Basic notes on Physics.
In the series: Note 3.

Subject: Some keypoints of Heat/Energy in relation to matter.

Date : 23 August, 2020
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.

This text is more or less on the level of Highschool Physics.

It describes Energy and heat transfer in relation with matter (solid, liquid, gas), and
touches upon some elementary subjects from Thermodynamics.


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.

1. Some basic concepts and theories.

1.1 The density, or Specific mass of materials:

With many sorts of materials, we can associate a measure of it's "density", or in other words,
it's "specific mass".

Take for example Iron. It weighs about 7.9 grams/cm3.
Do you notice that we say "so many grams per unit of volume"?
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/cm3". The unit
used here is in "Mass per Unit of Volume", like grams/cm3 or kg/m3.
So, what if you would have 5 cm3 of Iron, while you know that 1 cm3 has a mass of 7.9 grams?

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 cm3 * 7.9 grams/cm3 = 39.5 grams.
So, it's possible to define the "density" or "specific mass" ρ for some material as:

ρ = M
    (equation 1)

Where "M" is the mass (usually in g or Kg), and "V" is the volume (usually in cm3 or m3).

So, then it's also true that:

M = ρ * V

This was already demonstrated above: 5 cm3 * 7.9 grams/cm3 = 39.5 grams (for Iron),
where we had 5 cm3 of Volume, and ρ=7.9 grams/cm3

As another example:

Suppose substance "X" has a density of 5.0 grams/cm3.
Now suppose we have 10 cm3.

How much mass do we have here?

M = ρ * V = 10 * 5.0 = 50 grams (or 5.0 grams/cm3 * 10 cm3 = 50.0 grams).

It must be noted that ρ usually somewhat depends on the Temperature. Many substances expand with higher
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/m3 as the unit for the density ρ.
But, in many countries people still use g/cm3. How do you convert between those two?

-If we have x g/cm3, then what is that in kg/m3?
-If we have y kg/m3, then what is that in g/cm3?

Well: 1 kg = 1000 g, and 1 m3 = 1000000 cm3.
So, 1 kg/m3 = 1000 g / 1000000 cm3. So, the conversion factor is "1000" or "0.001".

Thus if Iron is 7.850 g/cm3, then it's equivalent to 7850 kg/m3.
This so, since 7850 kg/m3 = 7850000 g/1000000 cm3 = 7.850 g/cm3.

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

- of 1000 going from g/cm3 to in kg/m3 or
- use 1/1000 going from in kg/m3 to g/cm3

1.2 The Specific heat of materials:

Suppose we have a certain amount of some "substance" of some sort. And suppose we are able to transfer Energy
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 Tbefore to Tafter
of the amount "M" (mass) of that substance is:

Q = c*M*(Tafter - Tbefore) = c*M*ΔT     (equation 2)

where "c" is the "specific heat" of that specific substance.

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.


-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.

It must be noted that the Kelvin is the official SI unit to express temperatures.
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?


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

1.3 The Heat Capacitity of substances:

Different substances have different ways "to store" energy. For example, the atoms of a molecule
can stretch and rotate, but "the ways to do that" may be different, for different molecules.
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:

Q = C * ΔT     (equation 3)

Or you may write it as C = Q/ΔT. Thus the unit of "C" is Joules/Kelvin, J/K.
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


Q = c*M*ΔT


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

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

1.4 The "mole" in Chemistry and Physics:

A very interesting "entity" in Chemistry and Physics (but most prominently in Chemistry),
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 * 1023 "particles".
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"
in 12 gram pure 12C (Carbon).

It's remarkable, since the 12C atom, contains 6 protons and 6 neutrons, thus in total
there are 12 particles (protons/neutrons) in the nucleus of that atom.
They practically form the total mass of one 12C atom (neglecting electrons).

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 * 1023).

-If we have 12C, with 12 protons/neutrons in the nucleus of the atom,
which corresponds to 6.022 * 1023 atoms, then it is 12 grams.

-If we have H2O, with in total 18 protons/neutrons in the molecule,
which again corresponds to 6.022 * 1023 molecules, it is 18 grams.

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

To a high degree of accuracy (neglecting binding energies, electrons etc..), if you know
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 CO2? How many grams is one mole?
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.

It's important that isotopes of elements (like C, O, etc..) have a slightly different number
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:

N2 + 3 H2 -> 2 NH3

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

1 mole N2 + 3 mole H2 -> 2 mole NH3

Each term has a certain mass in grams. For example 1 mole N2 is (about) 28 grams.
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 N2. How many grams per term
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.

A further example:

Suppose we have the reaction:

2Mg + O2 -> 2 MgO

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

You may also read it as:

2 mole Mg + 1 mole O2 -> 2 mole MgO

Now, suppose 0.7 mole O2 is used in the reaction.
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=mc2. But for practical reasons, Δm is zero).

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.

1.5 The "idealized" Gas laws:

The "idealized" Gas laws are rather intuitive, meaning that they all sound rather logical.
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.

=> Boyle's law:

Suppose we have any sort of container, with some sort gas. This gas is completely confined, since the
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:

p1 * V1 = p2 * V2     (equation 4)

or, which is the same:

p * V = Constant

Figure 1: Illustrating Boyle's law:

The SI unit of pressure, is the Pascal (Pa). This corresponds to: 1 Pa = 1 Newton/m2.
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 m3 (SI), but many folks still stick to the cm3.

Example 1:

A barrel contains 1 m3 of gas. The current pressure is 500 Pa.
At a certain moment, the volume is increased to 2 m3.
What is the pressure p2 after increasing the Volume?


p1 * V1 = p2 * V2

500 * 1 = p2 * 2 => p2 = 250 Pa.

Example 2 :

A small barrel contains 1 cm3 of gas. The current pressure is 500 Pa.
At a certain moment, the volume is increased to 2 cm3.
What is the pressure p2 after increasing the Volume?
(note the cm3 this time).


p1 * V1 = p2 * V2

500 * 1 = p2 * 2 => p2 = 250 Pa.

(if you keep the units the same, like cm3, it works just as well).

=> Gay-Lussac's law:

Just like above, suppose we have any sort of container, with some sort gas. This gas is completely confined, since the
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:

= P2
    (equation 5)

Here P1 and V1 represents the pressure and temperature before
we add heat-energy to the gas. And P2 and V2 are the pressure and temperature
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:

= Constant


Suppose P1 = 500 Pa and T1 = 300 K.
We heat up the gas until the temperature is 900 K, while keeping the Volume constant.
What is P2?


500/300 = P2/900 => P2= 1500 Pa.

=> The General gas law:

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.

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
= 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 m3, and T is
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:

p1 * V1
= p2 * V2

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

1.6 Conservation of Energy:

If you would imagine "heat-energy" to be like a fluid, we know that heat flows from hot to cold.
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...".


Qlost = Qgained

So, if we have a system with mass M1, initial Temperature T1, and specific heat c1, and we have
another system mass M2, initial Temperature T2, and specific heat c2, and we mix them (or bring them
into close contact) then:

m1 * c1 * (Tfinal - T1) = m2 * c2 * (Tfinal - T2)     (equation 7)

After the mixing up, a final Temerature Tfinal is reached.


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



m1 * c1 * (Tfinal - T1) = m2 * c2 * (Tfinal - T2) =>

0.05 * c * (Tfinal - 30) = 0.1 * c * (60 - Tfinal)

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

0.05 * (Tfinal - 30) = 0.1 * (60 - Tfinal) =>

0.05 Tfinal - 1.5 = 6 - 0.1Tfinal =>

0.15Tfinal = 7.5 =>

Tfinal = 50 C.

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.

1.7 Heat-energy and phase transitions:

If you transfer a constant amount (in time) of heat-energy to some substance, the temperature will
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 Lmel, and Leva. Again, note that for every sort of substance, that this substance
has it's own Lmel and Leva parameters.

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

Q = M * Lmel     (equation 8)

A similar equation exists ofcourse for the evaporation process.

- 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.

1.8 Heat transfer:

I often said that "heat flows from hot to cold". That's an abstraction ofcourse.

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 combination of several mechanisms.

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 of the rate of heat flow of one place to another.

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 the heat transfer rate.

-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
= k*A*(T2-T1)
    (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.

That's it. Hope you liked it.