introductory thermal physics - california state...

Introductory Thermal Physics

Gregory G. WoodDepartment of Applied Physics

California State University Channel Islands

December 4, 2005

Contents

Abstract v

1 Temperature 11.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2.1 Absolute Zero . . . . . . . . . . . . . . . . . . . . . . . 21.3 Ideal Gas Approximation . . . . . . . . . . . . . . . . . . . . 3

1.3.1 Definition a Mole . . . . . . . . . . . . . . . . . . . . . 41.3.2 Examples of Ideal Gas Law Calculations . . . . . . . . 51.3.3 Ideal Gas Law Problems . . . . . . . . . . . . . . . . . 61.3.4 An Improved Ideal Gas Law: Van der Waal’s Equation 6

1.4 Equipartition Theorem . . . . . . . . . . . . . . . . . . . . . . 71.5 Effect of Temperature on Solids and Liquids: Thermal Ex-

pansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.6 Thermal Volume Expansion Coefficient . . . . . . . . . . . . . 9

1.6.1 Thermal Area Expansion Coefficient . . . . . . . . . . 91.6.2 Thermal Expansion Examples . . . . . . . . . . . . . . 10

1.7 Physical Underpinnings of Thermal Expansion . . . . . . . . 111.8 Ideal Solid Equation . . . . . . . . . . . . . . . . . . . . . . . 111.9 Crystal Structures . . . . . . . . . . . . . . . . . . . . . . . . 12

2 Thermal Energy 152.0.1 Specific Heat . . . . . . . . . . . . . . . . . . . . . . . 152.0.2 Specific Heat Examples . . . . . . . . . . . . . . . . . 172.0.3 Latent Heat and Phase Transitions . . . . . . . . . . . 182.0.4 Latent Heat Examples . . . . . . . . . . . . . . . . . . 19

2.1 Conduction/Convection/Radiation: Methods of Heat Transfer 19

iii

iv CONTENTS

3 Thermodynamics 213.1 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . 213.2 2nd law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 223.3 Third Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Abstract

Thermal physics is an extension of classical mechanics to the microscopicenvironment where complete specification of the position and velocity ofeach particle is impractical (in the extreme) and has been discovered to beimpossible (by quantum physics). The level of the text is at the calculusbased freshman physics course level, similar to that of many texts by thetitle of “Physics for Scientists and Engineers”. The text follows the familiarpattern of discussion, example problems, and problems (to be solved by thestudent). More examples can be found in Explicit Physics Solutions by thesame author - available on the internet - for free at this time.

v

vi ABSTRACT

Chapter 1

Temperature

1.1 Introduction

From the macroscopic, the world of objects on the order of one meter in size,we turn our attention to the world of the atom, on the order of 10−10 metersin size. Upon examining the air in the room, although there is no wind,and thus no bulk motion of gas in any direction, we discover each atom ismoving - in fact moving very fast. The atoms do not share a single speed,nor a single direction. They collide with the walls and with each other andwith everything in the room.

Thus these atoms have kinetic energy, 12mv2, and although the mass is

quite small, the velocity is high and the energy in, say, a cubic meter of gasis considerable. Momentum is carried by each atom, but momentum is avector and thus if we sum the total momentum of the (still, at rest) air in aroom, we must find zero (or a number very close thereto). The air movingleft cancels the momentum of the air moving right, same for up and down,same for back and forth. To quantify the situation further, we need to definetemperature - despite the disparate velocities and speeds, the gas does havea single temperature, the one your thermometer would read.

1.2 Temperature

If the air in the room is heated, we find the speed of the atoms increases.Intuitively, we know to use a thermometer to quantify this phenomenon asan increase in temperature, but if we are going to use a temperature, weshould define exactly what it is. There are three different temperature scalesyou may have heard of: the Fahrenheit, Celsius (also called Centigrade), and

1

2 CHAPTER 1. TEMPERATURE

Kelvin scales. The names are listed in the order of historical discovery. Allare still used, and all are linearly related to one another by simple equationslisted below:

T (C) =59T (F )− 32 (1.1)

T (K) = T (C)− 273.15 (1.2)

Equipped with one thermometer calibrated according to each scale, wetake careful measurements. We find the average speed squared is propor-tional to temperature in Kelvin: double the temperature in Kelvin and theaverage speed squared doubles. The other thermometers yield no such clearcorrelation. The temperature in Kelvin is referred to as the absolute tem-perature and this we will define.

Note that room temperature, about 750F or about 250C is about 300K.Examining the atoms carefully, we find the more massive atoms move

slower, in fact atoms of twice the mass have half the speed squared. Thusthe mass times the average speed squared is a constant, for the heavy as wellas the light gasses. This product, mv2, is exactly twice the kinetic energy,and thus we define absolute temperature as:

The absolute temperature is proportional to the average kinetic energy ofatoms in a classical gas.

The one caveat in the statement above is classical. We are discussingphysics before 1900, so called classical physics. At low enough or high enoughtemperature, this definition will not hold.

The constant of proportionality is a key constant of the universe. For us,it relates temperature to average kinetic energy, but we shall find it does farmore. It is called Boltzmann’s constant and it’s symbol is kB and it’s unitsare Joules per Kelvin. It is given by: kB = 1.38× 10−23J/K. For historicalreasons, there is a factor of 3/2 in the equation:

12mv2 =

32kBT (1.3)

where the bar denotes average.

1.2.1 Absolute Zero

At low enough temperature, the above equation suggests that the randommotion of atoms would stop altogether. This point, the zero of the Kelvinscale, is known as absolute zero. It is the coldest temperature possible and

1.3. IDEAL GAS APPROXIMATION 3

although it is possible experimentally to get very close to absolute zero,within a thousandth of a Kelvin, absolute zero itself is not attainable. The(good) reason for this is discussed much later.

What does happen near absolute zero? All free energy that can be lib-erated from an atom has been. There is motion left, due to non-classicalphysics, and this motion cannot be stopped by further reduction in temper-ature. In some sense, it is motion without energy. It is beyond the scope ofthis text.

1.3 Ideal Gas Approximation

An ideal, or perfect, gas has no collisions with itself, only with the wallsof it’s container, all collisions are perfectly elastic, and it has no rotationor vibration, thus a monatomic gas is typically envisioned. The life of sucha gas atom would be spent pinging from one wall to another at constantspeed. Using the above equation and solving for average speed, v, we find:

v2 = 3kBT/m (1.4)

and in a closed container of volume V, the dimension of which is thecube root of V, collisions will occur once every time t, where t is expressed(using distance is rate times time for constant speed travel) by:

t = V 1/3/v. (1.5)

Such a gas exerts pressure on the walls of the container via these colli-sions. The force of a single particle of gas would be an erratic function: verylarge for small times, but then zero as the particle travels to the far side ofthe container. However, a very large number of particles will create a con-stant stream of collisions and the pressure will be very nearly unchanging intime. To compute this, we will average the force of the single particle overit’s travel time between walls, then multiply by the total number of particlesto form up the total pressure.

What is the force due to a single particle? The result of the force ofthe wall on the particle is to change the particle’s momentum. The usualexpression is Σ~F = m~a but the right hand side can be rewritten as ∆~p/∆t.The change in momentum during a perfectly elastic collisions is 2mv sinceinitially the particle is moving one way with speed v and in the final stateis moves the other way with the same speed. Substituting we find:

F = 2mv/t = 2mv2/V 1/3 (1.6)


and substituting from Eq. (1.4) above we find:

F =6mkBT

mV 1/3. (1.7)

To total the force, we multiply by N the total number of particles present,but we divide by six because only one sixth will strike any given wall duringthe time. (We have assumed a cubic container; cubes have six sides.) Fur-ther, we want to find the pressure instead of the force. Pressure is force perunit area thus force is pressure times area, and the area of each side of thecontainer is V 2/3. The total pressure on any given wall of the container isgiven by:

PV 2/3 =NkBT

V 1/3(1.8)

Simplifying we find the ideal gas law:

PV = NkBT. (1.9)

1.3.1 Definition a Mole

The mole (abbreviated mol) is the number of atoms in 12 grams of Carbon-121 This number is 6.022×1023, known as Avogadros Number, NA, and isvast beyond imagining. Written without scientific notation it is 602200000000000000000000.One mol of housecats would correspond to the entire mass of the Earth, con-verted to felines2 - not just coating the surface of the Earth but the entirething, down to the core, converted to pure felis silvestris catus. In termsof money, one mol of dollars has never been spent, nor accumulated. Atthe present, about 4×1013 dollars are spent (exchanged from one hand toanother) each year on planet Earth. If our economy had been running sincethe Earth cooled about $0.3 mol would have been spent thus far and by thetime the Earth is enveloped by the Sun as it turns into a red giant, about$0.6 mol would have been spent. You have never had a mol of anything thatyou could hold in your hand (and recognize it was there).

Instead of writing the ideal gas law in terms of the enormous numberof particles N , and the tiny boltzmann constant, kB, we can rewrite itby multiplying Boltzmann’s constant by NA and dividing the number of

1Carbon-12 has six neutrons, six protons - the most common isotope, or “form” ofCarbon.

2Said cats would have to be slightly overweight to account for the entire mass of theEarth. Two mol of standard housecats would be more then sufficient, however.

1.3. IDEAL GAS APPROXIMATION 5

particles by the same. The number of particles divided by NA is the numberof mols, n, thus:

n =N

NA(1.10)

and the so-called gas constant R is given by:

R = NAkB. (1.11)

The gas constant, R, is 8.315 J/(mol K) (Joules per mol-Kelvin).

1.3.2 Examples of Ideal Gas Law Calculations

Volume of one mole of ideal gas At standard temperature and pres-sure, we can find the volume of one mole of ideal gas. Standard tempera-ture is zero centigrade, or 273 K and standard pressure is one atmosphere or1.013×105 Pa. Solving for V we find: V = nRT/P = 1×8.315×273/1.013×105m3 = 0.024 cubic meters of gas, or a sphere of 17.4 centimeters in radius(using v = 4πR3

3 for a sphere).

Balloon of Ideal Gas A balloon filled with ideal gas has a radius of 5cm at STP is submerged to a depth 2.2 m under water (density 103kg/m3).Find the final volume of the balloon assuming the temperature remainsunchanged. The extra pressure under a depth of fluid is given by ρgd,where d is the depth under the fluid. This pressure is on top of the oneatmosphere of standard pressure. Since n, R, and T are fixed, the ideal gaslaw can be reduced to: PfVf = PoVo. The final radius is 4.69 cm.

Factors of Two (a)Given an ideal gas, if the volume and temperature aredoubled, and the number of mols of gas remains unchanged, find the newpressure in terms of the initial pressure Po. The pressure is unchanged.

(b)Given an ideal gas, if the volume and pressure are doubled, and thenumber of mols of gas remains unchanged, find the new temperature interms of the initial temperature To. The new temperature is 4To, since bothV and P are doubled.

(c)Given an ideal gas, if the volume, number of mols and temperatureare all doubled, find the new pressure in terms of the initial pressure Po.The new pressure is 2Po.

(d)Given an ideal gas, if the pressure is doubled, and the number of molsand volume are both dropped in half, find the new temperature in terms ofthe initial temperature, To. The new temperature is half To.


1.3.3 Ideal Gas Law Problems

1. A child’s balloon has a diameter of 30 cm and is at room temperature, 25C. If the pressure in the balloon is 10% higher then atmospheric pressure,find the number of mols of gas in the balloon, assuming it is an ideal gas.

2. At what depth under water will an ideal gas assume one half it’svolume in atmosphere?

3. A typical oven has dimensions of 80 cm wide, 35 cm high and 40 cmdeep. At STP, how many mols of gas fit in the oven? If the temperatureis raised to 200 C, and pressure remains unchanged, find the new numberof mols of gas in the oven. What percentage of the gas has left? (Assumeideal gas).

1.3.4 An Improved Ideal Gas Law: Van der Waal’s Equation

In constructing the ideal gas law, several unrealistic approximations had tobe made. Principally that particles never interacted with one another (onlythe sides of the container). This is valid only at low density of particles andwith particle species which interact very weakly with one another (such asnoble gasses).

An improvement was posited by Johannes Van der Waals in 1873 whichdesignates each particle as having a volume b and introduces a parametera which relates to the strength of interaction between particles. This aparameter resists simple direct physical interpretation. The Van der Waalequation is:

(P − an

V

2)(V − bn) = nRT. (1.12)

For Helium gas, a ≈ 0.003J m3/mol2 and b ≈ 2 × 10−5m3/mol. ForNitrogen gas, a ≈ 0.1 and b ≈ 5 × 10−5 in the same units3. Helium is veryweakly interacting thus the very low a value compared with Nitrogen, andmost other common gasses, which have a values on the order of one half(from one tenth to one) in units of J m3/mol2.

Exercise with the Van der Waal Equation Compare the volume ofone mole of ideal gas (from above) with the volume of one mol of Van derWaals Helium and Van der Waals Nitrogen at STP.

3A table of such parameters can be found at http://www.ac.wwu.edu/vawter/ Physic-sNet/Topics/Thermal/ vdWaalEquatOfState.html among other places

1.4. EQUIPARTITION THEOREM 7

1.4 Equipartition Theorem

We defined temperature by using ideal (classical) gasses as an example. Indoing so, we wrote that: 3

2kBT = 12mv2. The average kinetic energy is three-

halves of kBT . In fact, each of the three components of velocity vector have,on average, one half a kBT of energy.

Energy is equally distributed between these three perpendicular direc-tions. Inherently, this is because we live in a three dimensional world thatthere are three independent directions and any point in space can be spec-ified by three coordinates and any velocity vector has three rectangularcomponents: the x, y and z components.

Yet what if there were other forms the energy could take, for example po-tential energy or rotational kinetic energy. How would energy be distributedinto these forms?

The answer is the equipartition theorem: that an equal amount of energygoes into each possible form of energy, on average, without bias.

Each form of energy is called a degree of freedom.Why should energy be evenly distributed? In a series of elastic collisions,

energy will be transferred more frequently from the high energy particle tothe lower then the reverse. (The reverse is possible.)

Imagine the extreme case: consider a two dimensional “gas” of friction-less ice hockey pucks initially at rest on the surface of an ice rink. Now shoottwo new pucks (from opposite directions to conserve the net zero momen-tum) at very high speeds into the rink. After many collisions, all pucks willhave been struck and thus will have some energy - although it is possible inprinciple for one to be totally at rest it is essentially impossible.

1.5 Effect of Temperature on Solids and Liquids:Thermal Expansion

All gasses (real and ideal) expand and contract with temperature, as indi-cated by the ideal gas law. This is a dramatic phenomena. Liquids andsolids also expand and contract with temperature, but not nearly so dra-matically. Your house (or car or clothes), for example, is not much smallerduring winter or larger during summer. Yet we know thermometers - filled,say, with alcohol or mercury - reflect changes in temperature. The marksfor various temperatures are nearly linearly spaced on the side. Although itis an approximation, most materials expand linearly in temperature over asmall range of temperatures. To formalize this, we need an equation and a


Material α x10−61/0CAluminum 23Brass 19Copper 17Lead 29Steel 12Iron 12Mercury 60Glass 8.5Pyrex 3.3Diamond 1Quartz, fused 0.59Silicon 3

Table 1.1: All numerical values from: “Coefficient of Thermal Expansion”,Wikipedia: The Free Encyclopedia http://en.wikipendia.org/wiki/ Coeffi-cient of thermal expansion (Accessed 14 October 2005).

coefficient, as follows:

L(T ) = Lo(1 + α∆T ). (1.13)

where Lo is the length at the initial temperature and ∆T is the changein temperature and α is the coefficient of thermal expansion for a givenmaterial. This coefficient subsumes the material dependant properties of thegiven material: all the microscopic interactions dictating the exact averagespacing between nuclei. On average, the atoms grow further apart uponheating - for most materials over most ranges of temperature. Exceptionsto this will be discussed below.

Typical coefficients of thermal expansion are small: on the order of10−51/K. The unit used is most commonly 1/0C or 1/K, which are identicalsince only changes in temperature are considered above, and the change intemperature in Celsius and Kelvin are identical. Some typical coefficientsare listed in table 1.5.

Two long thin metal films affixed together (back to back) with differentcoefficients of thermal expansion will bend upon a rise or fall in temperature.This so called bimetallic strip has been used traditionally in thermostatsto control temperature. A small bulb is affixed to the end of the stripwith a conductive liquid (often mercury) and two metal probes. When the

1.6. THERMAL VOLUME EXPANSION COEFFICIENT 9

temperature increases, the angle of the bulb changes and the liquid willspread over both probes, completing a circuit and allowing current to flow,turning on the air conditioning (or turning off the heating).

1.6 Thermal Volume Expansion Coefficient

The coefficient of volume expansion, denoted by β, is closely related to thecoefficient of linear expansion, α. The formula dictating the volume definesthe volume coefficient and is given below.

V (T ) = V0(1 + β∆T ) (1.14)

To understand the relation between α and β it is enlightening to con-sider area expansion. Fundamentally, the understanding of area is the un-derstanding of the factor of two in the derivative of the function x2.

1.6.1 Thermal Area Expansion Coefficient

Consider a square of side length L. Upon a small rise in temperature4 thereis an increase in length ∆L given by α∆T . Yet each side is increased by thesame ∆L, and thus the total new area is (L + ∆L)2 which expands in fullto:

A + ∆A = L2 + 2L∆L + ∆L2. (1.15)

where A is the original area, and ∆A is the increase. Following in thefootsteps of equations 1.13 and 1.14, we expect the change in area to begiven by:

∆A = γA0∆T. (1.16)

Identifying the original area with length squared, we find:

∆A = 2L0∆L + ∆L2. (1.17)

Yet by supposition, the change in length is far less then the originallength, thus if we can throw away the ∆L2 term, and using equation 1.13to find ∆L = αL0∆T , we find:

∆A = 2αA0. (1.18)

4In this case, small means such that the extra length is much less then L


By comparing the above with Eq. 1.16, we find the area expansion coeffi-cient is (approximately) twice the linear expansion coefficient. The approx-imation is particularly good when the change in length times length is farless then length squared (e.g. change in length far less then length) yet thiscondition is necessary to talk about a coefficient at all. None will remainexactly linear over a vast range of temperature. Indeed, upon achieving suf-ficient temperature, the material will pass through a phase transition andall bets are off.

Similarly, the coefficient of volume expansion, β = 3α, at least for smallchanges in volume.

Consider a cube, expanding in all three dimensions. Three faces expandby an amount αL2

0∆T , each face is bordered by two expanding edges, eachof which expands by a volume of α2L0∆T . Yet three of these edges areshared by two faces, thus there are a total of five such expanding edges.Further, there is one expanding corner which expands by the (assumedlytiny) volume α3∆T . Thus the total volume change is given by:

∆V = (3αL20 + 5α2L0 + α3)∆T (1.19)

of which we are keeping only the leading term. Thus the volume coef-ficient is a “worse” approximation then the linear coefficient, and accurateover a smaller range of temperatures (e.g. it becomes a worse approximationmore quickly).

1.6.2 Thermal Expansion Examples

1. A washer is a ring of inner diameter 0.500 cm and outer diameter 0.750cm at room temperature, made of aluminum. At what temperature will thering pass over a bolt of outer diameter 0.501 cm? (The bolt is not heated).

2. A brass bar is one millimeter shorter then an iron bar of length exactlyone meter at room temperature. At what temperature will they be equallylong? (Both are heated, equally).

3. A 300 mL Pyrex glass is full of boiling water, at 1000C. If the volumecoefficient of water is 102 x10−61/0C, find how much water could be addedto the glass at room temperature if: (a) the Pyrex does not expand at all,and (b) considering the contraction of Pyrex.

1.7. PHYSICAL UNDERPINNINGS OF THERMAL EXPANSION 11

1.7 Physical Underpinnings of Thermal Expansion

Most materials expand when heated. Exotic materials can be found whichcontract over some ranges of temperature increase. The question why?Solids can be approximated as a set of balls (the atoms) connected by springs(the chemical bonds holding the atoms together). Later we will discuss theexact geometries of bonds and the different lattice types which atoms can bearranged in. Yet true springs expand symmetrically with the application ofenergy: the energy is proportional to the spring constant and the distancestretched (or compressed) squared. This harmonic potential (proportionalto x2), favors neither expansion nor contraction upon application of moreenergy - the amplitude of oscillation would increase and nothing else.

Real solids have a more difficult time contracting then they do expanding.At long range, atoms attract one another yet at very short distances theymust repel as the electron clouds begin to overlap too greatly. This givesrise to the Lennard-Jones (or 6-12) potential:

U(r) = 4ε((σ

r)12 − (

σ

r)6) (1.20)

as indicated in the figure below. This is the potential energy landscapeon which one atom of a solid is allowed to move. At low temperature (lowenergy) it will reside in the minimum of the valley, and upon heating thetemperature rises and the energy is lifted out of the valley and the atom canoscillate back and forth along a line of constant energy. Yet the valley is notsymmetric: as the energy rises, the valley opens up more to the left then tothe right (e.g. toward larger radius). Every pair of atoms gets further apartupon heating (on average).

Conceptual Question: Imagine a rectangular sheet of metal, with a holecut in the middle. The sheet is heated to the point where it is now 10%larger in each dimension. Each pair of metal atoms is that much farterapart. Will the hole expand or contract?

1.8 Ideal Solid Equation

We can combine our knowledge of expansion of a solid with temperaturewith the change in length of a solid upon application of pressure to make ananalogous equation to the ideal gas equation, but for solids. Unlike gasses,however, this equation requires the use of two physical parameters: B, theBulk modulus and β, the coefficient of volume expansion.


Recall from Knight, chapter 15, equation (15.39), that the pressure givesrise to a change in volume of a solid via:

∆P = −B∆V

V(1.21)

and let us combine this with our equation for thermal expansion that:

∆V = βV ∆T. (1.22)

If we assume the two phenomena do not conflict with one another, andthat only small pressure and temperature changes are allowed, such that wecan take B and β to be roughly constant, we can simply sum the changesin volume due to each process and arrive at:

∆V = V (β∆T −∆P/B). (1.23)

The minus sign indicated increases in pressure act to shrink the solid,whereas increases in temperature act to expand said solid. This is a farless satisfying and ambitious equation then the ideal gas law because itrelies on initial knowledge of a particular situation (volume, at pressure andtemperature) and expands about that point linearly.

Ideal Solid Equation Example Consider a cubic meter of Aluminumat standard temperature and pressure. The temperature increases by 100C,find the increase in pressure needed to keep this cube exactly one meter inside length. Given B = 7× 1010 N/m2 and α = 23× 10−6 1/0C.

Setting ∆V to zero, the volume itself divides out of the equation and wefind: β∆T = ∆P/B. Solving for ∆P , we find: ∆P = 4.83× 107 N/m2.

Comment: this is about 477 atmospheres of pressure and thus we see thatby heating the solid an immense pressure is needed to keep it from thermallyexpanding. This increases is so vast, it is doubtful that this ideal solidequation is valid any longer. Thermal expansion is often a far larger effectthen pressure, simply by the units of the constants involved we can see this.Be sure when computing not to confuse the linear and volume coefficientsof thermal expansion. The volume coefficient is β and it is roughly threetimes the linear coefficient, α.

1.9 Crystal Structures

Although at first glance the possibilities seem endless, there are only fourteenpossible, distinct classes of crystal lattices - the fourteen Bravais lattices.

1.9. CRYSTAL STRUCTURES 13

The vast majority of the known elements on the periodic table fall into justa few of these categories. Each of these lattices are an infinite set of repeatedsimple units - and several of these units will be described here.

Simple Cubic - a cube with an atom on each corner. This is actuallyquite a rare structure, however a phase of Polonium does exhibit simplecubic (sc) symmetry.

Body Centered Cubic (bcc) - a cube with atoms at all corners and oneatom in the center. There is nothing special about the central atom - indeedit is a corner atom from a different point of view. Each corner atom has fournearest neighbors - which are body center atoms (from it’s point of view,with it at a corner) but from another view, it is the body center, itself.It has four more distant second nearest neighbors. Iron, Barium, Lithium,Sodium, Cesium, Tungsten, all adopt bcc structure.

Face Centered Cubic (fcc) - a cube with atoms at all corners and atomsalong each face. A single fcc cell resembles a hollow cage waiting for perhapsa foreign atom to become entrapped inside. Just like bcc, however, all atomsin the fcc crystal are equivalent. It is more difficult to see, but a plane cut ata 450 angle through the crystal reveals a face center atom to be the cornerfrom this point of view. Calcium, Nickel, Copper, Silver, Gold, Platinum,Palladium and most all the noble gasses (except Helium) adopt fcc structure.

There are many other structures and the differences are important tothe fundamentals of physical properties. A good introduction to this topicis available online at the International Union of Crystallographers website,IUCr.org. Click on teaching pamphlets for a variety of excellent resources.

Explain Why Expansion 6-12 potential (graph) avg. position of atomsCrystal Lattice Types (fig)Solid Phase equation

Chapter 2

Thermal Energy

When we push our textbook and impart kinetic energy to it, and eventuallyit slides to rest, where did this energy go? It has gone into the so called ther-mal (or internal) energy of the atoms. Energy is conserved. Occasionally,it goes into forms hard to measure; thermal energy would be one. Rubbingyour hands together to warm them is a direct sensation of friction creatingwarmth.

Let us begin again with ideal gasses and move on to a more general case.Ideal gasses have only kinetic energy which is their thermal energy. It

is the microscopic motion of each atom banging around the container. Wealready have a formula for this:

Ek = Eth =12mv2 =

32kBT. (2.1)

Thus to raise the temperature a certain amount, ∆T , the energy requiredper atom is 3kB∆T/2. Per atom is inconvenient since the number is quitesmall, thus we convert to per mole by multiplying by NA yielding: Ek =Eth = 3

2NAkB∆T , yet we know what Avogadros number times Boltzmann’sconstant equals: the gas constant, R = NAkB. Putting this together, wefind the energy needed to raise the temperature of one mole of ideal gas byone Kelvin is 3

2R. This quantity is called the specific heat.

2.0.1 Specific Heat

Specific heat is the energy required to heat a particular amount of a material(say, one mole or one kilogram) by one degree in temperature (say, oneKelvin). To distinguish the energy per mole from the energy per kilogram(aside from in units) the energy per mole is called the molar specific heat.

15

16 CHAPTER 2. THERMAL ENERGY

Although we have found the molar specific heat for the ideal gas, in gen-eral, we cannot know the specific heat of a new material without measuringit. However, there are a couple of cases in which the molar specific heat isknown (approximately) which are worth discussing to give a deeper insightinto the nature of specific heat and why some materials require more energythen others to raise the temperature.

Diatomic Ideal Gas In the case of the diatomic ideal gas, there are moredegrees of freedom available in the gas then simple translation. Two atomsare connected by a chemical bond which, like a spring, can compress andexpand. Further, the entire assembly can rotate. The equipartition theormsays that an equal ammount of energy, exactly 1

2kBT , is in each mode, onaverage. By counting the number of degrees of freedom, we can find thespecific heat (per atom, or per mole). There are three translational degreesof freedom, plus two for vibration: a kinetic and a potential, plus two non-trivial rotations: the cneter of mass can rotate about either of the two axesperpendicular to the bond. Rotation parallel to the bond requires no energy(and thus no energy can be stored in this rotation). This is the same reasonwe we neglected rotations altogether in the monatomic case. This is a totalof seven degrees of freedom (modes of energy) thus we expect the specificheat per atom to be 7

2kB and the molar specific heat to be 72R - and this is

indeed true, but only at high temperature.

Quantum mechanics and Available Degrees of Energy Quantummechanics is called quantum because various physical variables are quan-tized: that is they take on only certain values. Energy is one. There area large but innumerable number of possible energy configurations of a sys-tem. A system can rotate a little or a lot, but there is some integer numberenumerating each state. This means there is some lowest energy state calledthe ground state, which, in the case of rotation, is non-rotating and the nextstate up (the so called first excited state) requires some energy to attain.

For translation, the first excited state requires very little energy to attain;all the states are extremely close together. But for rotation and vibrationof the diatomic ideal gas, in any real gasses such as hydrogen (H2), there isa huge gap to the first excited state of rotation and vibration. This directlyeffects the specific heat.

Thus if we plot the specific heat of any diatomic gas (say Hydrogen), wefind that at high temperature our simple model is correct and the specificheat (per mole) is 7

2R, yet at lower temperature it falls to 52R and at still

17

much lower temperatures it falls to 32R. For Hydrogen gas (to be plotted)

the former occurs at about 1000 K and the latter at about 100 K, meaningnear room temperature (300 K), the specific heat is about 5

2R. This eithermeans either vibration or rotation is occuring, but not both. By experiment,we find it rotating.

Specific Heat of Solids: The Dulong Petit Law Each atom in a solid(in some kind of crystal lattice) can vibrate in any of the three cardianldirections (x,y,z). Each mode has both a potential and a kinetic energy.Thus there are a total of six degrees of freedom and we expect the specificheat to be 3R (one half per degree of freedom). This is the so-called Dulong-Petit Law. It works surprisingly well for metals near room temperature. Thedetails of the particular metal such as crystal lattice and density do not effectthis result since it relies only on the three independent directions (x,y,z) inour world. In reality, the atom may not be vibrating along these threedirections, in particular, it may move in a far more complex way and alongmany differet bond directions. However, any such motion can be brokenup into three and only three perpendicular directions. At one moment, wemay find an atom vibrating along a line at some angle θ above the x axis inthe x-y plane. Yet this can be broken up into vibration along the x and ydirections.

The law of Dulong and Petit works so well in metals that it is commonlydone in undergraduate labs. Yet why does it work so well for metals? Cer-tainly there is more to metals then simply vibrations in various directions.Metals are conducters of electricity and heat because of the presence of freeelectrons, dissociated from the nucleai. Experimentally, we find the specificheat of metals to be only 3R and thus heating the metal seems to requireonly enough energy to heat the atoms and counts nothing for the sea of freeelectrons at all.

2.0.2 Specific Heat Examples

Note: for brevity, temperatures in 0C are listed simply as C.1. Liquid water has a specific heat of 4186 J/kg C. How much cold 0

C water need be added to 350 grams of hot 100 C water to produce a finaltemperature of a nice, drinkable 45 C?

Use conservation of energy: heat flowing from the hot water is gainedby the cool. Since the final state is known, we actually know the energy:cm∆T = 4186J/kgC ∗ 0.4kg ∗ 55C = 92100J = cmc∆Tc of the cool. The


mass of the cool is thus: 92100/(4186∗45) = 0.49kg. This makes sense sincethe hot water needs to be cooled through a greater change in temperature.

2. Copper has a specific heat of 385 J/kg C. If a 55 gram piece of hot(234 C) copper is dropped into 105 grams of chilly 10 C water. Find thefinal temperature.

From conservation of energy, cwmw∆Tw = −ccmc∆Tc. Note the minussign. Heat flows from the copper into the water, thus heat flowing out ofthe copper requires a minus sign. The changes in temperature are: ∆Tw =Tf−10C which will be positive, and ∆Tc = Tf−234C, which will be negativesince the copper is cooling off.

Plugging our ∆T ’s in, we get four terms: 440Tf − 4400 = −21Tf + 4950units are all Joules. Solving we find: Tf = 20.3C.

2.0.3 Latent Heat and Phase Transitions

We cannot go on endlessly heating (or cooling) any substance without con-sequence. Take the cool liquid water in our earlier examples. Eventuallywe reach a point at which further heating ceases raising the temperatureand contributes to changing the liquid to a vapor. The energy required perkilogram is immense, order of 105J/kg, and this quantity is called the latentheat. Latent in this context means hidden. The heat is not doing what isusually does (change the temperature) and thus is hidden, or latent. Thiscessation of further temperature increase upon heating, and thus the latentheat is the character of a so called first order phase transition. The specificheat during the transition essentially goes to infinity since any applicationof heat doesn’t change the temperature at all (until enough has been appliedto finish the transition).

Continuous Phase Transitions The term second order phase transitionhas become rather outdated, and has been replaced with the more descrip-tive term: continuous phase transition. It refers to transitions wihtout aclear cut point at which the change in temperature ceases and the phasetransition commences. Instead of the specific heat going to infinity (as iteffectively does in first order phase transitions), the specific heat may sim-ply become rather large and if plotted as a function of temperature simplylook like a large bump. Thus by plotting the specific heat as a function oftemperature, such bumps are a sign that some kind of fundamental physicalchange is occuring. It may not be a traditional phase transition: e.g. fromgas to liquid to solid, it may be a strucutral phase transition from, say, a

2.1. CONDUCTION/CONVECTION/RADIATION: METHODS OF HEAT TRANSFER19

bcc to fcc solid phases. This can be accompanied by an actual bulk physi-cal property change such as color. Tin (element number 50, known to theGreeks as Snellium, abbreviation Sn) undergoes such a structural transitionfrom Grey to White Tin, however this is a first order phase transition.

2.0.4 Latent Heat Examples

1. To find the latent heat of copper, 155 grams of liquid copper at 1084C (it’s melting point) is solidified by passing tiny puffs of steam over it.The steam is freshly boiled at thus at 100 C. Since it is already steam, weassume no further phase transitions at let the specific heat of this steam is1950 J/kg C. If 38 grams of steam are required, what is the latent heat ofcopper?

The heat required to solidify the copper is Lmc and this equals the heatfrom the steam, csms∆Ts, which is known: 1950 ∗ 0.032 ∗ (1084 − 100) =6.14× 104J . Dividing by the mass of copper, we find: L = 3.96× 105J/kg,which is about right.

2.1 Conduction/Convection/Radiation: Methodsof Heat Transfer

There are three main methods of heat transfer between substances. Themethod we have been discussing is conduction: that of placing one substancenext to another and allowing the difference in temperature to drive heat fromone to another. The rate at which heat flows through a substance is calledthe thermal conductivity of the material. Metals, for example generally havegood thermal conductivity: heat flows into them quickly. Thus a hot metalburns faster then, say, a hot piece of wood. Further the metal spoon insoup will get hot faster then the wooden spoon. Convection is the mixingof one substance into another, such as the mixing of hot and cold water.This is generally the fastest method. Radiative heat transfer is usually onlyimportant if the other two are absent since it is quite slow. The Earth isheated by the Sun by radiation.

Chapter 3

Thermodynamics

Thermodynamis is a vast fied dealing with heat flow and work done by vari-ous systems. It has been of great concern in producing more efficient enginesand refridgerators. However it has been applied widely to microscopic sys-tems such as chemical reactions. This simple introduction will focus on themacroscopic.

3.1 First Law of Thermodynamics

Energy is conserved in any thermodynamic proceses and there are threemain constituients involved: the internal energy of hte system: U , the heatflowing into the system Q, and the work done by the system W . These arerelated by:

∆U = Q−W (3.1)

Any change in internal energy is reflected in some physical change suchas a change in phase or temperature. If we restrict ourselves to temperaturechanges, the change in interal energy will simply be cm∆T .

Any work done can be expressed as a force applied over a distance, butfor bulk systems, this can be shown to be equivalent to a pressure times achange in volume, for constant pressure systems. In general, the work is theintegral of P(V)dV.

With this we can analyze various thermodynamic processes some ofwhich are described below.

Isochoric The isochoric or isovolumetric process is one in which the vol-ume change is zero. Thus the work, P∆V is zero.

21

22 CHAPTER 3. THERMODYNAMICS

Isobaric The work done is exactly P∆V since the pressure is constant.

Isothermal The change in temperature is zero, thus the pressure andvolume are both changing, in general. We can find the work done, however,for an ideal gas system. Since we know the ideal gas law PV=nRT, with thetemperature constant, the pressure as a function of volume is given by: P(V)= nRT/V. Integrating this, we find the work done: W = nrT ln(Vf/Vo).

Adiabatic In this process, no heat is exchanged. For an ideal gas, ananalysis far more sophisticated then is suitable for this course shows thatthe expression PV 2/3 =const in adiabatic processes. This expression is validfor monatomic ideal gasses only and the exponent (2/3) of the volume willchange for diatomic ideal gasses.

3.2 2nd law

The second law can be stated in many forms. Experimentally, it is that theheat always flows from hot to cold (and not the reverse. This is obviousto us from personal experience. If we put a cup of cold water in the oven,the water warms and the ambient temperature of the oven must be loweredsomewhat. However, if we only had the first law of thermodynamics, all itsays is that the total energy is conserved and this could be accomplishedby the heat flowing out of the cold water into the oven: the water couldfreeze to ice and the oven would be even hotter! It never happens, of course,although we can cause heat to flow out of a colder area into a hotter, wehave to plug such a device into the wall and take power from somewhere.In this “somewhere” power is generated and in doing so, heat will flow frommhot to cold, or the equivalent will occur.

Another statement of the second law is that something called the entropyalways increases. The definition of entropy is the Boltzmann’s constanttimes the log of the number of available states of the system. An evenmixture of salt in water is far higher entropy then the salt sitting placidlyat the bottom of the water in a lump. Even though the salt at the bottomis energetically favorable, the salt will dissolve because of entropy.

In isothermal processes, entropy can be computed by: S = Q/T. Thisrelation reveals heat flowing from hot to cold always increases entropy - andthus occurs spontaniously (requires no external work). For a more involveddiscussion of this topic, see www.secondlaw.com.

3.3. THIRD LAW 23

Gibb’s Free Energy The defining quantity to determine the spontanityof a process is the Gibb’s Free Energy, G. This quantity is technically definedin general, but useful in particular when your system is in the presence ofa constant pressure and temperature environment: that is, an environmentwhich will not significantly change in pressure if the system gets a bit biggeror smaller (small changes in volume) and the environment will not signifi-cantly change in temperature if the system gives or takes a bit of heat. TheGibb’s Free energy is defined as: G = U + PV - TS. If during a process, theGibbs free energy would be lowered, said process will occur spontaniously.Note that the entropy is multiplied by the temperature (in Kelvin).

A third definition of the second law is: no engine can be constructedwith the sole effect of converting heat to work.

Efficiency The efficiency is defined to be the ratio of the work done to theheat input to an engine. Symbolically, the efficiency is e and must be lessthen one to satisfy the 2nd law. There is a theoretical maximum efficiencyproved by Carnot in the 19th century. Carnot developed a thermodynamiccycle which bares his name which combines two adiabatic processes withtwo isothermal processes and has an efficiency given by: e = 1− TL/TH .

3.3 Third Law

The third law of thermodyanmics is that a system cannot attain absolutezero. Were this possible, a Carnot cycle could be produced which had anefficiency of one, since the low temperature resovior could have been reducedpreviously to zero Kelvin. Experimentally, temperatures to within tiny frac-tions of absolute zero are routinely achieved yet with phenominal effort, toactually reach zero Kelvin remains illusive.

Time Reversibility The second law indicates an arrow of time: timealways runs toward a state of increasing entropy. No earlier laws of physicswe have encountered thus far have distinguished past and future. Simply,entropy will go on incresing, heat will continue flowing, until the universeis totally uniform and no further heat will flow. The ultimate fate of theuniverse: to be totally uniform and with maximum entropy. Unable to drivefurther heat engines, there would be no available work to drive entropy back.This is the so called heat death of the universe. This fate may be avoidableif the energy density of the universe is great enough to contract the entiremess back into a single supermassive black hole: the so called big crunch.

introductory thermal physics - california state...

Documents