Temperature• Temperature (T): Measure of the average kinetic energy of the particles composing an object or system due to

–Vibrations–Rotations–Random translations

•Does a system’s temperature depend on your relative velocity to it?

–No: The center of mass (CM) motion of the system does not contribute to the system’s temperature.

Temperature Scales• Water based: freezing and boiling points

– Fahrenheit: 32 oF to 212 oF– Celsius (Centigrade): 0 oC to 100 oC– Which scale is more precise?

• Kelvin scale– Directly proportional to the average kinetic energy of

the system.– 0K is the unreachable absolute zero of temperature

where all particle motion stops.– Increments are the same as Celsius: 1K = 1 Co

– Conversion between the two: T(K) = T(oC)+273 Co

Internal Energy of a System (U)

• Includes:– Internal translational K.E. (no CM motion)– Vibrational and Rotational K.E. of atoms about

their bonds in molecules.– P.E. associated with bonds within molecules.– P.E. due to attraction between molecules.

• Processes that change Internal Energy:– Work (Force applied through a distance)– Heat (Differences in Temperature)

Heat • Heat: Process of energy transfer due to

temperature differences.

• Measured in calories or kilocalories

• Mechanical Equivalent of heat: 1 cal = 4.186 J– Work and heat are both processes of energy

transfer.– Discovered by James Joules

Joule's Experiment: Falling mass (work) used to heat water.

Examples• How many reps of pressing 50 kg must you

do to burn one pound of body fat?– One of gram of fat is burned by 9 kilocalories– There are 454 grams in a pound.– Assume 1 meter long arms.

• A 3.0 gram bullet passes through a tree, slowing down from 400 m/s to 200 m/s. How much K.E. is lost to heating the tree in calories?

Calorimetry• Deals with change in a system’s energy due to

heating or cooling.• Specific Heat of a substance (c): measure of the

amount of energy per unit mass needed to cause a rise of one degree in Temperature. (Q = m c ∆T)

• Latent heat of fusion (Hf): amount of energy per unit mass needed to melt a solid. (Qf = mHf)

• Latent heat of vaporization (Hv): amount of energy per unit mass needed to vaporize a liquid. (Qv = mHv)

• No Temperature changes occur during the phase transitions.

Calorimetry Example• We initially have 2 kg of ice at -20°C. How

much energy is needed to turn the ice into steam at 100°C?

• Given:– cice = 2100 J/kg°C

– cwater = 4186 J/kg°C

– Hf = 3.33 ×105 J/kg

– Hv = 22.6 ×105 J/kg

Heat Transfer• Thermodynamics: deals with properties of

different states and amounts of heat or work to go between them.

• Heat Transfer: deals with the rate at which energy (as heat) is transferred.

• Three modes of heat transfer:– Conduction: propagation of vibrations through

adjacent atoms.– Convection: energy carried by actual transfer of mass.– Radiation: energy carried by electromagnetic waves.

Conduction• How fast does heat flow through a wall with the

ends kept at different temperatures?

• Fourier’s Law of Heat Conduction:

(∆Q/∆t) = k A (TH-TC)/l

• ∆Q/∆t = energy transferred per unit time.

• A = area; more atoms available to vibrate.

• TH – TC ; higher K.E. differences

• l = length or thickness of wall.

• k = thermal conductivity; atomic structure.

• Hydraulic and Electric Analogs exist.

Example 1: Heat Loss through a Window

• The window has an area of 3 m2, a thickness of 3.2 mm, and is made of glass with a thermal conductivity of .84 W/m °C . The temperature inside is 15 °C and outside it is 14 °C.

• How fast does the window lose heat?

• An equilibrium is reached at which there is a steady and constant heat flow.

• A temperature profile can be drawn, showing the steady state temperature as a function of distance from one end.

Ex2: Composite Material

• Realistic Situation: Walls with insulation between them.

• Inside Temperature: TH= 25 °C

• Outside Temperature: TC=5 °C

• Insulation (inner wall): l1=1m, k1=.048 W/m °C

• Brick (outer wall):l2=.5m, k2=.84 W/m °C

• Find Temperature at junction and temperature profile at steady state.

• Key Point: At steady state: ∆Q/∆t = constant throughout the thickness of the walls.

Other Forms of Heat Transfer

• Convection: (combines with fluid dynamics)– Free: natural currents due to temperature and density differences.

– Forced: accelerated through fans or turbines.

• Radiation:– Needs no medium to travel.

– An object emits radiation at a rate dependent on temperature (in Kelvins): (∆Q/∆t) = AσeT4

• Emissivity (e) = 0 to 1; measure of surface’s ability to emit.

• Stefan-Boltzmann Constant: =5.67×10-8 W/(m2/K4)

• Second Sound: – Very fast heat transfer in superfluid (cold < 2.7K) helium.

– Travels as waves of varying temperature and specific entropy

– Analogous to sound waves of varying pressure and density.

How Does a Solid Expand When Heated?

• We know solids expand when heated. Experimentally: ∆L = α L0 ∆T

• ∆L = change in length of object (+ or -)• α = coefficient of linear expansion• L0 = original length of the object• ∆T = change in temperature (+ or -)• L = L0 + ∆L = (1+ α ∆T) L0 = final length.• Why should elongation depend on

temperature change?• Why should elongation depend on original

length of object?

Examples

• Ex1: A plate with a hole cut out. When we heat the plate how does the hole behave?

• Ex2: A ring of material with α1 stuck around a tube with α2 and (α2 > α1 ). How can I remove the ring from the tube?

More Examples

• Uniform rectangular plate of area: A = l×w

• Show ∆A = 2αlw∆T (neglecting small quantities)

Thermal Stress (ref. ex 10.5 pg 358)

• Ex: A beam bridges two walls. The walls cannot expand or contract compared to the beam (αwall << αbeam)

• What will happen if I heat the beam?– Beam wants to expand by ∆L from L0

– Inability to expand causes a compressive strain: ∆L / L0

– and a resulting compressive stress: F/A = Y (∆L / L0 ) (where Y = Elastic or Young’s Modulus; Hooke’s Law)

– Since: ∆L = α L0 ∆T,

– We have: F/A = Y α ∆T = thermal stress.– Can be compressive or tensile. Usually enormous.

How do we describe large numbers of atoms and molecules?

• New unit of mass: – Atomic mass unit (amu) or (μ) – 1 μ = 1/12 the mass of a carbon-12 atom (definition)– Atomic masses of other elements are often given in

amu, expressed relative to the mass of 12C atom. (example: the periodic table).

• A mole = the number of 12C atoms that has a mass of 12 grams. (definition)

The Mole• A “mole” is convenient number, similar to

“dozen” eggs or a “gross” of nails.

• A mole of atoms of an element has a mass in grams equal to the element’s atomic mass.

• A mole of molecules of a compound has a mass in grams equal to the compound’s molecular mass.

• The number to which a mole is equal is: NA= 6.022×1023 (Avogadro’s Number)

Examples • Atomic Hydrogen (H):

– What is the mass of an H atom in μ?– What is the mass of a mole of H atoms in grams?– What is the mass of one H atom in grams?

• Molecular Hydrogen (H2):

– What is the mass of one H2 molecule in μ?

– What is the mass of a mole of H2 molecules in grams?

– What is the mass of one H2 molecule in grams?

More Examples

• What is the mass of two moles of H2 O?

• A copper penny has a mass of 3.4 grams.– How many moles of copper atoms are in

the penny?

– How many atoms of copper are in the penny?

How is a gas described by its properties?• How is a gas different from a liquid or solid?

– Fills its container– The K.E. of its motion is large compared to P.E.

between particles.

• How does a gas fill its container?– How is volume for a gas defined?– How is this different from solids and liquids?– If we put a few atoms in a container, how do they

“fill the container”?– Gas atoms bounce off walls; resulting impulses on

walls exert pressure.

Summarizing: Properties of a gas system

• Volume (V): The extent of space through which the gas particles can move. (m3 or L )

• Temperature (T): describes the kinetic energy associated with molecular and atomic motions. (K)

• Particle Number – In moles (n) or– In particles (N)

• Pressure (P): measure of forces involved in hitting walls and bouncing off. (N/m2 or atm)

Equation of State

• Relates the properties of a system to each other.• For an “ideal” gas: PV = nRT

– R = ideal gas constant (.0821 L-atm/mol-K) = (8.31 J/mol-K) (mol=mole)– For a set n; fixing two properties sets the other two.– P = absolute pressure, T must be in kelvins.

• Boyle’s Law: If T is constant: V~ 1/P• Charles’ Law: If P is constant: V ~ T• Gay-Lussac’s Law: If V is constant: P~T• STP: “Standard Temperature and Pressure

T=273K, P= 1 atm (sea level)

Examples• What is the volume (in liters) of one mole of a

gas at STP?

• Suppose we have a piston in a cylinder and we introduce at STP:

1 mole of nitrogen gas (N2)

3 moles of hydrogen gas (H2)

This reaction occurs (still at STP):

N2(g) + 3H2(g) 2 NH3(g)

What happens to the volume?

Examples (cont’d)

• An auto tire is inflated to a gauge pressure of 200 kPa at 10oC. After driving a bit, the temperature rises to 40oC.

What is the new gauge pressure?

Ideal Gas Law (in particles instead of moles)

• PV = nRT

• Let N = number of atoms or molecules

N = (# moles) (# particles/mole)

= n NA

• So : n = N/ NA

• PV = (N/NA) = N(R/NA)T = NkbT

• kb= Boltzmann’s constant = 1.38×10-23J/K

What makes an ideal gas “ideal”?

• Why does PV=nRT hold for all ideal gases?

• Why is R the same for all ideal gases?

• How is an ideal gas characterized?

• The ideal gas law makes no use of the identity of the particles.

What makes an ideal gas “ideal”?

• There must be a large number of particles in the system.

• The particles in the gas must be much smaller than their separations.

• The attractive P.E. between particles must be much lower than their K.E.

• Collisions of particles with container walls and with each other must be elastic. (No energy loss through friction.)

Departures from Ideal Gas Behavior• Modifications can be made to the Equation of State for non ideal

gases. • Most common is the “Van der Waals” gas, where a and b are

parameters of the particular gas :

2( )( )

aP V b nRT

V

•What is the physical significance of these constants?

• b: decreases effective volume of the container. Why?

• a/V2 : increases effective pressure. Diminished effect with larger volume. Why?

Thermodynamic Systems

• A system is a part of the universe under consideration. The rest of the universe is called the “environment” or the “surroundings”.

• Isolated system: No matter or energy is exchanged with the environment. (ex: thermos)

• Closed system (or “control mass”): no matter is exchanged with the environment. (ex: gas in a cylinder with a piston.)

• Open system (or “control volume”): Allows exchange of both matter and energy with the environment. (ex: animal cell surrounded by a membrane)

Properties of a Thermodynamic System

Two types of properties are used to describe systems:

• Extensive Properties:– Ex: mass (M), volume (V), internal energy (U), heat

capacity (C)– Depend on system size

• Intensive Properties (also called “point functions”):

– Ex: pressure (P), temperature (T), density (ρ)– Independent of system size– Have a value at every point in the system.– An extensive property can be made intensive by dividing

by the mass of the system (v=specific volume, u=specific energy, c=specific heat)

Internal Energy of a System (U)

• Includes:– Internal translational K.E. (no CM motion)– Vibrational and Rotational K.E. of atoms about

their bonds in molecules.– P.E. associated with bonds within molecules.– P.E. due to attraction between molecules.

• Processes that change Internal Energy:– Work (Force applied through a distance)– Heat (Differences in Temperature)

Kinetic Theory of an Ideal gas• How is Temperature related to K.E. and U in an

ideal gas?

• Suppose we have a cubic box:– Side length (l)– N atoms of mass (m)– Average K.E. of the atoms: – Speeds in all directions are the same on average:

• We see: T~ (K.E.)AV and U depends only on T

– (K.E.)AV = (3/2)kbT

– U = (3/2)NkbT = (3/2)nRT

First Law of Thermodynamics• Internal Energy is

– Increased by “adding heat”– Decreased by “removing heat”– Increased when positive work is done on the system– Decreased when positive work is done by the system

• Two conventions1. ∆U = Q + W ; W = work done on the system2. ∆U = Q – W; W = work done by system

In both cases: Q = heat absorbed by system• An explicit question would ask for the work

done on or by the system.

How is a system affected by processes of heat and work?

• We characterize a system by its properties.

• Usually, specifying two properties will define the state of the system. – All other properties are then determined.– Ex: PV=nRT; for a system of n moles, specifying P

and V determines T.

• Processes involving heat and/or work – take us from one state to another– Result in changes of the energy of the system governed

by the First Law: ∆U = Q + W

How is a system affected by processes of heat and work? (cont’d)

• We can draw a diagram of P vs. V and represent – States as points on the diagram.

– Processes as lines or curves between points (states) with an arrow specifying direction.

• Properties fix a system’s state without regard to the process taking us there.– Think of an initial state A as a starting point.

– Think of a final state B as an ending point.

– Different routes on a P-V diagram can be taken from A to B.

• We will study various processes in which heat is absorbed or expelled from the system and work is done on or done by the system.

How can we apply processes of heat and work to take a system to a desired state?

• Types of processes we’ll study:– Isothermal (constant Temperature: ∆T = 0)– Adiabatic (no heat absorbed or expelled: Q=0)– Isobaric (constant Pressure: P = const)– Isochoric (constant Volume: V=const)

• (also called “isometric” or “isovolumetric”)

• Relationships we’ll use frequently – PV=nRT, U= (3/2) nRT, ∆U = Q + W

• Questions we’ll be asking:– What are the properties at a state?– What work is done on or by the system for a process?– What heat is absorbed or expelled by the system for a

process?

Isothermal Process(constant Temperature)

• ∆T = 0, ∆U = 0 (for ideal monatomic gas) – Heat absorbed = work done by the system– Isothermal expansion: must add heat– Isothermal compression: must remove heat.

Adiabatic Process(Heat is neither gained nor lost)

• Q = 0; – ∆U = work done on system– System is thermally insulated or– Work is done too quickly for heat to transfer

• How does an adiabatic expansion compare to an isothermal expansion?

Isobaric Process(Pressure is constant)

• P is constant– Heat is added (expansion) or removed (compression) to

keep pressure constant.– Expansion: ∆V > 0; W < 0 (work done on system)– Compression: ∆V < 0; W > 0 (work done on system)– Work is easy to calculate:

W = -P∆V = work done on system

• In general: – Work for a process equals the area under its PV-curve– The sign depends on the direction of the process

(expansion or compression)

Isochoric Process(Volume is constant)

• Also called isovolumetric or isometric

• W=0; no work is done on or by the system.

• System neither expands nor contracts; closed rigid container.

• Change in internal energy equals heat absorbed: ∆U = Q

Thermodynamic Cycle• A series of processes that begin and end at the same state:

– (add arrows)

• The area enclosed equals the net work done on or by the cycle.

• Goal of an Engine: to extract useful work done by the system.

Example: Adiabatic Expansion

• In an engine cylinder, .25 moles expands rapidly and adiabatically against a piston. The temperature drops from 1150K to 400K.– Does something like this really happen?– How much work does the gas do?

Example: Change of State

• One liter of water (mass= 1kg) is boiled at 100°C into 1671 liters of steam at 100°C. The process occurs at 1 atm.– What is the heat absorbed by the water?

– What is the work done by the steam in expanding?

– What is the change in internal energy of the system?

Schematic of an Engine• Goal: To extract useful

work from the engine.• Price: Energy input as

heat (QH) obtained by burning fuel.

• Wasted Energy: Energy lost through heat (QC)

• Efficiency (e):

e = W(OUT)/ QH

= (QH-QC)/ QH

Operation of a Heat Engine

• A “working substance” (recall the control mass or closed system) is taken through a thermodynamic cycle.

• Since Initial State = Final State:– ΔUCYCLE = 0 = QNET - WNET(BY)

– QNET = WNET(BY)

• The net heat absorbed by the system equals the net work done by the system.

Analyzing a Thermodynamic Cycle

• How do we analyze a cycle for its state properties and usefulness?

• Example: One mole of an ideal gas goes through a cycle with the following processes:– AB: Isobaric expansion at 2P0 from a volume of V0 to

2V0.

– BC: Adiabatic expansion from 2V0 to 3V0 and P0.

– CD: Isobaric compression from 3V0 to V0.

– DA: Isometric compression from P0 to 2P0.

• First Step: Draw the cycle on a PV-diagram.

Analyzing a Thermodynamic Cycle

• For the preceding cycle, find the following in terms of P0, V0, and R:

a) Find T at every state.

b) For each process, find Q, W(BY), ΔU

c) Find: QH, QC, WNET(BY) , ΔUCYCLE

d) What is the thermal efficiency?

Heat Capacity of an Ideal Gas• Heat Capacity of a system (C) is the amount of

energy needed to cause a rise in temperature of one degree.

• Heat Capacity per unit mass is specific heat.• For an ideal gas: Heat capacity depends on the

process during which the heat is added.

• CV = (Q/ΔT) V=constant = (3/2)nR

• CP= (Q/ΔT) P=constant = (5/2)nR

• CP= CV +R; γ= CP / CV = (5/3)

How Efficient can we make an Engine?

• Carnot worked to increase engine efficiency

• Ideal Engine: all processes occur reversibly

• Reversible Process: Carried out in infinitesimally small steps:– No friction– Equilibrium achieved at each step.

Carnot Engine• Carnot (Ideal) Efficiency achieved with the following cycle:

• The Carnot Efficiency depends only on the temperatures between which engine is operated:e (Carnot) = [(QH-QC)/ QH] (Carnot) = [(TH-TC)/TH] = 1 - TC/TH

How do We Describe Disorder in a System?

• Second Law of Thermodynamics– Heat flows sponraneously from hot to cold.

(Clausius)– Heat can’t be converted completely into work.

(Kelvin-Planck)– A system spontaneously tends towards a state of

higher disorder.

• Examples: Liquids mixing, gases mixing, gases spontaneously expand, solids melting.

Two Views of a System• Systems can be described microscopically or

macroscopically.• Rolling Two Dice:

– 36 possible combos; each is a “microstate”(µ-state), specified by two numbers (1-6 and 1-6).

– Each microstate is equally probable– 11 possible “macrostates” (each specifed by one number: 2-12)– 7 is most probable macrostate; most disordered.

• Ideal Gas of N particles:– Specify a microstate by listing positions and momenta

of all particles (6N numbers)– Specify a macrostate by two numbers (e.g. P,V)

Entropy: Quantitative Measure of Disorder

• Microscopically: S = kb ln Ω (Ω = number of possible microstates)

• Macroscopically (for an ideal gas): Consider an isothermal expansion over a very small ∆V:Q = W(by system) = P∆V =(nRT/V) ∆V

Then : n R (∆V/V) ~ Q/T (both reflect disorder)(n ∆V/V) = relative increase in V and in n reflect

increase in disorder as process progresses.Q/T = relative increase in K.E. of system also reflects

increasing disorder.

Change in Entropy

• ∆S = (Q/T)(REVERSIBLE PATH)

• By the Second Law of Thermodynamics

(For processes)

– For an isolated system: ∆S >= 0

– For other systems and their environments:

∆S (TOT) = ∆S (SYSTEM) + ∆S(ENVIRON) >= 0

For Real Processes: ∆S (TOT) > 0