg + 1 mole O

Now, suppose 0.7 mole O

How much MgO is produced? It must be 0.7 * 2 mole MgO.

The ratio of moles is dictated by the equation of atoms/molecules.

Then if one number changes, the others change too according to the ratio.

If you know how much grams the substances are in moles, then you can do similar

calculations in grams. Here too, the ratio's are fixed. Besides that, the total mass

on the left of "->" is equal to the mass on the right of "->". Mass will not be created

or destroyed in Chemical reactions.

(besides extremely low changes in mass, due to changes

in energy in the reactions, since E=mc

We don't do Chemistry here, so no need to go into this on a deeper level.

Next, let's go to some basic Gas laws, which are pretty important in Highschool Physics.

That's great ofcourse.

Below we will see some mathematical relations between "p" (pressure), "V" (volume), and "T" (temperature),

which we will reckognize from daily life as well.

For example, suppose we have a airtight container, containing some gas. Also, some piston is mounted.

If we now reduce the Volume, we will not be very surprised that the pressure increases.

The idealized Gas laws, are pretty close to reality. However, some assumptions are in place.

I only mention a few of them here:

-We consider the Kinetic energy as leading in this closed system.

Any "Potential" energies between the particles, and environment, are disregarded.

-The collisions between the "particles" (like molecules) are elastic.

-We do not have any phase transitions, like going from gas to liquid.

container is fully airtight. We further assume that during our experiments, the Temperature remains constant.

Suppose that we are able to increase or decrease the Volume, then:

or, which is the same:

p * V = Constant

The SI unit of pressure, is the Pascal (Pa). This corresponds to: 1 Pa = 1 Newton/m

Thus it's really 1 Newton of force, excerted on 1 square meter.

It's rather tiny. Historically, other units are quite common too, like the "bar".

However, it's increasingly common to use the "hecto Pascal", that is 100 Pa, notated as hPa, which is 1 milliBar.

100 Pa = 1 hPa = 1 milli Bar = 0.001 Bar. 100000 Pa = 1 Bar.

But, in all sorts of Physics, using the SI units, is recommended (in most countries).

Ofcourse, you can always convert from one unit to the other, for example from Pa to Bar or the otherway around.

Volume is expresses in m

A barrel contains 1 m

At a certain moment, the volume is increased to 2 m

What is the pressure p

Answer:

p

500 * 1 = p

A small barrel contains 1 cm

At a certain moment, the volume is increased to 2 cm

What is the pressure p

(note the cm

Answer:

p

500 * 1 = p

(if you keep the units the same, like cm

container is fully airtight. We further assume that during our experiments, the Volume remains constant.

Suppose, at a certain moment, that we are able to add Heat-energy to the container, so that the

the temperature of gas will increase. Again, the Volume is kept the same as it was before.

Figure 2: Illustrating Gay-Lussac's law:

If we add heat to the container, the temperature will increase. It's only logical

that the pressure increases too. From a molecular model, the particles will increase

in speed and will collide harder and more often to the container's wall.

But it's rather remarkable that the quotient between P/T is constant. If T increases, then so

does P, both in the same amount. This, we can capture in the formula below:

P_{1}--- T _{1} |
= | P_{2}--- T _{2} |
(equation 5) |

Here P

we add heat-energy to the gas. And P

after heating the gas.

This time, we need to represent the temperature in Kelvin (the "absolute" temperature).

You may also read the equation above as:

P -- T |
= Constant |

Suppose P

We heat up the gas until the temperature is 900 K, while keeping the Volume constant.

What is P

Answer:

500/300 = P

was constant, or we had the constraint that the volume needed to be constant.

Ofcourse, we also have a "universal" or "general" gas law.

This time, we do not have those constraints.

If you look at the formula below, you certainly will reckognize components

from the former two laws.

p * V ------ T |
= Constant = n*R | (equation 6) |

However, the term "n*R" relates to the amount of gas in mole, and (after some re-arragements)

also to Boltzmann constant.

Indeed, in the former two paragraphs, we had either the constraint that the temperature

was constant, or we had the constraint that the volume needed to be constant.

This time, we have a general equation, which is even more close to "reality".

As usual, "p" is the pressure in Pa, "V" is the volume in m

the temperature in Kelvin.

In the formula, "n" is the amount of gas in mole, and "R" is a general gas constant,

where R = (about) 8.31 Joule/Kelvin . mole.

Using these values, you can perform actual calculations. However, I need to add

some additional information, which we will see in later sections.

But, calculations using:

p_{1} * V_{1}--------- T _{1} |
= | p_{2} * V_{2}--------- T _{2} |

are pretty straightforward, and resemble the calculations we have seen above.

This is ofcourse not an exact representation of reality. However, the "picture" works.

So, if you would for example, mix a hot gas with a colder gas, the hot gas looses heat-energy, while

the colder gas gains heat-energy. The mixed system, sort of "levels out".

From modern insights, we can see that the faster molecules of the hotter gas, collides with the slower

molecules of the colder gas, thereby transferring energy.

This will go on and on, until we have a "heat" equilibrium.

"What one system looses is gained by the other...".

or:

another system mass M2, initial Temperature T2, and specific heat c2, and we mix them (or bring them

into close contact) then:

Suppose we mix 2 glasses of water into one bowl.

Glass 1: 50 g water, T1=30C.

Glass 2: 100g water, T22=60C.

Calculate the final temperature T

We assume that it is a closed system, with no heat loss to the environment.

So:

m1 * c1 * (T

0.05 * c * (T

Since the specific heat "c" is the same on both sides, we can eliminate it:

0.05 * (T

0.05 T

0.15T

T

If it were two different liquids, then you simply needed to find the specific heats c1 and c2,

and use those in the equation above. It does not make any fundamental difference.

increase linearly, according to equation 2.

There is however, an exception. At a phase transition, things are a little bit different.

I will not bore you to death with the different phases a certain substance may in.

Some well-known phases are solids, liquids, and gasses.

(there are more "sorts" of phases, like a "plasma" at very high temperatures).

Different sorts of compounds have different phases at a certain temperature. Furthermore, you know

that a certain compound will turn into a solid if the temperature goes below a treshold, or may turn in gasseous state

if the temperature goes above another treshold (at a certain pressure).

Figure 3: Diagram Temperature change - supplied heat:

In the intervals, where the line is blue, you can certainly use equation 2, for various calculations.

However, at the interval where the phase transition takes place (e.g. from solid to liquid), we need

another calculation.

Why is that? At such transition interval, the Temperature does not increase, while we still are feeding in energy.

This might be viewed as rather strange at first.

No, it is not strange. For example, when a solid turns into liquid (or liquid into gas), the former potential energy

of all of the molecules (or atoms) need to be taken into account.

When the molecules were "free", they contained a certain potential energy among them. Once the got closer and closer,

they went into a "potential well", loosing that energy.

Now, if a solid goes into a liquid again, the supplied energy is mainly used to get out of the potential well again.

This is why, you still feed in energy, but the temperature does not increase, until the whole substance is liquid again.

For each sort of substance, this energy used for melting, or for evaporation, is different.

Like the "specific heat" we saw above, we have a specific heat-energy per utit of mass, needed for melting or evaporation.

Let's call those L

has it's own L

Then, for example, the energy needed in the melting process for a certain substance, is:

- When you need to do calculations, and the temperatures do not cross any phase transition, simply use equation 2.

- When you need to do calculations and the temperature crosses a phase transition, then you have two parts

in the calculation (or even more parts in the calculation), which you calculate independently, and add up,

to find the total heat-energy "Q". So, then use equation 2, and equation 8, and calculate the sum of those.

The "L's" per substance, typically need to be looked up in a sort of repository which collects such values.

In reality, three main mechanisms are:

- radiation (via electromagnetic waves, like infra red etc..)

- convection (often with fluids but also other substances)

- conduction (most often with solids)

However, the "kinematic model" still stands. Above we saw that faster molecules in a gas may collide with slower

molecules, transferring energy. But in a solid we can see something similar: more energetic atoms or molecules

in a solid, may occilate in a higher energetic state, transferring energy to neigboring molecules,

which on their turn do the same, and so on, and so on.

The latter case may be interpreted as "conduction". But it also falls in the "kinematic model".

But in case of some metals, free electrons may act als conducting electrons, transferring energy.

Except for pure radiation, both convection and conduction are related to the kinematic model.

But it also depends quite a bit on the sort of substance.

You know, in many real cases it will even be a

You can easily obtain more information about these mechanisms, if you like, using the internet,

articles, textbooks etc..

That's interresting information, for sure. However, I like to keep this note rather consise,

and a bit in line with Highschool Physics.

Here I like to show one important formula for the mechanism of conduction, and an example thereof.

We already did a few heat calculations, like mixing two glasses of water, of different temperatures.

However, this time, we like to see an example

A classic example is this one:

Suppose we have two adjecent, isolated, rooms, both with different temperatures. A window is part

of the wall connecting those rooms. You may think of a glass window, but in this example,

I like to use a flat, square, copper window, to illustrate conduction of heat.

Bye the way, copper conducts heat much better than concrete walls.

Say, the copper window has a surface area "A", and a thickness of "d".

Every solid material has a "thermal conductivity" parameter, often named "k" in the literature.

Again, for any material, the corresponding "k" needs to be looked up in a table, or database etc..

So, in our case, we need to find "k" for copper in some table.

Would the window be of Aluminium, then we would need the "k" of Aluminium.

We want to calculate the "heat transfer". Here we do not mean the total heat transferred over a whole period,

like one day, or a week etc.. We want it per time unit (which is the second).

So, we did a bit better, if we would have called it

-Note the term "rate", which here means "Q/second".

- But even if you would have, for example, a Total Q = 10000 Joules, over say, 150 seconds,

then we simply calculate Q / t = 10000 / 150 = 66.6 J/s. Now we have the "rate per second".

So, you can calculate the heat transfer rate with any Q and or t.

The general formula for heat rate transfer for conduction is:

rate = | Q -- t |
= | k*A*(T2-T1) ------------ d |
(equation 9) |

- (T2-T1) is the temperature difference between both rooms, thus also between

both sides of the window.

- A is the surface area of the copper window (or the area of other material).

- d is the thickness of the window.

- k is the thermal conductivity of copper (or the k of other material).

This formula is quite easy to grasp:

- The greater "A" is, the higher Q/t will be, and the other way around.

- The larger "k" is, the higher Q/t will be, and the other way around.

- The larger "d" is, the more difficult it becomes to transfer heat between both rooms.

- The larger (T2-T1) is, the higher Q/t will be, and the other way around.