review for test 3 introduction section 0 lecture 1 slide 1 lecture 27 slide 1 introduction to modern...

Review for Test 3

Introduction Section 0 Lecture 1 Slide 1

Lecture 27 Slide 1

INTRODUCTION TO Modern Physics PHYX 2710

Fall 2004

Physics of Technology—PHYS 1800

Spring 2009

Physics of Technology

PHYS 1800

Lecture 27

Review for Test 3

Review for Test 3


Lecture 27 Slide 2


Fall 2004


Spring 2009

PHYSICS OF TECHNOLOGY Spring 2009 Assignment Sheet

*Homework Handout

PHYSICS OF TECHNOLOGY - PHYS 1800 ASSIGNMENT SHEET

Spring 2009 Date Day Lecture Chapter Homework Due Feb 16 17 18 19 20

M Tu W H F*

Presidents Day Angular Momentum (Virtual Monday) Review Test 2 Static Fluids, Pressure

No Class 8 5-8 5-8 9

-

Feb 23 25 27

M W F*

Flotation Fluids in Motion Temperature and Heat

9 9 10

6

Mar 2 4 6

M W F*

First Law of Thermodynamics Heat flow and Greenhouse Effect Climate Change

10 10 -

7

Mar 9-13 M-F Spring Break No Classes Mar 16 18 20

M W F*

Heat Engines Power and Refrigeration Electric Charge

11 11 12

8

Mar 23 25 26 27

M W H F*

Electric Fields and Electric Potential Review Test 3 Electric Circuits

12 13 9-12 13

-

Mar 30 Apr 1 3

M W F

Magnetic Force Review Electromagnets Motors and Generators

14 9-12 14

9

Apr 6 8 10

M W F*

Making Waves Sound Waves E-M Waves, Light and Color

15 15 16

10

Apr 13 15 17

M W F*

Mirrors and Reflections Refraction and Lenses Telescopes and Microscopes

17 17 17

11

Apr 20 22 24

M W F

Review Seeing Atoms The really BIG & the really small

1-17 18 (not on test) 21 (not on test)

No test week 12

May 1 F Final Exam: 09:30-11:20am * = Homework Handout

Review for Test 3


Lecture 27 Slide 3


Fall 2004


Spring 2009

Notes on Test

1. Covers Chapters 9-11

2. ~8 short answer problems or questions (5 point each)

3. 3 Numerical problems based heavily on the material from the homework and Lab/Demo sessions (20 points each). One problem each from Chapters 9, 10 and 11.

4. You will have a formula sheet just like the one in the handout.

5. Test is Thursday March 26 1:30-2:45 in ESLC 46.

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Lecture 27 Slide 4


Fall 2004


Spring 2009


PHYS 1800

Lecture 27

Review for Test 3

Introduction and Review

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Lecture 27 Slide 5


Fall 2004


Spring 2009

What Do We Need To Measure?

• What is the minimum about things we need to know?

• Where things are—a length, L• When things are there—a time, t

• How thing interact with gravity—a mass, M• How things interact with E&M—a charge, Q

• How thing inter act with weak nuclear force• How things interact with strong nuclear force

• Random collections of objects—a temperature, T

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Lecture 27 Slide 6


Fall 2004


Spring 2009

Describing Motion and InteractionsPosition—where you are in space (L or meter)

Velocity—how fast position is changing with time (LT-1 or m/s)

Acceleration—how fast velocity is changing with time (LT-2 or m/s2)

Force— what is required to change to motion of a body (MLT-2 or kg-m/s2 or N)

Inertia (mass)— a measure of the force needed to change the motion of a body (M)

Energy—the potential for an object to do work. (ML2T-2 or kg m2/s2 or N-m or J)

Work is equal to the force applied times the distance moved. W = F dKinetic Energy is the energy associated with an object’s motion. KE=½ mv2

Potential Energy is the energy associated with an objects position.Gravitational potential energy PEgravity=mghSpring potential energy PEapring= -kx

Momentum— the potential of an object to induce motion in another object (MLT -1 or kg-m/s)

Angular Momentum and Rotational Energy— the equivalent constants of motion for rotation (MT-1 or kg/s) and (MLT-2 or kg m/s2 or N)

Review for Test 3


Lecture 27 Slide 7


Fall 2004


Spring 2009

Newton’s Laws in Review

• 1st Law —a special case of the 2nd Law for statics, with a=0 or Fnet=0 • An objects velocity remains unchanged, unless

a force acts on the object.

• 2nd Law (and 1st Law)—How motion of a object is effected by a force.– The acceleration of an object is directly

proportional to the magnitude of the imposed force and inversely proportional to the mass of the object. The acceleration is the same direction as that of the imposed force.

• 3rd Law —Forces come from interactions with other objects.• For every action (force), there is an equal but

opposite reaction (force).

F ma

units : 1 newton = 1 N = 1 kgm s2

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Lecture 27 Slide 8


Fall 2004


Spring 2009

Energy: The potential to do work.Conservation of Energy: The total

energy of a closed system remains constant.

– Energy can be converted from one form to another.

– Not all forms of energy can be fully recovered.

Conservation of Energy

Time

Ene

rgy

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Lecture 27 Slide 9


Fall 2004


Spring 2009

Momentum and Impulse• Multiply both sides of Newton’s second law by the time interval over which the force acts:

• The left side of the equation is impulse, the (average) force acting on an object multiplied by the time interval over which the force acts.

• How a force changes the motion of an object depends on both the size of the force and how long the force acts.

• The right side of the equation is the change in the momentum of the object.

• The momentum of the object is the mass of the object times its velocity.

vF

vaF

mt

tmm

net

net

p mv

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Lecture 27 Slide 10


Fall 2004


Spring 2009

Impulse-Momentum Principle

The impulse acting on an object produces a change in momentum of the object that is equal in both

magnitude and direction to the impulse.

impulse = change in momentum

= p

In analogy,

work = change in energy= ΔE

Review for Test 3


Lecture 27 Slide 11


Fall 2004


Spring 2009

Formulas We Know and Love

Formulas as They Will Appear on the Test Sheet

221 tatvd o 2

21 tto

r

vac

2

t

ddv of

t

of

2r

mmGF ba

gravity

t

vva of

tof

22

3

2

3

4a

b

b

a

a Gm

T

R

T

R

amF Inet G=6.67·10-11 N-m2/kg2

dFW lF g=9.8 m/s2 2

21 mvKEtrans 2

21 IKErot mghPEgravity

tFp tL

mvp IL xkFelastic

t

WPower

2mrI 2

21 xkPEelastic

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Lecture 27 Slide 12


Fall 2004


Spring 2009


PHYS 1800

Lecture 27

Review for Test 3

Review for Test 3


Lecture 27 Slide 13


Fall 2004


Spring 2009

Test 3 Review ConceptsConcepts and Terms to Be Familiar With

Know what pressure and density are and how this relates to fluids.Know Pascal’s Principle and how to apply it to hydraulics problems.Know how buoyant force is related to pressure and Archimedes’ Principle.Know what an ideal gas is and what the ideal gas law says about pressure volume and temperature of an ideal gas.

Understand how conservation of mass is related to flow rate.Understand the difference between laminar and turbulent flow.Understand Bernoulli’s Principle as a fluid form of the conservation of energy.

Be able to state the four laws of thermodynamics.Be able to define heat and temperature and explain how they are different.Understand heat capacity, heat of fusion (melting), and heat of vaporization (boiling).Be able to do simple calorimitry problems.Be able to qualitatively explain the difference between the three forms of heat transfer: conduction, convection and radiation.

Be able to explain what a heat engine is and what the components of work, high temperature reservoir and low temperature reservoir.What is efficiency of a heat engine? Of a Carnot engine?

Review for Test 3


Lecture 27 Slide 14


Fall 2004


Spring 2009

Formulas We Know and Love

New Formulas as They Will Appear on the Test Sheet

AFP / Vm / TNkPV B kB=1.38 10-23 J/K

Avrateflow constant221 hgvP VPW

TcmQ H

CHH Q

QQQWe

/

H

CHCarnot T

TTe

Review for Test 3


Lecture 27 Slide 15


Fall 2004


Spring 2009


PHYS 1800

Lecture 27

Review for Test 3

Fluids and Pressure

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Lecture 27 Slide 16


Fall 2004


Spring 2009

Test 3 Review Concepts

Concepts and Terms to Be Familiar With

Know what pressure and density are and how this relates to fluids.

Know Pascal’s Principle and how to apply it to hydraulics problems.

Know how buoyant force is related to pressure and Archimedes’ Principle.

Know what an ideal gas is and what the ideal gas law says about pressure volume and temperature of an ideal gas.

Understand how conservation of mass is related to flow rate.

Understand the difference between laminar and turbulent flow.

Understand Bernoulli’s Principle as a fluid form of the conservation of energy.

Fluids and Pressure

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Lecture 27 Slide 17


Fall 2004


Spring 2009

States of Matter

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Lecture 27 Slide 18


Fall 2004


Spring 2009

Pressure

• The man weighs more, so he exerts a larger force on the ground.

• The woman weighs less, but the force she exerts on the ground is spread over a much smaller area.

• Pressure takes into account both force and the area over which the force is applied.– Pressure is the ratio of the force to

the area over which it is applied:

– Units: 1 N/m2 = 1 Pa (pascal)– Pressure is the quantity that

determines whether the soil will yield.

P F

A

Review for Test 3


Lecture 27 Slide 19


Fall 2004


Spring 2009

Dennison’s Laws of Fluids

• When push comes to shove, fluids are just like other stuff.

• Pascal’s Principle: Pressure extends uniformly in all directions in a fluid.

• Boyle’s Law: Work on a fluid equals PΔV

• Bernoulli’s Principle: Conservation of energy for fluids

Review for Test 3


Lecture 27 Slide 20


Fall 2004


Spring 2009

Pascal’s Principle

• Fluid pushes outward uniformly in all directions when compressed.

• Any increase in pressure is transmitted uniformly throughout the fluid.

• Pressure exerted on a piston extends uniformly throughout the fluid, causing it to push outward with equal force per unit area on the walls and the bottom of the cylinder.

• This is the basis of Pascal’s Principle:– Any change in the pressure of a

fluid is transmitted uniformly in all directions throughout the fluid.

Review for Test 3


Lecture 27 Slide 21


Fall 2004


Spring 2009

Pascal’s Principle for Gases

• Gas molecules lack strong interactions.• Pressure is understood as resulting from

momentum transfer to the container walls through unbalanced collisions

• Pressing on one surface adds force and hence imparts impulse to the gas

• That impulse is taken up as added collisons (pressure) on other surfaces

• The random nature of the motion of gas particles assures that the force is distributed evenly to all surfaces

• For fixed walls, a decrease in V results in an increase in P

• For expandable walls (like a balloon) the volume “appears elsewhere to make up for the lost volume

Review for Test 3


Lecture 27 Slide 22


Fall 2004


Spring 2009

Pascal’s Principle for Liquids

• Liquid molecules have strong interactions.• Liquids do not compress much• Pressure is understood as resulting from

momentum transfer to the container walls through unbalanced spring forces

• Pressing on one surface adds force that is transferred to other springs

• The network nature of the forces on the particles assures that the force is distributed evenly to all surfaces

• For expandable walls (like a balloon) the volume “appears elsewhere to make up for the lost volume

• For fixed walls, a small decrease in V (a compression) results in a large increase in P

• For solids, you can think of the strong forces holding the atoms in there equilibrium positions, equivalent to fixed walls

+

+ ++

+ ++

+ + +

+ +

+

+ +

+

+ ++

+ ++

+ +

+

+ + +

+ +

++ +

+

+ +

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Lecture 27 Slide 23


Fall 2004


Spring 2009

Archimedes’ Principle• The average density of an object compared to a fluid determines

whether the object will sink or float in that liquid.• The upward force that pushes objects back toward the surface in

liquids is called the buoyant force.• Archimedes’ Principle: The buoyant force acting on an object

fully or partially submerged in a fluid is equal to the weight of the fluid displaced by the object.

Review for Test 3


Lecture 27 Slide 24


Fall 2004


Spring 2009

Archimedes’ Principle

• For example, consider a block submerged in water, suspended from a string.– The pressure of the water pushes on the block from all sides.– Because the pressure increases with depth, the pressure at the bottom of the block

is greater than at the top.– There is a larger force (F = PA) pushing up at the bottom than there is pushing

down at the top.– The difference between these two forces is the buoyant force.

The buoyant force is proportional to both the height and the cross-sectional area of the block, and thus to its volume.

The volume of the fluid displaced is directly related to the weight of the fluid displaced.

Weight mg VdgVolume Ah

Excess Pressure P W

A

dgAh

adgh

Review for Test 3


Lecture 27 Slide 25


Fall 2004


Spring 2009

Flow Rate

• The volume of a portion of water of length L flowing past some point in a pipe is the product of the length times the cross-sectional area A, or LA.

• The rate at which water moves through the pipe is this volume divided by time: LA / t.

• Since L / t = v, the rate of flow = vA.

Review for Test 3


Lecture 27 Slide 26


Fall 2004


Spring 2009

• Laminar flow is smooth flow, with no eddies or other disturbances.– The streamlines are roughly parallel.

– The speeds of different layers may vary, but one layer moves smoothly past another.

• Turbulent flow does have eddies and whorls; the streamlines are no longer parallel.

Laminar vs Turbulent Flow

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Lecture 27 Slide 27


Fall 2004


Spring 2009

Bernoulli’s Principle

• How does a large passenger jet manage to get off the ground?

• What forces keep it in the air?• How is a ball suspended in mid-air by

a leaf blower?• What happens if we do work on a

fluid?• Bernoulli’s principle applies

conservation of energy to the flow of fluids:

• The sum of the pressure plus the

• kinetic energy per unit volume of

• a flowing fluid must remain constant.

constantE/V 2

1 2 ghvP

constantE 2

1

constant

2

mghmvPV

EPEKEWork Total

Review for Test 3


Lecture 27 Slide 28


Fall 2004


Spring 2009

How does pressure vary in pipes and

hoses?

• Will the pressure be greatest in the narrow section or the wide section?

• The speed will be greater in the narrow section.• To keep the sum P + 1/2 dv2 constant, the pressure must be

larger where the fluid speed is smaller (h is fixed).• If the speed increases, the pressure decreases. (This goes

against our intuition.)• This can be shown using vertical open pipes as pressure

gauges.• The height of the column of water is proportional to the

pressure.

Pressure Changes with Area

Review for Test 3


Lecture 27 Slide 29


Fall 2004


Spring 2009


PHYS 1800

Lecture 27

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Lecture 27 Slide 30


Fall 2004


Spring 2009


PHYS 1800

Lecture 27

Review for Test 3

Temperature and Heat

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Lecture 27 Slide 31


Fall 2004


Spring 2009



Be able to state the four laws of thermodynamics.

Be able to define heat and temperature and explain how they are different.

Understand heat capacity, heat of fusion (melting), and heat of vaporization (boiling).

Be able to do simple calorimitry problems.

Be able to qualitatively explain the difference between the three forms of heat transfer: conduction, convection and radiation.


Review for Test 3


Lecture 27 Slide 32


Fall 2004


Spring 2009

Dennison’s Laws Thermal Poker(or How to Get a Hot Hand in Physics)

• 0th Law: Full House beats Two Pairs

• 1st Law: We’re playing the same game (but with a wild card)

• 2nd Law: You can’t win in Vegas.

• 3rd Law: In fact, you always loose.

• 0th Law: Defines Temperature

• 1st Law: Conservation of Energy (with heat)

• 2nd Law: You can’t recover all heat losses (or defining entropy)

• 3rd Law: You can never get to absolute 0.

Review for Test 3


Lecture 27 Slide 33


Fall 2004


Spring 2009

• What is heat?• What is the relationship between quantity of heat

and temperature?• What happens to a body (solid, liquid, gas) when

thermal energy is added or removed?

Thermal Energy

Heat

Solid: Atoms vibrating in all directions about their fixed equilibrium (lattice) positions. Atoms constantly colliding with each other.

Liquid: Atoms still oscillating and colliding with each other but they are free to move so that the long range order (shape) of body is lost.

Gas: No equilibrium position, no oscillations, atoms are free and move in perpetual high-speed “zig-zag” dance punctuated by collisions.

gas

liquid

solid

Review for Test 3


Lecture 27 Slide 34


Fall 2004


Spring 2009

• When two objects at different temperatures are placed in contact, heat will flow from the object with the higher temperature to the object with the lower temperature.

• Heat added increases temperature, and heat removed decreases temperature.

• Heat and temperature are not the same.

• Temperature is a quantity that tells us which direction the heat will flow.

Heat is a form of energy.(Here comes conservation of energy!!!)


Review for Test 3


Lecture 27 Slide 35


Fall 2004


Spring 2009

Gas Behavior and The First Law

Consider a gas in a cylinder with a movable piston. If the piston is pushed inward by an external force, work is done on

the gas, adding energy to the system.

• The force exerted on the piston by the gas equals the pressure of the gas times the area of the piston: F = PA

• The work done equals the force exerted by the piston times the distance the piston moves:

W = Fd = (PA)d = PV

Review for Test 3


Lecture 27 Slide 36


Fall 2004


Spring 2009

• In an isothermal process, the temperature does not change.– The internal energy must be constant.– The change in internal energy, U, is zero.– If an amount of heat Q is added to the gas, an equal amount of work

W will be done by the gas on its surroundings, from U = Q - W.

• In an isobaric process, the pressure of the gas remains constant.– The internal energy increases as the gas is heated, and so does the

temperature.– The gas also expands, removing some of the internal energy.– Experiments determined that the pressure, volume, and absolute

temperature of an ideal gas are related by the equation of state:

PV = NkT where N is the number of molecules

and k is Boltzmann’s constant.

Ideal Gas Behavior

Review for Test 3


Lecture 27 Slide 37


Fall 2004


Spring 2009

• The specific heat capacity of a material is the quantity of heat needed to change a unit mass of the material by a unit amount in temperature.– For example, to change 1 gram by 1 Celsius degree.– It is a property of the material, determined by experiment.– The specific heat capacity of water is 1 cal/gC: it takes 1

calorie of heat to raise the temperature of 1 gram of water by 1C.

• We can then calculate how much heat must be absorbed by a material to change its temperature by a given amount:

Q = mcT where Q = quantity of heatm = massc = specific heat capacityT = change in temperature

Heat and Specific Heat Capacity

Review for Test 3


Lecture 27 Slide 38


Fall 2004


Spring 2009

If the specific heat capacity of ice is 0.5 cal/gC°, how much heat would have to be added to 200 g of ice, initially at a temperature of -10°C, to raise the ice to the melting

point?

a) 1,000 calb) 2,000 calc) 4,000 cald) 0 cal

m = 200 gc = 0.5 cal/gC° T = -10°C

Q = mcT = (200 g)(0.5 cal/gC°)(10°C) = 1,000 cal

(heat required to raise the temperature)

Review for Test 3


Lecture 27 Slide 39


Fall 2004


Spring 2009

• When an object goes through a change of phase or state, heat is added or removed without changing the temperature. Instead, the state of matter changes: solid to liquid, for example.

• The amount of heat needed per unit mass to produce a phase change is called the latent heat.– The latent heat of fusion of water corresponds to the amount of heat

needed to melt one gram of ice.– The latent heat of vaporization of water corresponds to the amount of

heat needed to turn one gram of water into steam.

Phase Changes and Latent Heat

+

+ +

+

+ +

++ +

Solid

Review for Test 3


Lecture 27 Slide 40


Fall 2004


Spring 2009

If the specific heat capacity of ice is 0.5 cal/gC°, how much heat would have to be added to 200 g of ice, initially

at a temperature of -10°C, to completely melt the ice?

a) 1,000 calb) 14,000 calc) 16,000 cald) 17,000 cal

Lf = 80 cal/g Q = mLf = (200 g)(80 cal/g) = 16,000 cal

(heat required to melt the ice)

Total heat required to raise the ice to 0 °C and then to melt the ice is: 1,000 cal + 16,000 cal = 17,000 cal = 17 kcal

Review for Test 3


Lecture 27 Slide 41


Fall 2004


Spring 2009

A hot plate is used to transfer 400 cal of heat to a beaker containing ice and water; 500 J of work are also done on the contents of the beaker by stirring. What is the increase in internal energy of the ice-water mixture?

a) 900 Jb) 1180 Jc) 1680 Jd) 2180 J

W = -500 JQ = 400 cal

= (400 cal)(4.19 J/cal) = 1680 J

U = Q - W = 1680 J - (-500 J)= 2180 J

Review for Test 3


Lecture 27 Slide 42


Fall 2004


Spring 2009

A hot plate is used to transfer 400 cal of heat to a beaker containing ice and water; 500 J of work are also done on the contents of the beaker by stirring. How much ice melts in this process?

a) 0.037 gb) 0.154 gc) 6.5 gd) 27.25 g

Lf = 80 cal/g = (80 cal/g)(4.19 J/cal) = 335 J/g

U = mLf

m = U / Lf = (2180 J) / (335 J/g) = 6.5 g

Review for Test 3


Lecture 27 Slide 43


Fall 2004


Spring 2009

The Flow of Heat

• There are three basic processes for heat flow:

– Conduction

– Convection

– Radiation

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Lecture 27 Slide 44


Fall 2004


Spring 2009

The Flow of Heat

– In conduction, heat flows through a material when objects at different temperatures are placed in contact with one another. L

TAk

t

Q

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Lecture 27 Slide 45


Fall 2004


Spring 2009

– In convection, heat is transferred by the motion of a fluid containing thermal energy.

• Convection is the main method of heating a house.• It is also the main method heat is lost from buildings.

The Flow of Heat

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Lecture 27 Slide 46


Fall 2004


Spring 2009

The Flow of Heat

– In radiation, heat energy is transferred by electromagnetic waves.

• The electromagnetic waves involved in the transfer of heat lie primarily in the infrared portion of the spectrum.

• Unlike conduction and convection, which both require a medium to travel through, radiation can take place across a vacuum.

• For example, the evacuated space in a thermos bottle.

• The radiation is reduced to a minimum by silvering the facing walls of the evacuated space.

428

4

/107.5 KmW

TA

B

BtQ

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Lecture 27 Slide 47


Fall 2004


Spring 2009


PHYS 1800

Lecture 27

Review for Test 3

Heat Engines and the Second Law

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Lecture 27 Slide 48


Fall 2004


Spring 2009



Be able to explain what a heat engine is and what the components of work, high temperature reservoir and low temperature reservoir.

What is efficiency of a heat engine? Of a Carnot engine?

Heat Engines and the Second Law

Review for Test 3


Lecture 27 Slide 49


Fall 2004


Spring 2009

Heat Engines

It is a device that uses input heat to generate useful work.

From the 1st Law (Conservation of Energy)

In cyclic engines we return to the original state every cycle so

What is a heat engine?

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Lecture 27 Slide 50


Fall 2004


Spring 2009

Heat Engines

All heat engines share these main features of operation:– Thermal energy (heat) is

introduced into the engine.– Some of this energy is

converted to mechanical work.

– Some heat (waste heat) is released into the environment at a temperature lower than the input temperature.

What is a heat engine?

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Lecture 27 Slide 51


Fall 2004


Spring 2009

Efficiency

Efficiency is the ratio of the net work done by the engine to the amount of heat that must be supplied to accomplish this work.

Or from the 1st Law

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Lecture 27 Slide 52


Fall 2004


Spring 2009

A heat engine takes in 1200 J of heat from the high-temperature heat source in each cycle, and does 400 J of work in each cycle. What is the efficiency of this engine?

a) 33%b) 40%c) 66%

QH = 1200 JW = 400 Je = W / QH

= (400 J) / (1200 J)= 1/3 = 0.33= 33%

Review for Test 3


Lecture 27 Slide 53


Fall 2004


Spring 2009

Carnot Efficiency

• The efficiency of Carnot’s ideal engine (one using an ideal gas with PV=NkBT) is called the Carnot efficiency and is given by:

• This is the maximum efficiency possible for any engine taking in heat from a reservoir at absolute temperature TH and releasing heat to a reservoir at temperature TC.

• This provides a useful limiting case.• Even Carnot’s ideal engine is less than 100% efficient.

Review for Test 3


Lecture 27 Slide 54


Fall 2004


Spring 2009

Second Law of Thermodynamics

• You can’t recover all heat losses .

• You can’t win in Vegas.

• No engine, working in a continuous cycle, can take heat from a reservoir at a single temperature and convert that heat completely to work.

• Therefore, no engine can have a greater efficiency than a Carnot engine operating between the same two temperatures.

• Define entropy (something that measures randomness or disorder in an object) to take account of this.

Heat (random motion) is a special form of energy that cannot be fully (with complete efficiency) transformed to other forms of energy.

This leads to various forms of the Second Law of Thermodynamics.

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Lecture 27 Slide 55


Fall 2004


Spring 2009

A Third Statement of The Second Law of Thermodynamics

• Entropy remains constant in reversible processes but increases in irreversible processes.

• The entropy of a system decreases only if it interacts with some other system whose entropy is increased in the process.

– This happens, for example, in the growth and development of biological organisms.

• The entropy of the universe or of an isolated system can only increase or remain constant. Its entropy can never decrease.

Review for Test 3


Lecture 27 Slide 56


Fall 2004


Spring 2009

A heat pump uses 200 J of work to remove 300 J of heat from the lower-temperature reservoir. How

much heat would be delivered to the higher-temperature reservoir?

a) 100 Jb) 200 Jc) 300 Jd) 500 J

W = 200 JQC = 300 JQH = W + QC

= 200 J + 300 J= 500 J

review for test 3 introduction section 0 lecture 1 slide 1 lecture 27 slide 1 introduction to modern...

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