Thermal Physics (8-12%) (P. 203-221)

Front 1

Heat and Temperature (p. 204-210)

Back 1

PART 1/4 (4%)

Front 2

Temperature

Back 2

-measure the concentration of thermal energy in an object in much the same way that density measures the concentration of matter in an object.
-as a result, a large object will have a much lower temperature than a small object w/ the same amount of thermal energy. As we shall see shortly, different materials respond to changes in thermal energy w/ more or less dramatic changes in temperature.
-is a measure of the average KE of the molecules that make up that material.
-Absolute Zero is defined as the temperature at which the molecules have zero KE, which is why it is impossible for anything to be colder.

Front 3

Degrees Celsius
y°F= (9/5)x°C+32

Back 3

-The SAT II won't ask you convert b/w Fahrenheit and Celsius, but if you have a hard time thinking in terms of degrees Celsius, it may help to know how to switch back and forth b/w the two.

Front 4

Kelvins

Back 4

-the Kelvin scale (K) is a measure of absolute temperature, defined so that temperatures expressed Kelvins are always positive. ABSOLUTE ZERO, 0K, which is equivalent to -273°C, is the lowest theoretical temperature a material can have.
-Water freezes at 273K and boils at 373K.

Front 5

Heat

Back 5

-a measure of how much thermal energy is transmitted from one body to another.
-While both work and heat can be measured in terms of joules, they are not measures of energy but rather of energy transfer.
-A hot water bottle has a certain amount of thermal energy; when you cuddle up with a hot water bottle, it transmits a certain amount of heat to your body.

Front 6

calories (cal)
1 cal= 1 g/°C= 4.19J

Back 6

-unlike joules, calories relate heat heat to changes in temperature, making them a more convenient unit of measurement for the kinds of thermal physics problems you will encounter on the SAT II.
-Be forewarned, however, that a question on thermal physics on the SAT II may be expressed either in terms of calories or joules.
-calories is the amount of heat needed to raise the temperature of one gram of water by one degree Celsius. One Calorie is equivalent to 4.19 J.
-Food calories are not quite the same as what we're discussing here: they are actually Calories, with a capital "C," where 1 Calorie= 1,000 calories. Also, these Calories are not a measure of thermal energy, but rather a measure of the energy stored in the chemical bonds of food.

Front 7

Specific Heat
Q=mcΔT
•Q is the heat transferred to a material
•m is the mass of the material
•c is the specific heat of the material
•ΔT is the change in temperature

Back 7

-measure how much heat is required to raise the temperature of a certain mass of a given substance.
-is measured in units of J/kg.°C or cal/g.°C. Every substance has a different specific heat, but specific heat is a constant for that substance.
-Substances that are easily heated, like copper, have a low specific heat, wile substances that are difficult to heat, like rubber, have a high specific heat.

Front 8

Specific Problem
-4190 J of heat are added to 0.5kg of water w/ an initial temperature of 12°C. What is the temperature of the water after it has been heated?

Back 8

-By rearranging the equation above, we can solve for ΔT:
•ΔT=Q/mc
= 4190J/ (0.5kg)(4190 J/kg.°C)
=2C°
-The temperature goes up by 2C°, so if the initial temperature was 12°C, then the final temperature is 14°C. Note that when we talk about an absolute temperature, we write °C, but when we talk about a change in temperature, we write C°.

Front 9

Thermal Equilibrium

Back 9

-heat flows spontaneously from a hotter object to a colder object, but never from a colder object to a hotter object.
-This is one way of stating the Second Law of Thermodynamics, to which we will return later in this chapter.
-when both objects have the same temperature we say they are in THERMAL EQUILIBRIUM.
-With this in mind, it is possible to calculate the temperature two objects will reach when they arrive at thermal equilibrium.

Front 10

Latent Heat of Transformation, q

Back 10

-tells us how much heat it takes to change the phase of a substance.
-Latent Heat of Vaporization: tell us how much heat is gained or lost in transforming a liquid into a gas or a gas into a liquid, is a different value from the latent heat of fusion.
-For instance, the latent heat of vaporization for water is 2.3x10(6) j/kg, meaning that you must add 2.3x10(6) J to change one kilogram of water into steam, or remove the same amount of heat to change one kilogram of steam into water. Throughout this phase change, the temperature will remain constant at 100°C.
-To sublimate a solid directly into a gas, you need an amount of heat equal to the sum of the latent heat of fusion and the latent heat of vaporization of that substance.

Front 11

Latent Heat Example:
-How much heat is needed to transform a 1kg block of ice at -5°C to puddle of water at 10°C?

Back 11

-How much heat does it take to raise the temperature of the ice to 0°C:
Q=mcΔT= (1kg)(4.19x10(3) J/kg.C)(5C)
= 2.1 x 10(4) J

-How much heat it takes to melt the ice into the water?
Q= mq(fusion)= (1kg)(3.3x10(5)j/kg)
= 3.3 x 10(5) J

-How much heat it takes to warm the water up to 10C. (2 x 5C)
4.2 x 10(4) J

-Add all three values and you get: 3.9 x 10(6) J

-Note that far more heat was needed to melt the ice into liquid than was needed to increase the temperature.

Front 12

Thermal Expansion
ΔL=αL(i)ΔT
ΔV=βV(i)ΔT

Back 12

-Any given substance will have a COEFFICIENT OF LINEAR EXPANSION, α, and a COEFFICIENT OF VOLUME EXPANSION, β. We can use these coefficient to determine the change in a substance's length, L, or volume, V, given a certain change in temperature.

Front 13

Thermal Expansion Example:
-A bimettalic strip of steel and brass of length 10cm, initially at 15°C, is heated to 45°C. What is the difference in length b/w the two substances after they have been heated? The coefficient of linear expansion for steel is 1.2x10(-5)/C°, and the coefficient of linear expansion for brass is 1.9x10(-5)/C°.

Back 13

-First, let's see how much the steel expands:
ΔL=αL(i)ΔT =(1.2x10(-5))(0.1m)(30C) =3.6x10(-5)m

-Next, let's see how much the brass expands:
ΔL=αL(i)ΔT =(1.9x10(-5))(0.1m)(30C) =5.7x10(-5)m

-The difference in length is 5.7-3.6= 2.1x10(-5)m. Because the brass expands more than the steel, the bimetallic strip will bend a little to compensate for the extra length of the brass.
-Thermostats work according to this principle: when the temperature reaches a certain point, a bimetallic strip inside the thermostat will bend away from an electric contact, interrupting the signal calling for more heat to be sent into a room or building.

Front 14

Methods of Heat Transfer

Back 14

-There are 3 different ways heat can be transferred from one substance to another or form one place to another.
-THIS MATERIAL IS MOST LIKELY TO COME UP ON THE SAT II AS A QUESTION ON WHAT KIND OF HEAT TRANSFER IS INVOLVED IN A CERTAIN PROCESS. YOU NEED ONLY HAVE A QUALITATIVE UNDERSTAND OF THE 3 DIFFERENT KINDS OF HEAT TRANSFERS.

Front 15

Conduction (1/3 Methods of Heat Transfer)

Back 15

-transfer of heat by intermolecular collisions.
-Ex: when you boil water on a stove, you only heat the bottom of the pot.
-most common way of transferring heat b/w two solids or liquids, or w/in a single solid or liquid. Is also a common way of transferring heat through gases.

Front 16

Convection (2/3 Methods of Heat Transfer)

Back 16

-While conduction involves molecules passing their kinetic energy to other molecules, CONVECTION involves the molecules themselves moving from one place to another.
-For example, a fan works by displacing hot air w/ cold air.
-usually takes place w/ gases traveling form one place to another.

Front 17

Radiation (3/3 Methods of Heat Transfer)

Back 17

-molecules transform heat into electromagnetic waves, so that heat is transferred not by molecules but by the waves themselves.
-takes place when the source of heat is some form of electromagnetic wave, such as a microwave or sunlight.

Front 18

Kinetic Theory & Ideal Gas Laws (p. 211-215)

Back 18

PART 2/4 (2-3%)

Front 19

Kinetic Theory of Gases

Back 19

-makes the transition b/w the microscopic world of molecules and the macroscopic world of quantities like temperature and pressure.
-THe kinetic theory projects a picture of gases as tiny balls that bounce off one another whenever they come into contact. This is, of course, only an approximation, but it turns out to be remarkably accurate approximation for how gases behave in the real world.
-These assumptions allow us to build definitions of temperature and pressure that are based on the mass movement of molecules.

Front 20

Kinetic Theory of Gases (4 Basic Postulates)

Back 20

1.) GASES ARE MADE UP OF MOLECULES.

2.) MOLECULES ARE IN CONSTANT RANDOM MOTION.

3.) THE MOVEMENT OF MOLECULES IS GOVERNED BY NEWTON'S LAWS.

4.) MOLECULAR COLLISIONS ARE PERFECTLY ELASTIC.

Front 21

Temperature

Back 21

-the kinetic theory explains why temperature should be a measure of the average kinetic energy of molecules. As we said, molecules in any system move at a wide variety of different velocities, but the average of these velocities reflects the total amount of energy in that system.

Front 22

Pressure, P
P= F/A

Back 22

-is the measure of the force exerted over a certain area.
-something exerts a lot of pressure on an object if it exerts a great amount of force on that object, and if that force is exerted over a small area.
-is measured in unites of PASCALS (Pa), where 1 Pa=1 N/m(sq)
-The kinetic theory tells us that gas molecules obey Newton's Laws: they travel w/ a constant velocity until they collide, exerting a force on the object w/ which they collide.

Front 23

Ideal Gas Law
PV=nRT
•R is 8.31 J/mol.K

Back 23

-relates temperature, volume, and pressure, so that we can calculate any one of these quantities in terms of the others.

•Temperature, T: is directly proportional to volume, V, and pressure, P.
•n (moles) (6.023x10(23)): is the number of hydrogen atoms in a gram of hydrogen. Because we deal w/ a huge number of gas molecules at any given time, it is usually a lot easier to count them in moles rather than counting them individually.
•R is a constant of proportionality called the UNIVERSAL GAS CONSTANT. This constant effectively related temperature to kinetic energy. If RT is the KE of an average molecule, then nRT is the total KE of all the gas molecules put together.

Front 24

Boyle's and Charles Law

Back 24

-The SAT II will only ask you to consider how changes in one of those values affects another of those values. These 2 equations are simplifications of the ideal gas laws.

Front 25

Boyle's Law (Simplification of Ideal Gas Law)
PiVi=PfVf

Back 25

-deals w/ gases at a constant temperature. Tells us that an increase in pressure is accompanied by a decrease in volume, and vice versa.
-Aerosol canisters contain compressed (low-volume) gases, which is why they are marked w/ high-pressure warning label. When you spray a substance out of an aerosol container, the substance expands and the pressure upon it decreases.

Front 26

Charles's Law (Simplification of Ideal Gas Law)
•Vi/Ti=Vf/Tf

Back 26

-deals with gases at a constant pressure.
-volume and temperature are directly proportional.
-this is how hot-air balloons work: the balloon expands when the air inside of it is heated.

Front 27

Gases in a Closed Container

Back 27

-another way of saying that the volume remains constant.
-pressure and temperature are directly proportional: Pi/Ti=Pf/Tf.

Front 28

Example 1:
-A gas in a cylinder is kept at a constant temperature while a piston compresses it to half its original volume. What is the effect of this compression on the pressure the gas exerts on the walls of the cylinder?

Back 28

-Questions like this come up all the time on the SAT II. Answering it is a simple matter of applying Boyle's Law, or remembering that pressure and volume are inversely proportional in the ideal gas low. If volume, is halved, pressure is doubled.

Front 29

Example 2:
-A gas in a closed container is heated from OC to 273C. How does this affect the pressure of the gas on the walls of the container?

Back 29

-Temperature goes from 273K to 546K (temperature doubles).
-Closed Container: we know the volume remains constant. Because pressure and temperature are directly proportional, we know that if the temperature is doubled, then the pressure is doubled as well. This is why it's a really bad idea to heat an aerosol canister.

Front 30

Heat Engines (219-220)

Back 30

PART 3/4 (2-3%)

Front 31

Heat Engines

Back 31

-operates by taking heat from a hot place, converting some of that heat into work, an dumping the rest in a cooler heat reservoir.
-For example, the engine of a car generates heat by combusting gasoline. Some of that heat drives pistons that make the car do work on the road, and some of that heat is dumped out the exhaust pipe.

Front 32

Internal Energy U

Back 32

-assume that a heat engine starts w/ a certain internal energy U, intakes heat ΔQ(in) from a heat source at temperature T(in), does work ΔW, and exhausts heat ΔQ(out) into the cooler heat reservoir w/ temperature T(out). With a typical heat engine, we only want to use the heat intake, no the internal energy of the engine, to do work, so ΔU=0. The First Law of Thermodynamics tells us:
ΔU=0=ΔQ(in)-ΔQ(out)-ΔW

Front 33

Efficiency, e
e=ΔW/ΔQ(in)= ΔQ(in)-ΔQ(out)/ΔQ(in)= 1-(ΔQ(out)/ΔQ(in))

Back 33

-to determine how effectively an engine turn heat into work, we define the efficiency, as a ratio of work done to heat input.
-Because the engine is doing work, we know that ΔW>0, so we can conclude that ΔQ(in)<ΔQ(out). Both ΔQ(in) and ΔQ(out) are positive, so the efficiency is always b/w 0 and 1:
0≤e≤1.

Front 34

Heat Engine Problem:
•80 J of heat are injected into a heat engine, causing it to do work. The engine then exhausts 20 J of heat into a cool reservoir. What is the efficiency of the engine?

Back 34

-If we know our formulas, this problems is easy. The heat into the system is ΔQ(in)=80J and the heat out of the system is ΔQ(out)=20J. The efficiency, then, is: 1-20/80= 0.75= 75%.

Front 35

The Laws of Thermodynamics (p.216-218)

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PART 4/4 (1%)

Front 36

The Laws of Thermodynamics

Back 36

-the study of what compels heat to move in the way that it does. The Laws of Thermodynamics give us the whats and whys of heat flow.
-The SECOND and FIRST LAWs are the most important.
-Questions on the Laws of Thermodynamics will probably be qualitative: as long as you understand what these laws mean, you probably won't have to do any calculating.

Front 37

Zeroth Law (Not Too Important)

Back 37

-If system A is at thermal equilibrium w/ system B, and B is at thermal equilibrium w/ system C, then A is at thermal equilibrium with C.
-Same Temperature means Thermal Equilibrium.
-the significant consequence of the Zeroth Law is that, when a hotter object and a colder object are placed in contact w/ one another, heat will flow from the hotter object to the colder object until they are in thermal equilibrium.

Front 38

FIRST LAW (SAT II-WILL COME UP)
ΔU=ΔQ+ΔW
-U signifies internal energy
-Q signifies heat
-W signifies Work

Back 38

-tells us that the internal energy of a system increases if heat is added to the system or if work is done on the system, and decreases if the system gives off heat or does work.
-Isolated System: has INTERNAL ENERGY U, which is related to the KE of the molecules in the system, and therefore to the system's temperature. Internal energy is similar to PE in that it is a property of a system that is doing no work, but has the potential to do work.
-just another way of stating the law of conservation of energy. Both heat and work are forms of energy, so any heat or work that goes into or out of a system must affect the internal energy of that system.

Front 39

First Law Problem:
-Some heat is added to a gas container that is topped by a movable piston. The piston weighed down w/ a 2kg mass. The piston rises a distance of 0.2m at a constant velocity. Throughout this process, the temperature of the gas in the container remains constant. How much heat was added to the container?

Back 39

-The temperature of the container remains constant. ΔU=0
-1st Law: the force the expanding gas exerts to push the piston upward must be equal and opposite to the force of gravity pushing the piston downward.

•Piston is Weighted down by a 2kg mass:
F=mg= (2kg)(9.8 m/s(sq))= 19.6N

-Gas Exerts a Force of 19.6N Upwards
•W=Fd= (19.6N)(0.2m)= -3.92J
-The 3.92 is negative since work is done by the system rather than on the system.
-Because ΔU=0, we can conclude that ΔQ= 3.92 J, so 3.92 J of heat must have been added to the system to make the piston rise as it did.

Front 40

The Second Law in Terms of Heat Flow

Back 40

-heat flows spontaneously from a hotter object to a colder one.

Front 41

The Second Law in Terms of Heat Engines

Back 41

-no machine can work at 100% Efficiency: all machines generate some heat, and some of that heat is always lost to the machine's surroundings.

Front 42

The Second Law in Terms of Entropy

Back 42

-entropy is a technical term for talking about disorder.
-ordered systems are liable to fall into disorder, but disordered systems are not liable to order themselves spontaneously.
-drop a jar breaks it but no action that simple can fix it.

Front 43

Third Law (Not Too Important)

Back 43

-Though the universe tends toward maximum entropy, this entropy state is never actually reached.