Thermal Physics

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Thermal Physics

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Thermal Physics · Temperature, Thermal Equilibrium and Thermometers · Thermal Expansion · Heat and Temperature Change · Thermal Equilibrium : Heat Calculations · Phase Transitions · Heat Transfer · Gas Laws Temperature, Thermal Equilibrium · Kinetic Theory Click on the topic to go to that section · Internal Energy and Thermometers · Work in Thermodynamics · First Law of Thermodynamics · Thermodynamic Processes · Second Law of Thermodynamics · Heat Engines · Entropy and Disorder Return to Table of Contents Slide 5 / 163 Slide 6 / 163

Here are some definitions of temperature: In everyday language, many of us use the terms temperature · A measure of the warmth or coldness of an object or substance with and heat interchangeably reference to some standard value. But in physics, these terms have very different meanings. · Any of various standardized numerical measures of this ability, such Think about this… as the Kelvin, Fahrenheit, and Celsius scale. · When you touch a piece of metal and a piece of wood both · A measure of the ability of a substance, or more generally of any resting in front of you, which feels warmer? · When do you feel warmer when the air around you is 90°F physical system, to transfer heat energy to another physical system. and dry or when it is 90°F and very humid? · A measure of the average kinetic energy of the particles in a sample · In both cases the temperatures of what you are feeling is of matter, expressed in terms of units or degrees designated on a the same. Why then are you feeling a difference? standard scale.

In this unit, we will learn about temperature, heat, and the laws of thermodynamics that relate heat, mechanical work We'll consider each of these definitions in this unit. and other forms of energy. Slide 7 / 163 Slide 8 / 163

Thermometers and Thermal Equilibrium Thermometers

To measure temperature of a substance, we need...

1. A measuring device (Thermometer) that changes visibly and is calibrated to a scale. Thermometers relate the change in a physical property of substance to temperature. Examples include: > The change of volume of a gas or liquid > The change in length of a metal strip or wire > The light or infrared radiation emitted by an object

2. To bring the Thermometer into contact with the substance > When the thermometer has settled on a value, we say that the thermometer and the substance are in Thermal Equilibrium

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Temperature Scales Temperature Scales

Recall that temperature can be defined as...

· a measure of the warmth or coldness of an object or substance with reference to some standardized numerical measures or scales

Three common scales are: CO2 Solidifies · Fahrenheit (°F) - used mainly in the USA O2 Liquifies · Celsius (°C) - used in most of the world · Kelvin (K) - used in the physical sciences > also known as the Absolute Temperature Scale Kelvin Celsius Fahrenheit

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Temperature Conversions 1 Which temperature scale does not have negative values?

A Fahrenheit Celsius # Fahrenheit B Celsius

C Kelvin

D All the above Celsius # Kelvin E None of the above Slide 12 (Answer) / 163 Slide 13 / 163

1 Which temperature scale does not have negative values? 2 Water freezes at 32°F. What temperature would this be on the Celsius scale?

A Fahrenheit A 32 °C B Celsius B 0° C C Kelvin C 25° C D All the above C Kelvin D 212°C

E None of the above Answer E 100° C

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2 Water freezes at 32°F. What temperature would this be on the 3 Water boils at 100° C. What temperature would this be on the Celsius scale? Fahrenheit scale?

A 32 °C A 32 °F

B 0° C B 100° F

C 25° C C 0° F B 0°C D 212°C D 212° F Answer E 100° C E 180° F

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3 Water boils at 100° C. What temperature would this be on the 4 “Room temperature” is often taken to be 68 °F; what is this on the Fahrenheit scale? Celsius scale?

A 32 °F A 34 ° C

B 100° F B 37.78° C

C 0° F C 5.78° C

D 212° F D 212°F D 20° C Answer E 180° F E 52° C

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4 “Room temperature” is often taken to be 68 °F; what is this on the 5 The coldest temperature recorded on earth was −89.2 °C at the Celsius scale? Soviet Vostok Station in Antarctica, on July 21, 1983. What would a Fahrenheit scale thermometer have measured?

A 34 ° C

B 37.78° C

C 5.78° C D 20°C D 20° C Answer use TC = (TF - 32) /1.8 E 52° C

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5 The coldest temperature recorded on earth was −89.2 °C at the 6 The coldest temperature recorded on earth was −89.2 °C at the Soviet Vostok Station in Antarctica, on July 21, 1983. What would a Soviet Vostok Station in Antarctica, on July 21, 1983. What would a Fahrenheit scale thermometer have measured? Kelvin scale thermometer have measured?

TF = 1.8TC + 32 = 1.8(89.2) + 32 Answer = - 128.6 °F

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6 The coldest temperature recorded on earth was −89.2 °C at the Thermal Equilibrium: The Zeroth Law of Soviet Vostok Station in Antarctica, on July 21, 1983. What would a Kelvin scale thermometer have measured? Thermodynamics

Two objects placed in thermal contact will eventually come to the same temperature. When they do, we say they are in thermal equilibrium.

The zeroth law of thermodynamics says that if two objects are each in equilibrium with a third object, they are also in thermal

TK = TC + 273 equilibrium with each other. = -89.2 + 273 Answer = 183.8 K

T3 T1 T2

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That is if T1 = T3 and T2 = T3 then T1 = T2 Slide 19 / 163 Slide 20 / 163

Thermal Conductors and Insulators 7 Three objects A, B, and C initially have different temperatures TA>TB>TC. Objects A and B are separated by an insulating plate but they are in contact with object C through a conducting plate. Which of the following is true when objects A and B reach thermal equilibrium with object C? Conductors - materials that allow heat to flow easily (metals)

Insulators - materials that slow or block heat flow (wood, plastic, fiberglass)

A The temperatures of all three objects do not change

B Object A has a higher temperature than Object B and Object C

C Object C has a higher temperature than Object A and Object B

D Object B has a higher temperature than Object A and Object C

E All three objects have the same temperature

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7 Three objects A, B, and C initially have different temperatures TA>TB>TC. Objects A and B are separated by an insulating plate but they are in contact with object C through a conducting plate. Which of the following is true when objects A and B reach thermal equilibrium with object C?

E All three objects have the same Thermal Expansion A The temperatures of all three objectstemperature do not change Answer

B Object A has a higher temperature than Object B and Object C

C Object C has a higher temperature than Object A and Object B [This object is a pull tab]

D Object B has a higher temperature than Object A and Object C

E All three objects have the same temperature Return to Table of Contents Slide 22 / 163 Slide 23 / 163

Thermal Expansion Linear Expansion

Suppose a rod composed of some substance has a length L0 at · Most materials expand when their temperatures increase. an initial temperature of T0. · Liquids expand in a thermometer. If the temperature is changed by ∆T, the length changes by ∆L. · A tight metal jar lid can be loosened by running it in hot water. If ∆T is not too large, ∆L is directly proportional to ∆T. · These are examples of thermal expansion. L0

T0 · We consider two types of thermal expansion: T0+∆T

> Linear L = L0+∆L > Volume The change in length is ∆L = α L0 ∆T

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Linear Expansion 8 A steel rod measures 10 meters at 0° C. Given that the coefficient of linear expansion of steel is 1.2 x 10-5 per °C, what will the rod measure at 75°C? Substance Coefficient of Linear Exansion α (×10-6 / °C) Aluminum 23.1 Diamond 1

Copper 17 Glass 8.5 Iron 11.8

Gold 14

Steel 13.2 Ice 51

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8 A steel rod measures 10 meters at 0° C. Given that the coefficient 9 A simple pendulum is made of a steel string supporting a brass sphere. The of linear expansion of steel is 1.2 x 10-5 per °C, what will the rod temperature in a room with the pendulum is increased from 15° C to 30° C. measure at 75°C? Which of the following is true about the period of oscillations?

A the period doubles

B the period does not unchanged

L = L0 + ∆L C the period slightly increases = L0 (1+α∆T)

Answer = 100(1+0.000012*75) =100.09 m D the period slightly decreases

E the period increases by √2

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9 A simple pendulum is made of a steel string supporting a brass sphere. The Volume Expansion temperature in a room with the pendulum is increased from 15° C to 30° C. Which of the following is true about the period of oscillations? Suppose a volume of some substance (gas, liquid or solid) has a length V0 at an initial temperature of T0. A the period doubles If the temperature is changed by ∆T, the volume changes by ∆V. B the period does not unchanged If ∆T is not too large, ∆V is directly proportional to ∆T. C C the period slightly increases · The absolute T increases from D the period slightly decreases288K to 303K. The change in volume is Answer · So L increases slightly ∆V = βV0 ∆T E the period increases· Whichby √2 means the period 2π√(L/g) increases slightly where β is the coefficient of volume expansion.

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Volume Expansion 10 A glass flask is filled to the brim with glycerin, both at the same initial temperature. When the temperature of the flask and glycerine is increased by a few degrees, which of the following occurs? -6 -5 -1 Substance Coefficient of Volume Exansion β (×10 / °C) The coefficients of volume expansion are: βglycerin = 49x10 K and -5 -1 βglass = 2x10 K . Gasoline 950

Glycerine 485 A the level of the glycerine in the flask goes down Water 207 B the level of the glycerin in the flask remains unchanged Aluminum 69 Diamond 3 C the glycerine overflows

Glass 9.9 D there isn't enough information given to answer this question Steel 32.4 Gold 42

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10 A glass flask is filled to the brim with glycerin, both at the same Volume Expansion of Water initial temperature. When the temperature of the flask and glycerine is increased by a few degrees, which of the following Water above 4°C water expands occurs? -5 -1 when heated. The coefficients of volume expansion are: βglycerin = 49x10 K and -5 -1 βglass = 2x10 K . BUT in the temperature range from to 0°C to 4°C, the volume decreases as temperature increases. A the level of the glycerine in the flaskC goes down This means that 4°C water is more dense than 0°C water. B the level of the glycerin· The glycerinin the flask has remains a much greaterunchanged Hence water has its greatest density coefficient of volume expansion at 4°C.

C the glycerine Answer overflowsthan the glass. Consequently when the temperature rises, the 4°C water will sink to the glycerine will overflow. bottom of a lake, so ice will D there isn't enough information given to answer this question form in the 0°C water floating on top. [This object is a pull tab] This protects life in bodies of freshwater since the water on the bottom will be at worst 4°C; not freezing! Slide 31 / 163 Slide 32 / 163

Heat

When you pour hot water into a cold cup, the water cools down and the cup warms up as they approach thermal equilibrium. The reason for these temperature changes is that... Energy flows from the the higher temperature object to the lower temperature object. This flow of energy is called HEAT. Heat and Temperature Change Q is the symbol for HEAT

Heat and Temperature are Different · Temperature is a quantitative measure of an object's hotness or coldness. · Heat is the energy that moves from one object to another because of a temperature difference. Return to Table of Contents Slide 33 / 163 Slide 34 / 163

Units of Heat Units of Heat

Other common units are:

· calorie (cal) Because heat is a form of energy... > the amount of heat required to raise the temperature of 1 gram of water by 1°C. > 1 cal = 4.186 J

· kcal or Calorie (food calorie) the SI unit for heat is the joule (J) > 1 kcal = 1000 cal = 4186 J

· BTU (British Thermal Unit) > the quantity of heat required to raise the temperature of 1 pound of water by 1°F > 1 BTU = 252 cal = 1055 J

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Specific Heat Specific Heat

Consider the the cup of water shown below. Consider the cups below: #1 contains half the mass of water as #2.

T =10°C 10°C 10°C

#1 #1 #2

If it takes an amount of heat Q to raise the temperature of the Cup #1 by ∆T, If it takes an amount of heat Q to raise the temperature of the water to 20°C (a change of temperature ∆T = 10°C) It will take 2Q to raise the temperature of Cup #2 by the same ∆T. It will take 2Q to raise its temperature 30°C (by ∆T = 20°C) Heat required is directly proportional to the change in mass m as well as ∆T Heat required is directly proportional to the change in temperature Q ~ ∆T Q ~ m∆T

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Specific Heat Units for Specific Heat

Consider the objects below: a cup containing a mass m of water and a piece of copper of mass m

rearrange this for c: 10°C 10°C m m water Copper

Experiment tells us that it takes less heat to raise the temperature of so it follows that the units for specific heat are the copper than the water by the same amount

Heat required depends on the nature of each substance. The quantity that adjusts for substance is call the Specific Heat (c). So... Q = mc∆T Slide 39 / 163 Slide 40 / 163

Specific Heat 11 How much heat is required to raise the temperature of 0.5 kg of aluminum (c = 837 J/kg°C) from 15°C to 40°C? Specific Heat Substance (J/kg°C)

water (H2O) 4186 ethylene glycol (anti-freeze) 2386

ice (H2O) 2093 aluminum (Al) 837 copper (Cu) 419 gold (Au) 126

Notice that water has the highest heat capacity in the table. Challenge: Can you find a material with a higher heat capacity? Note: Metals have much lower heat capacities than water, so... > It takes less heat to raise the temperature of a mass of metal by ∆T than it takes to raise the same mass of water by ∆T. Slide 40 (Answer) / 163 Slide 41 / 163

11 How much heat is required to raise the temperature of 0.5 kg of 12 It takes 2 minutes to raise the temperature of 1 liter of water by aluminum (c = 837 J/kg°C) from 15°C to 40°C? 50°C with a hot plate. How much time would it take to raise the temperature of 2 liters of water 50°C using the same hot plate?

A ½ minute

Q = mc∆T B 1 minute =(0.5)(837)(40-15) =10,462 J C 2 minutes Answer

D 4 minutes

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12 It takes 2 minutes to raise the temperature of 1 liter of water by 13 The ocean temperature doesn't change drastically because of 50°C with a hot plate. How much time would it take to raise the temperature of 2 liters of water 50°C using the same hot plate? A Water is a good heat conductor B Water has a high boiling point

A ½ minute C Water has a high specific heat C 2 minutes D Water has a low melting temperature B 1 minute

Qoriginal = m c ∆T C 2 minutes Qnew = 2m c ∆T Answer D 4 minutes Qnew = 2Qoriginal same hotplate ➜ same power (energy/time) 2Q takes[This object twice is a pullthe tab] time Slide 42 (Answer) / 163 Slide 43 / 163

13 The ocean temperature doesn't change drastically because of 14 A solid copper ball, a solid silver ball and a solid aluminum ball, all having the same mass and at room temperature, are placed in a A Water is a good heat conductor 300°C oven at the same time. Which of the three will increase in temperature fastest? (hint: look up the specific heats in the table B Water has a high boiling point provided earlier) C Water has a high specific heat D Water has a low melting temperature A the aluminum ball C Water has a high specific heat B the copper ball Answer

C the silver ball

[This object is a pull tab] D they all increase temperature at the same rate

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14 A solid copper ball, a solid silver ball and a solid aluminum ball, all 15 A solid copper ball, a solid silver ball and a solid aluminum ball, all having the same mass and at room temperature, are placed in a having the same mass and a temperature of 200°C are placed on a 300°C oven at the same time. Which of the three will increase in huge block of ice at 0°C. Which ball will melt the most ice? temperature fastest? (hint: look up the specific heats in the table provided earlier)

A The silver ball

A the aluminum ball B The copper ball C B the copper ball Silver has the lowest specific heat, C The aluminum ball Answer so it takes the least amount of C the silver ball energy of the three materials to D They all melt the same amount of ice increase its temperature D they all increase temperature at the same rate [This object is a pull tab]

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15 A solid copper ball, a solid silver ball and a solid aluminum ball, all having the same mass and a temperature of 200°C are placed on a huge block of ice at 0°C. Which ball will melt the most ice?

A The silver ball

B The copper ball C Thermal Equilibrium : Heat Aluminum has the greatest heat C The aluminum ball capacity. Therefore has the most Answer heat energy to give up per degree Calculations D They all melt the same amount of ice

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Thermal Equilibrium and Conservation of Heat Calculations

Energy Example 1: Two equal masses of water at different temperatures are When two objects (isolated from their surroundings) are brought in mixed together. What is the equilibrium temperature TF? contact with one another, we know that heat will flow from the hotter object to the colder object until the objects reach thermal equilibrium. T0=90°C T0=10°C heat TF =?

hot cold heat same T before contact contact thermal equilibrium

before contact contact thermal equilibrium

Because there is no where else for the heat to go, and Because heat is energy and energy is conserved... Heat Lost from hotter object + Heat Gained by colder object = ZERO

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Heat Calculations 16 A styrofoam cup containing 400 g of 60°C water is poured into another styrofoam cup containing 800 g of 15°C water. What is the Example 1: Two equal masses of water at different temperatures are temperature of the combination? mixed together. What is the equilibrium temperature TF? A 75.0°C

T0=90°C T0=10°C heat TF =? B 45.0°C T0=60°C T0=15°C before contact contact thermal equilibrium before contact C 37.5°C

Q Lost from 90°C water + Q Gained by 10° water = 0 D 30.0°C

(mc∆T)90°C water + (mc∆T)10°C water = 0 E 15.0°C TF =? (TF - 90) + (TF - 10) = 0 thermal equilibrium 2TF - 100 = 0

TF = 50°C

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16 A styrofoam cup containing 400 g of 60°C water is poured into 17 Two vials of mercury sit on a chemist's desk. One vial contains another styrofoam cup containing 800 g of 15°C water. What is the 80 g of mercury at 20°C while the other vial contains 20 g of temperature of the combination? mercury at 60°C. She then pours one vial into the other. What will the final temperature of the mercury be? (The specific heat of mercury is 0.14 J/g) A 75.0°C D 30 °C T0=60°C T0=15°C B 45.0°C Q1 + Q2 = 0 A 40 °C Q = mc∆T before contact C 37.5°C since materials are the same c can B 67 °C be eliminated D 30.0°C Answer m1(Tf - T1) + m2(Tf - T2) = 0 C 21 °C TF =? E 15.0°C 400(Tf - 60) + 800(Tf - 15) =0 D 28 °C 1200Tf - 36000 = 0 thermal equilibrium [ThisTf object= 30 is a°C pull tab] Slide 50 (Answer) / 163 Slide 51 / 163

17 Two vials of mercury sit on a chemist's desk. One vial contains Calorimetry 80 g of mercury at 20°C while the other vial contains 20 g of mercury at 60°C. She then pours one vial into the other. What Often, a calorimeter (heat measuring will the final temperature of the mercury be? (The specific heat apparatus) is used to find the initial of mercury is 0.14 J/g) temperature, specific heat or other thermal properties of a substance. D 28°C A simple Calorimeter is composed of a Q1 + Q2 = 0 A 40 °C container that is insulated from the outside Q = mc∆T environment (so practically no heat can B 67 °C since materials are the same c can enter or leave). be eliminated

Answer For our simple "experiments", the C 21 °C m1(Tf - T1) + m2(Tf - T2) = 0 calorimeter's inner cup may be made of water 80(Tf - 20) + 20(Tf - 60) =0 metal (Aluminum). The calorimeter will be Aluminum D 28 °C inner cup 100Tf - 2800 = 0 filled with a quantity of water. We will drop [ThisTf object= 28 is a°C pull tab] our sample substance from the top before Styrofoam inserting the stopper. outer cup

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Calorimetry 18 A 0.1 kg aluminum calorimeter is filled with 0.5 kg of water, both at Example 2: A 0.4 kg Aluminum calorimeter is filled with 0.8 kg of 0°C. A 0.1 kg sample of an unknown substance at 210°C is dropped into the filled calorimeter. When the combination reaches water. The calorimeter and water come to thermal equilibrium at a thermal equilibrium, the temperature is 15°C. What is the specific temperature of T0 = 20°C. 0.1 kg of a material (U) with a specific heat heat of the unknown substance? of 628 J/(kg°C) and an initial temperature of 300°C is dropped into the calorimeter. Find the equilibrium temperature of the combination. Substance (U) Aluminum can (Al) Water (w) m = 0.1 kg m = 0.4 kg m = 0.8 kg U T0 = 300°C T0 = 20°C T0 = 20°C c = 628 J/(kg°C) c = 837 J/(kg°C) c = 4186 J/(kg°C) ∑Q=0 Al QU + Qw + QAl = 0 substitute Q=mc∆T for each component: (mc∆T)U + (mc∆T)w + (mc∆T)Al = 0 w substitute ∆T = TF - T0 for each component: 0.1(628)(TF–300) + 0.8(4186)(TF–20) + 0.4(837)(TF–20) = 0 solve for TF: 3746.4TF = 92,511 or TF = 24.69°C

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18 A 0.1 kg aluminum calorimeter is filled with 0.5 kg of water, both at 0°C. A 0.1 kg sample of an unknown substance at 210°C is dropped into the filled calorimeter. When the combination reaches thermal equilibrium, the temperature is 15°C. What is the specific heat of the unknown substance?

Q1 + Q2 + Q3 = 0 Q = mc∆T c3 is unknown Phase Transitions m1c1(Tf - T1) + m2c2(Tf - T2) + m3c3(Tf - T3) = 0

0.1(837)(15-0) + 0.5(4186)(15-0) + 0.1c3 Answer (15-210) = 0

c3 = 1674.4 J/(kg•K)

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Phase Transitions Latent Heat of Fusion

The term phase refers to a specific state if matter, such as... We will focus on phase transitions in water. · solid · liquid Ice melts at 0°C. For 1 kg of 0°C ice to change completely to water at · gas 0°C requires the addition of 334 kJ of heat.

When a substance undergoes a change from one state to another, it Similarly to freeze 1 kg of 0°C water into 0°C ice requires the removal is called a phase transition. of 334 kJ of heat.

Phase transitions (at a given pressure) take place at a constant The heat added to change a solid to liquid -or- removed to change a temperature, are accompanied by addition or removal of heat, and may involve a change in volume (density). Examples include liquid to a solid is called the latent heat of fusion (Lf). • water freezing • ice melting For water: Lf = 334 kJ/kg • water vaporizing • steam condensing

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Latent Heat of Vaporization Phase Transitions of Water

We will focus on phase transitions in water. T (°C)

steam Water vaporizes (boils) at 100°C. For 1 kg of 100°C water to change Lv = 2260 kJ/kg completely to steam at 100°C requires the addition of 2260 kJ of heat. water and steam

Similarly to condense 1 kg of 100°C steam into 100°C water requires water the removal of 2260 kJ of heat. c = 4.186 kJ/(kg°C)

The heat added to change a liquid to gas -or- removed to change a ice water and ice gas to a liquid is called the latent heat of vaporization ( v). Q (kJ/kg) L Lf = 334 kJ/kg

For water: Lv = 2260 kJ/kg · H2O transforms from ice to liquid water to steam as heat is added · During phase transitions the T-Q graph is horizontal - T is constant while heat is added · While in a single phase (ice or water or steam), the temperature rises as heat is added Slide 59 / 163 Slide 60 / 163

Phase Transitions and Internal Energy 19 As water vapor condenses

Notice that when a phase transition is taking place, the A The temperature increases Temperature remains constant. B The temperature decreases C Energy is absorbed Where is the energy going to or coming from? D Energy is released E None from the above The answer is that is going into making or breaking bonds between the atoms (or molecules) of the material.

For instance when ice melts, the bonds that hold the water molecules in place are broken with the addition of energy.

When that energy is removed from water at 0°C, the molecules bond again, crystalizing into ice. Slide 60 (Answer) / 163 Slide 61 / 163

19 As water vapor condenses 20 Ice is placed in an empty container in a room where the air temperature is 20°C and allow to melt. While the ice is melting the A The temperature increases temperature of the water is… B The temperature decreases A less than 0°C C Energy is absorbed D D Energy is released B 0°C Energy is released E None from the above (The temperature remains constant) C room temperature Answer D greater than 0°C but less than room temperature

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20 Ice is placed in an empty container in a room where the air 21 How much heat must be provided to melt a 0.3 kg chunk of ice, temperature is 20°C and allow to melt. While the ice is melting the then raise the temperature of the melt water to 40°C? temperature of the water is…

A less than 0°C

B 0°C B 0°C C room temperatureduring phase transition, the temperature remains constant D greaterAnswer than 0°C but less than room temperature

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21 How much heat must be provided to melt a 0.3 kg chunk of ice, Calorimetry then raise the temperature of the melt water to 40°C? Example 3: A 0.5 kg Aluminum calorimeter is filled with 1 kg of water. The calorimeter and water come to thermal equilibrium at a temperature of T0 = 80°C. A 0.1 kg ice cube (at 0°C) is dropped into the calorimeter. Find the equilibrium temperature of the combination.

Qtotal = Q to melt ice + Q to heat water Ice (I) Aluminum can (Al) Water (w) m = 0.1 kg m = 0.5 kg m = 1.0 kg Qtotal = mLf + mc∆T LF = 334,000 J/kg T0 = 80°C T0 = 80°C I = 0.3(334,000) + 0.3(4186)(40-0) c = 837 J/(kg°C) c = 4186 J/(kg°C) = 150,432 J The ice melts and then the melt water absorbs heat, Answer The calorimeter and the water it contains lose heat Al ∑Q=0 QI,melting + QI,as water + Qw + QAl = 0 w [This object is a pull tab] mILf + (mc∆T)I + (mc∆T)w + (mc∆T)Al = 0

0.1•334,000 + 0.1•(4186)(TF–0) + 1•4186(TF–80) + 0.5(837)(TF–80) = 0

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Heat Transfer

Earlier we defined thermal insulators and conductors. Now we are going to examine the mechanisms of heat transfer and the rates of heat transfer.

Mechanisms of Heat Transfer Heat Transfer There are three mechanisms of heat transfer: · conduction · convection · radiation

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Conduction Conduction

As indicated earlier. Temperature is directly related to the average kinetic energy of the the molecules in a substance.

The rate of heat transfer (∆Q/∆t) depends:

· Directly on the relative temperature at both ends (T2 – T1)

In a warmer region of an object (T2), the molecules have a higher > if no difference - no heat transfer average kinetic energy than in a cooler region (T1) · Directly on the cross-sectional area (A) available for collisions to Conduction is the flow of heat due to the transfer of kinetic occur energy in molecular collisions within the object. · Inversely as the length (L) along which the heat must pass · Directly on the conductivity (k) - ease of heat transfer - of the material

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22 An iron tube with a length of 1 m and a radius of 10 cm is heated 22 An iron tube with a length of 1 m and a radius of 10 cm is heated from the bottom by a lighter. If the bottom of the tube has a from the bottom by a lighter. If the bottom of the tube has a temperature of 35 degrees Celsius and the top of the tube has a temperature of 35 degrees Celsius and the top of the tube has a temperature of 20 degrees Celsius, what is the rate of heat transfer temperature of 20 degrees Celsius, what is the rate of heat transfer through the tube? (k=80) through the tube? (k=80)

A A 37.69 W A 37.69 W

B 376,991 W B 376,991 W ∆Q/∆t = 80(π 0.102)(35-20)/1 Answer C 84.26 W C 84.26 W = 37.69 W

D 842,661 W D 842,661 W

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23 Two objects of different temperatures are separated by a wall. If 23 Two objects of different temperatures are separated by a wall. If the thickness (L) of the wall is doubled, therate of heat transfer the thickness (L) of the wall is doubled, therate of heat transfer due will be... due will be...

A Doubled A Doubled B Quadrupled B Quadrupled D C Unchanged C Unchanged D Cut to one-half D Cut to one-half E Cut to one-fourth E Cut Ifto L one-fourth is doubled, ∆Q/∆t is halved Answer

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Convection 24 Convection can occur

Convection is the process where heat is A Only in solids transferred by the mass movement of molecules in gases and liquids from one B Only in liquids place to another. C Only in gasses Warm masses rise while cold masses fall. D Only in liquids and gasses This mechanism is important not only for E In solids, liquids, and gasses the examples shown below, but also

weather, ocean currents, and the transfer Convective heat transfer. (2015, June 14). In Wikipedia, The Free Encyclopedia. Retrieved 12:31, July 13, 2015, from https://en.wikipedia.org/w/index.php? of heat inside stars. title=Convective_heat_transfer&oldid=666858777

Convection in gasses Convection in liquids Slide 71 (Answer) / 163 Slide 72 / 163

24 Convection can occur Radiation

A Only in solids Radiation is energy transfer by electromagnetic waves. B Only in liquids C Only in gasses You have directly experienced it as the D warming you feel from the sun or sitting D Only in liquids and gasses close to a fire. E In solids, liquids,convection and gasses can only occur where the molecules of a substance are As you will learn later in the Electromagnetic Waves Unit, Answer free to move translationally (gasses and liquids) · radiation includes radio waves, microwaves, visible light and more, not just the infrared radiation you feel [This object is a pull tab] · does not matter to transfer heat - it can travel through a vacuum

Thermal radiation. (2015, May 22). In Wikipedia, The Free Encyclopedia. Retrieved 12:35, July 13, 2015, from https://en.wikipedia.org/w/index.php? title=Thermal_radiation&oldid=663541214 Slide 73 / 163 Slide 74 / 163

Radiation 25 A block of ice has a temperature of 0 °C and a surface area of 1 m2. What is the rate of heat transfer by radiation? (Ice has an emissivity of 0.97) The rate of heat transfer by radiation is given by the Stefan- Boltzmann equation.

A 0 W

where B 413 W · e - emissivity, is a number between 0 and 1 that increases with the darkness of the surface of an object C 305 W · σ - the Stefan-Boltzmann constant D 612 W · A - the surface area of the object

The rate at which an object radiates energy is proportional to the fourth power of the absolute temperature.

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25 A block of ice has a temperature of 0 °C and a surface area of 1 m2. What is the rate of heat transfer by radiation? (Ice has 26 When the temperature of a heater is doubled, by what factor does an emissivity of 0.97) the radiating power change?

A 2 B 4 A 0 W D C 8 B 413 W D 16 E 32 C 305 W∆Q/∆t = 0.97 (5.67x10-8)(1)(273)4 Answer = 305 W D 612 W

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26 When the temperature of a heater is doubled, by what factor does 27 Which of the following is responsible for raising the temperature of the radiating power change? water in a pot placed on a hot stove?

A 2 A Conduction B 4 B Convection D C 8 C Radiation D 16 D Vaporization

E 32 if T is doubled, E Condensation Answer ∆Q/∆t will increase by 24 = 16

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27 Which of the following is responsible for raising the temperature of water in a pot placed on a hot stove?

A Conduction B Convection C Radiation A and B D Vaporization the water in contact with the pot is E Condensation heated by conduction. Gas Laws

Answer this heat is transmitted to the rest of the water in the pot by convection.

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Gas Laws Boyle's Law

Robert Boyle performed a set of experiments measuring the Physical properties of gases include pressure (P), volume (V) and volume (V) of gases as the pressure (P) was changed. temperature (T). The relationships between these quantities were studied by a Boyle's Law... "The pressure exerted on an ideal gas is inversely proportional to progression of scientists: the volume it occupies (if the temperature and amount of gas remain unchanged within a closed system)"

- or -

Boyle (17th century) Charles (18th century) Gay-Lussac (19th century) Boyle's law. (2015, June 14). In Wikipedia, The Free Encyclopedia. Retrieved 14:48, July 9, 2015, from https://en.wikipedia.org/w/index.php?title=Boyle%27s_law&oldid=666852786 Slide 80 / 163 Slide 81 / 163

Boyle's Law 28 Which of the following graphs represents the isothermal process?

"The pressure exerted on an ideal gas is inversely proportional to A B C the volume it occupies (if the temperature and amount of gas remain unchanged within a closed system)"

constant

D E

Note Boyle's Law assumes an isothermal (constant T) process.

Different temperatures give different P-V diagrams. Slide 81 (Answer) / 163 Slide 82 / 163

28 Which of the following graphs represents the isothermal process? 29 A container filled with an ideal gas at pressure P is compressed to one-fourth of its volume while the temperature is kept constant. A B C What is the new pressure in the gas relative to its original pressure P?

A 2P E B 4P D E PV = nRT C P so P~1/V D 1/2P Answer E 1/4P

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29 A container filled with an ideal gas at pressure P is compressed to Charles's Law one-fourth of its volume while the temperature is kept constant. What is the new pressure in the gas relative to its original Jacques Charles performed a set of experiments measuring the pressure P? volume (V) of gases as the temperature (T) was changed. Charles's Law... A 2P "When the pressure of an ideal gas is kept constant, the volume is directly proportional to the absolute (Kelvin) temperature" B 4P B 4P Note this is an isobaric (constant P) process. same reason as last question C P PV = nRT D 1/2P so P~1/V

E 1/4P Answer - or -

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Charles's law. (2015, May 19). In Wikipedia, The Free Encyclopedia. Retrieved 15:40, July 9, 2015, from https://en.wikipedia.org/w/index.php?title=Charles%27s_law&oldid=663155562 Slide 84 / 163 Slide 85 / 163

Charles's Law 30 Which of the following graphs represents the isobaric process?

A B C "When the pressure of an ideal gas is kept constant, the volume is directly proportional to the absolute (Kelvin) temperature"

D E

Note Charles's Law assumes an isobaric (constant P) process.

Different pressures give different V-T diagrams. Slide 85 (Answer) / 163 Slide 86 / 163

30 Which of the following graphs represents the isobaric process? 31 An ideal gas is taken from one state at temperature T1=273 K to another state at temperature T2 = 546 K isobarically. What A B C happens to the volume of the ideal gas?

A It quadruples B It is cut to one-fourth B C It doubles D E D It is cut to a half isobaric means P = constant E It doesn't change during the isobaric process Answer

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31 An ideal gas is taken from one state at temperature T1=273 K to Gay-Lussac's Law another state at temperature T2 = 546 K isobarically. What happens to the volume of the ideal gas? Joseph Louis Gay-Lussac formulated the last three of the gas laws. A It quadruples "When the volume of an ideal gas is kept constant, the pressure is B It is cut to one-fourth C directly proportional to the absolute (Kelvin) temperature" C It doubles PV=nRT D It is cut to a half isobaric P= constant so V~T

E It doesn'tAnswer change during the isobaric process T doubles, so V doubles Note Gay-Lussac's Law assumes an isochoric (constant V) process.

[This object is a pull tab] Different pressures give different P-T diagrams.

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32 Which of the following graphs represents the isochoric process? 32 Which of the following graphs represents the isochoric process?

A B C A B C

C D E D E isochoric means V = constant Answer

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33 A sample of an ideal gas is enclosed in a container with rigid 33 A sample of an ideal gas is enclosed in a container with rigid walls. The temperature of the gas is raised from 20°C to 60°C. walls. The temperature of the gas is raised from 20°C to 60°C. What happens to the pressure of the gas? What happens to the pressure of the gas?

A It doubles A It doubles B It quadruples B It quadruples E C It triples C It triples D It is cut to one-third D It is cut to one-third PV=nRT E It is slightly increased E It is slightly increased rigid walls ➔ V = constant so P~T Answer T increases from 293K to 333K so P increases slightly

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The Ideal (or Combined) Gas Law 34 The number of moles of an ideal gas is doubled while the temperature and container volume remain the same. What happens to the pressure of the gas? The Boyle, Charles and Gay-Lussac Laws can be combined into a single general relationship among the pressure, volume, and A It doubles temperature a fixed quantity of gas. B It quadruples C It remains the same D -or- It is decreased to one-half E It is decreased to one-fourth The Ideal Gas Law where · n is the number of moles of gas · is the universal gas constant

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34 The number of moles of an ideal gas is doubled while the 35 An ideal gas is taken through a closed cycle A⇒ B⇒ C⇒ A. As temperature and container volume remain the same. What shown on the diagram. Which point is associated with the highest happens to the pressure of the gas? temperature?

A It doubles B It quadruples A C It remains the same PV=nRT A A D It is decreased to one-half T, V are constant B B E It is decreasedAnswer to one-fourth so P~n C C n doubles, so P doubles D The temperature is the same at A, B and C

[This object is a pull tab] E More information is required Slide 92 (Answer) / 163 Slide 93 / 163

35 An ideal gas is taken through a closed cycle A⇒ B⇒ C⇒ A. As shown on the diagram. Which point is associated with the highest temperature?

B A A

B B PV=nRT Kinetic Theory C C T is greatest when P x V is greatest Answer D The temperature is the same at A, B and C E More information is required

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Atoms and the Properties of Matter Kinetic Theory

The idea that all familiar matter is made up of atoms goes back to the ancient Greeks. From Dalton's principles, it followed that properties of materials are Democritus (460-370 BC) proposed that if one determined by the properties of the atoms in the material, the motion was to cut a piece of iron into smaller and of those atoms (or molecules), and their interactions. smaller portions, there would be a portion In fact, the properties of matter we have studied so far: thermal which could not be divided further. This expansion, melting, boiling, cooling, heating... can be explained smallest piece he called an atom (indivisible). based on the concept that matter is made up of atoms. Kinetic theory is the first of the theories of materials. It relates the Another of the scientists working with gases, motion of the atoms (or molecules) which comprise a gas to the John Dalton (1766-1844) reprised the theory of thermodynamic properties of the gas. atoms. His experiments led him to propose the Kinetic theory provides us with with a better understanding of: foundational postulates of modern chemistry. · temperature The first of these was that the elements · heat (fundamental materials) are made of atoms, and · internal energy that all atoms of a given element have the same · pressure physical properties (mass, size...).

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Kinetic Theory of Ideal Gases The Ideal Gas Model

Assumptions: The study of a real gas can be mathematically complicated. 1. The gas consists of a very large number of particles (atoms, molecules) in a container. We would have to consider: 2. The particles behave as point · the motion of the molecules in the gas particles; their size is small in · the motion of the atoms inside each molecule comparison to the average distance between particles and to the size of · whether the molecules repel or attract and even stick together the container. · whether they will combine to form different molecules. 3. The particles are in constant motion; they obey Newton's Laws of motion. Each particle collides occasionally with In our application of kinetic theory we will be using a set of a wall of the container. These simplifying assumptions called The Ideal Gas Model. collisions are perfectly elastic. 4. The walls of the container are rigid and very massive. Slide 98 / 163 Slide 98 (Answer) / 163

36 Which of the following is not included into the assumptions of the 36 Which of the following is not included into the assumptions of the ideal gas? ideal gas?

A The number of molecules in a container is verylarge A The number of molecules in a container is verylarge B The molecules interact when they collide with each other B The molecules interact when they collide with each other

C The molecules interact all the time during their motion because C The molecules interact all the time during their motion because of intermolecular forces of intermolecular forces C D The collisions between molecules are perfectly elastic D The collisions between molecules are perfectly elastic E The size of molecules can be ignored E The size of molecules can be ignored Answer

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Pressure Pressure

The first physical quantity we will model is with kinetic theory is Pressure. The momentum change during a single collision of a molecule with a wall of the Pressure is just the force per unit area on the walls of the container: container is just the momentum change perpendicular to the wall, ∆p x.

Recalling the impulse-momentum change theorem, we can find the force from the change of momentum: Since in the Ideal Gas Model, all collisions are elastic...

So we'll begin our derivation by finding ∆p and ∆t for a single molecule colliding with one of the very massive walls of its perfectly rigid container. Slide 101 / 163 Slide 102 / 163

Pressure Pressure

These collisions occur every time interval ∆t, The time between the collisions is very small. which is the time it takes for the molecule to travel from one side of the box to the other Combining what we've gathered so far... and back, a distance of 2L.

we find that... Slide 103 / 163 Slide 104 / 163

Pressure Pressure

To calculate the force due to all the The velocity in one direction can be related to the speed using molecules in the box, we have to add the the Pythagorean Theorem force contributions from each molecule.

Since all molecules move in random directions and there is no preference between x, y, and z we can write the following...

Defining the average value of the square of the x component of velocity The force on a wall is then

We find that the force on a wall due to all Recalling that P=F/A and that volume V=LA, we get the molecules is The Pressure in an ideal gas is directly proportional to the average square speed of the molecules Slide 105 / 163 Slide 106 / 163

The Ideal Gas Law and Average Kinetic Energy The Ideal Gas Law and RMS Velocity

Since Comparing and

we can rewrite our expression for pressure as We can find the "root-mean-square" velocity

Comparing this to the ideal gas law that emerged from experiments or · The higher the temperature, the faster Where k =1.38# 10-23 J/K, the Boltzmann constant. molecules move on average.

We find that:

· The average kinetic energy of molecules in a gas is directly proportional to the absolute temperature. · This is the most important result of kinetic theory. · The higher the temperature, the faster molecules move on the average. Slide 107 / 163 Slide 108 / 163

Kinetic Theory Summary 37 If the average kinetic energy of molecules is increased while the number of moles is kept constant, what happens to the pressure of · The pressure in the ideal gas is directly an ideal gas? proportional to the average square of the velocity of molecules. A It increases B It decreased · Temperature was explained on the microscopic C It remains constant level. · The average kinetic energy of molecules in a gas D It decreases and then increases is directly proportional to the absolute temperature. · The temperature can't be negative and can only E None from the above reach absolute zero when the average kinetic energy of molecules is zero.

· The average velocity of molecules depends on absolute temperature and molecular mass. · Increasing temperature causes molecules to move faster. · The lighter the molecule, the faster it moves. Slide 108 (Answer) / 163 Slide 109 / 163

37 If the average kinetic energy of molecules is increased while the 38 The average kinetic energy of molecules can be increased by number of moles is kept constant, what happens to the pressure of increasing which of the following? an ideal gas? A Pressure A It increases B Volume B It decreased A C Temperature C It remains constant D Number of moles D It decreases and then increases E All of the above E None from the above if N/V is kept constant but KE increases then P increases [This object is a pull tab]

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38 The average kinetic energy of molecules can be increased by 39 If the temperature of an ideal gas is increased from 300 K to 600 increasing which of the following? K, what happens to the average kinetic energy of the molecules?

A Pressure B Volume A it doubles C C Temperature B it quadruples D Number of moles C it reduced to 1/2 E All of the above D it is reduced to 1/4 if T is raised, KE is raised

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39 If the temperature of an ideal gas is increased from 300 K to 600 40 If the temperature of an ideal gas is increased from 25 C to 50 C, K, what happens to the average kinetic energy of the molecules? what happens to the average kinetic energy of the molecules?

A It doubles A it doubles B It quadruples A B it quadruples C It is cut to one-half D It is cut to one-fourth C it reduced to 1/2 E It slightly increases if theD itabsolute is reduced T is to doubled 1/4 KE is doubled

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40 If the temperature of an ideal gas is increased from 25 C to 50 C, 41 If the absolute temperature of an ideal gas is doubled, what what happens to the average kinetic energy of the molecules? happens to the average speed of the molecules?

A It doubles A It doubles B It quadruples B It quadruples C It is cutE to one-half C It increases by √2 D It is cut to one-fourth D It decreases by √2 E It slightly increases E It remains unchanged if T is increased from 298K to 323K KE is raised slightly

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41 If the absolute temperature of an ideal gas is doubled, what happens to the average speed of the molecules?

A It doubles B It quadruples C It increasesC by √2 D It decreases by √2 E It remains unchanged Internal Energy if absolute T is doubled

vrms is increased √(2)

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Internal Energy of an Ideal Gas

The internal energy of an ideal gas depends on temperature and the number of moles* (n) of gas.

An increase in temperature causes an increase in internal energy.

* A mole is a quantity of matter. It is defined as the 6.022 x 1023 molecules of a substance. (Defined as the number of atoms of carbon-12 that would have a mass of 12 grams). Slide 116 / 163 Slide 116 (Answer) / 163

42 The temperature of a monatomic ideal gas is increased from 42 The temperature of a monatomic ideal gas is increased from 35°C to 70°C. How does it change its internal energy? 35°C to 70°C. How does it change its internal energy?

A It doubles A It doubles B It quadruples B It quadruples C It is slightly increased C It is slightlyC increased D It is decreased to one-half D It is decreased to one-half E It is decreased to one-fourth E It is decreased to one-fourth

if T is increased from 308K to 343K U is increased slightly

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43 The state of an ideal gas is changed through the closed path 43 The state of an ideal gas is changed through the closed path 1⇒ 2⇒ 3⇒ 1. What happens to the internal energy of the gas 1⇒ 2⇒ 3⇒ 1. What happens to the internal energy of the gas between point 2 and point 3? between point 2 and point 3?

A It increases A It increases B It decreases B It decreases C It remains constant C It remainsC constant D It decreases and then increases D It decreases and then increases E It increases and then decreases E It increases and then decreases

2 ➔ 3 is isothermal U is constant

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Internal Energy 44 Which of the following is true about melting process?

A The energy is required to increase the average kinetic energy The concept of Internal Energy can be generalized to include the of molecules potential energy due to the forces between molecules as well as their individual kinetic energies. B The energy is required to decrease the average kinetic energy of molecules

C The energy is required to increase the potential energy between the molecules

D The energy is required to decrease the potential energy between the molecules

This more general equation enables physicists to model the thermal E No energy is required for this process it happens properties of complicated gases, liquids and solids. spontaneously Slide 119 (Answer) / 163 Slide 120 / 163

44 Which of the following is true about melting process?

A The energy is required to increase the average kinetic energy of molecules

B The energy is required to decrease the average kinetic energy of moleculesC C The energy is required to increase the potential energy between the molecules

D The energy is required to decrease the potential energy Work in Thermodynamics between the molecules the energy is required to increase the EpotentialNo energy energy is between required the for this process it happens moleculesspontaneously (moving them further apart) [This object is a pull tab]

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Internal Energy, Heat and Work Work in Thermodynamics

The state of any thermodynamic system can be described with A simple example of a thermodynamic the internal energy. system is a quantity of gas enclosed in a cylinder with a movable piston. The internal energy (U) of a thermodynamic system can be changed in two different ways: adding heat (Q) to the system or Consider the work done by the gas doing work (W) on the system. during its expansion. Note: The work done will be positive since the pressure (force) and the motion of the piston are in the same In 1845, James Joule presented his direction. paper "On the mechanical equivalent of heat". He was the first to discover that doing work on a system increases its temperature.

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Work in Thermodynamics Work in Thermodynamics

Suppose that the cylinder has a When the piston moves down, the cross-sectional area A and the gas volume decreases, and so the pressure exerted by the gas is Pgas. work W done by the gas is negative. The force exerted by the gas on the piston is F = PgasA. During the compression of the gas When the piston moves up a the work W' done by the external distance Δx and the pressure P is force Fext is positive. constant, the work W is

The relationship between work or done by the gas and work done on the gas can be presented by following: Slide 125 / 163 Slide 126 / 163

Work in Thermodynamics Work in Thermodynamics

This relationship can be The work done equals the represented as a graph of area under the curve on a P as a function of V on a PV-diagram. PV-diagram.

In a compression, the work The work done equals the done by the gas is negative. area under the curve on a PV-diagram.

In an expansion, the work done by the gas is positive.

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45 The state of an ideal gas is changed in a closed path 45 The state of an ideal gas is changed in a closed path 1⇒ 2⇒ 3⇒ 1. Which of the following is true about work done by 1⇒ 2⇒ 3⇒ 1. Which of the following is true about work done by the gas between point 1 and point 2? the gas between point 1 and point 2?

No work is done because V is Answer constant A Work done by the gas is positive A Work done by the gas is positive B Work done by the gas is negative B Work done by the gas is negative

C Work done by the gas is zero C Work done by [Thisthe object gas is ais pull zero tab] D Work done by the gas is greater than work done on the gas D Work done by the gas is greater than work done on the gas E Work done by the gas is less than work done on the E Work done by the gas is less than work done on the gas gas

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46 The state of an ideal gas is changed in a closed path 1⇒ 2⇒ 3⇒ 1. 46 The state of an ideal gas is changed in a closed path 1⇒ 2⇒ 3⇒ 1. Which of the following is true about work done by the gas Which of the following is true about work done by the gas between point 2 and point 3? between point 2 and point 3?

The volume is increasing along an Answer isotherm A Work done by the gas is positive A Work Thedone area by the under gas the is positiveline 2⇒ 3 is B Work done by the gas is negative B Work done by thepositive gas is negative [This object is a pull tab] C Work done by the gas is zero C Work done by the gas is zero D Work done by the gas is greater than work done on the gas D Work done by the gas is greater than work done on the gas E Work done by the gas is less than work done on the E Work done by the gas is less than work done on the gas gas Slide 129 / 163 Slide 129 (Answer) / 163

47 The state of an ideal gas is changed along a closed path 47 The state of an ideal gas is changed along a closed path X⇒ B⇒ Y⇒ A⇒ X. What is the total amount of work done on X⇒ B⇒ Y⇒ A⇒ X. What is the total amount of work done on the gas? the gas? P P A 2PV A 2PV C 2P X B 2P X B B -2PV B -2PV

C PV C PV The work done along (area under)

A Y Answer A Y P X⇒ BP is 2PV D -PV V D -PV The work done along (area under) V V 2V Y⇒ A is -PV V 2V

The net[This workobject is adone pull tab] is PV

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First Law of Thermodynamics

In previous sections of this chapter we defined the internal energy, heat, and work in thermodynamics.

Now we will combine them in one formula expressing conservation of energy in thermal processes. First Law of Thermodynamics

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First Law of Thermodynamics 48 150 J of heat is added to a system and 100 J of work done on the system. What is the change in the internal energy of the system?

We have learned that there are two ways to increase the Internal Energy (by ∆U) of a thermodynamic system: A 250 J · add Heat (Q) to the system B 150 J

· do Work (W' = Pext∆V) on the system C 100 J D 50 J E 0 J First Law of Thermodynamics Slide 133 (Answer) / 163 Slide 134 / 163

48 150 J of heat is added to a system and 100 J of work done on the 49 250 J of heat is added to a system and the system does 100 J of system. What is the change in the internal energy of the system? work on surroundings. What is the change in the internal energy of the system?

A 250 J A A 250 J B 150 J B 150 J ∆U = Q + W' C 100 J C 100 J = 150 + 100

D 50 J Answer D 50 J = 250 J E 0 J E 0 J

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49 250 J of heat is added to a system and the system does 100 J of work on surroundings. What is the change in the internal energy of the system?

A 250 J B B 150 J ∆U = Q + W' C 100 J = 250 - 100

D 50 J Answer Thermodynamic Processes = 150 J E 0 J

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Thermodynamic Processes Isothermal Processes

We will next look at 4 different types of thermodynamic processes In an isothermal process, the and their effect on the internal energy of a thermodynamic system. temperature is constant (∆T = 0). For an ideal gas, since the internal energy U depends on T, Each process takes place in a situation where a particular · the internal energy is constant thermodynamic quantity is fixed. · ∆U = 0 · isothermal - constant temperature (∆T = 0) · isobaric - constant pressure (P = constant) Since ∆U = Q + W', or Q = -W' · isochoric - constant volume (∆V = 0) · all heat added to the system is · adiabatic - no heat added or removed (Q=0) converted to work by the system

"Isothermal process" by Netheril96 - Own work. Licensed under CC0 via Wikimedia Commons - https://commons.wikimedia.org/wiki/ File:Isothermal_process.svg#/media/File:Isothermal_process.svg Slide 138 / 163 Slide 139 / 163

Isobaric Processes Isochoric Processes

In an isobaric process, the pressure is constant. In an isochoric process, the volume is constant (∆V = 0). Since W' = Pext∆V · work is done as soon as the volume This means that P and T (and as V changes therefore U) change. Since T is allowed to change · ∆U ≠ 0 Since W = P∆V = 0 Since ∆U = Q + W' can be rearranged · no work is done as Q = ∆U - W' · ∆U = Q · heat added to the system causes the system to do work and will increase · heat added to the system "Isothermal process" by Netheril96 - Own work. Licensed under CC0 its internal energy (and therefore T) increases the systems internal via Wikimedia Commons - https://commons.wikimedia.org/wiki/ File:Isothermal_process.svg#/media/File:Isothermal_process.svg energy "Isobaric process plain" by IkamusumeFan - Own work. Licensed under C BY-SA 3.0 via Wikimedia Commons - https://commons.wikimedia.org/wiki File:Isobaric_process_plain.svg#/media/File:Isobaric_process_plain.svg Slide 140 / 163 Slide 141 / 163

Adiabatic Processes Thermodynamic Processes

In an adiabatic process, the no Process Condition First Law heat enters or leaves the system isothermal ∆T = 0 (∆U = 0) 0 = Q + W' (Q = 0). isobaric ∆P = 0 ∆U = Q + W' Since Q = 0 isochoric ∆V = 0 (W' = 0) ∆U = Q · any work done on the system adiabatic Q = 0 ∆U = W' increases its internal energy · any work by the system decreases its internal energy · ∆U = W'

"Adiabatic" by User:Stannered - Image:Adiabatic.png. Licensed under CC BY- SA 3.0 via Wikimedia Commons - https://commons.wikimedia.org/wiki/ File:Adiabatic.svg#/media/File:Adiabatic.svg

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50 A sample of an ideal gas is taken through a closed cycle. Which 50 A sample of an ideal gas is taken through a closed cycle. Which of the following is true about the change in internal energy and of the following is true about the change in internal energy and work done on the gas between point 2 and point 3? work done on the gas between point 2 and point 3?

A ΔU =0, W' > 0 A ΔU =0, W' > 0 C B ΔU =0, W' = 0 B ΔU =0, W' = 0 C ΔU =0, W' < 0 C ΔU =0, W' < 0 ∆U = 0 since 2 ➙3 is an isotherm D ΔU > 0, W' > 0 D ΔU > 0, Answer W' >W' 0 < 0 since the gas is doing work E ΔU < 0, W' < 0 E ΔU < 0, W' < 0

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51 A sample of an ideal gas is taken through a closed cycle. Which 51 A sample of an ideal gas is taken through a closed cycle. Which of the following is true about the change in internal energy and of the following is true about the change in internal energy and work done on the gas between point 1 and point 2? work done on the gas between point 1 and point 2?

A ΔU =0, W' > 0 A ΔU =0, W' > 0 B ΔU >0, W' = 0 B ΔU >0, W' = 0 B C ΔU =0, W' < 0 C ΔU =0, W' < 0 D ΔU > 0, W' > 0 D ΔU > 0, W' > 0 ∆U > 0 since 1 ➙2 PV=nRT is Answer increasing E ΔU < 0, W' < 0 E ΔU < 0, W' < 0 W' = 0 since V is constant

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Second Law of Thermodynamics

· Many thermal processes proceed naturally in one direction but not the opposite. These are called irreversible. · For example, heat by itself always flows from a hot object to a cooler object, never the reverse. The reverse process would not violate the first law of thermodynamics; energy would be conserved. · In order to properly account for irreversible processes, the second Second Law of Thermodynamics law of thermodynamics was formulated.

Sadi Carnot is credited with having formulated the Second Law of Thermodynamics in 1824

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Second Law of Thermodynamics

Heat flows naturally from a hot object to a cold object; Heat never flows spontaneously from a cold object to a hot object.

Heat Engines

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Heat Engines Heat Engines

A heat engine is a thermodynamic How a heat engine works: system that converts heat into · The high-temperature reservoir transfers mechanical energy, which can then be an amount of heat QH to the engine used to do mechanical work. · In the engine, part of the heat is transformed into work W (during the Denis Papin (about 1690), inventor of expansion of gas) the pressure cooker, first described three basics components of any heat · The rest of the heat, QL, is exhausted to engine the low- temperature reservoir · a high-temperature reservoir · a low-temperature reservoir · an engine containing gas or steam

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Heat Engines 52 An engine has an efficiency of 20% and produces 2000 J of mechanical work during one cycle. How much energy is transferred out of the engine by heating the environment during one cycle? The efficiency e of any heat engine can be defined as a ratio of work W to the heat input QH. A 0 J

B 100 J

C 2,000 J

D 8,000 J

E 10,000 J

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52 An engine has an efficiency of 20% and produces 2000 J of Carnot Cycle mechanical work during one cycle. How much energy is transferred out of the engine by heating the environment during one cycle? One of the important challenges in engine design was that of A 0 J increasing efficiency. D B 100 J This question was answered in 1824 by the French engineer Sadi e = W/QH C 2,000 J Carnot. Answer QH = W/e = 2,000/0.2 = 10,000 J D 8,000 J heat to environment: Carnot proposed an idealized heat engine (The Carnot Engine) that QL = QH - W = 8,000 J would provide the maximum E 10,000 J [This object is a pull tab] possible efficiency consistent with the second law of thermodynamics. Slide 153 / 163 Slide 154 / 163

Carnot Cycle Carnot Theorem

The Carnot engine operates between a high temperature Carnot Theorem reservoir at TH and a low No heat engine operating between temperature reservoir at TL. two temperatures TH and TC engine can have a greater efficiency than a The Carnot engine consists of: Carnot engine operating between · two reversible isothermal the same two temperatures. processes A⇒ B and C⇒ D, · two reversible adiabatic processes B⇒ C and D⇒ A ideal efficiency

Note: temperatures are in Kelvin.

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53 A Carnot engine moves 1000 J of heat from a 500 K reservoir to a 53 A Carnot engine moves 1000 J of heat from a 500 K reservoir to a 300 K reservoir. 300 K reservoir. With what efficiency will the engine produce work during this With what efficiency will the engine produce work during this process? process?

D A 10% A 10%

B 20% B 20% e = (TH - TL)/TH

Answer e = (500-300)/500 = 0.40 C 30% C 30%

D 40% D 40% [This object is a pull tab] E 50% E 50%

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54 A Carnot engine moves 1000 J of heat from a 500 K reservoir to a 54 A Carnot engine moves 1000 J of heat from a 500 K reservoir to a 300 K reservoir. 300 K reservoir. How much work did the engine perform? How much work did the engine perform?

A 400 J A 400 J A

B 800 J B 800 J e = W/(QH-QL)

C 1000 J C 1000 J Answer 0.4 = W/1000 W = 400 J D 1600 J D 1600 J

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Entropy

Kinetic energy of macroscopic objects (a ball, pendulum...) is associated with organized, coordinated motions of many molecules.

In contrast, heat transfer involves changes in energy of random, disordered molecular motion.

Therefore conversion of mechanical energy into heat involves an Entropy and Disorder increase of randomness or disorder. The energy is conserved, its just no longer usable to do work in the system.

Entropy provides a quantitative measure of this disorder.

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Entropy and the Second Law Entropy and the Second Law

The total entropy of an isolated system never decreases An important statement of the Second Law of Thermodynamics:

· The entropy of an isolated system may increase but can never decrease.

· When a system interacts with its surroundings, the total entropy change of system and surroundings can never decrease.

· When the interaction involves only reversible processes, the total entropy is constant, ∆S=0.

· When there is any irreversible process, the total entropy increases and ∆S>0. The mixing of the two liquids cannot be undone.

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55 Which of the following is NOT true when two systems with unequal 55 Which of the following is NOT true when two systems with unequal temperatures come into contact with one another? temperatures come into contact with one another?

A Heat will flow until thermal equilibrium is reached A Heat will flow until thermal equilibrium is reached

B No net heat flow occurs once thermal equilibrium is reached B No net heat flow occurs once thermal equilibriumC is reached

C Entropy is reduced during the process C Entropy is reduced during the process Answer D The natural flow of heat will be from the warmer system to D The natural flow of heat will be from the warmer system to the colder system the colder system

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56 When ice freezes, its molecules become much more structured. 56 When ice freezes, its molecules become much more structured. Does this break the second law of thermodynamics? Does this break the second law of thermodynamics?

A Yes. The second law of thermodynamics is wrong. All A Yes. The second law of thermodynamics is wrong. All scientists should be fired right now. scientists should be fired right now. C B No, because the energy of the molecules increased B No, because the energy of the molecules increased

C No, because the entropy of the system as a whole (ice+ice's C No, because the entropy of the system as a whole (ice+ice's environment) increased environment) increasedAnswer

D No, because the density of water decreased D No, because the density of water decreased

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