Physics 231 Lecture 35

•  Main points of last lecture: •  Heat engines and efficiency:

•  Carnot cycle and Carnot engine.

–  T is in Kelvin.

•  Refrigerators

•  Ideal refrigerator

•  Entropy

h

c

h

ch

h

eng

TT

TTT

QW

−=−

== 1 e

h

c

h

ch

h

eng

QQ

QQQ

QW

−=−

== 1 e

TQS reversible =Δ

COP = Qc

Weng

=Qc

Qh − Qc

COPrev =Tc

Th −Tc

Some dates

•  Final Exam Thursday Dec. 15, 8-10 pm in E100 Vet. Medical Center –  The exam will be over the entire material for this term.

•  Attitude Survey Wed. 10 pm •  Concept Test Thu. 10 pm

•  Homework set 10 Fri. 10 pm.

quiz:

•  What is the work done by the gas going from (Pi ,Vi) to (Pf ,Vf) ? –  a) Pf(Vi-Vf) –  b) Pf(Vf-Vi) –  c) 0 –  d) Vi(Pi-Pf) –  e) Vf(Pi-Pf)

•  The work done by the gas is PΔV, which is the mathematical area under the curve Pf(Vf-Vi).

•  Note that the work done on the gas is the negative of the work do by the gas.

•  No work is done on the isovolumetric part of the path!

Computation of work:

•  What is the work done by the gas going from (Pi ,Vi) to (Pf ,Vf) ? –  a) Pi(Vi-Vf) –  b) Pf (Vi-Vf) –  c) 0 –  d) (Pi+ Pf)(Vf-Vf)/2 –  e) (Pi+ Pf)(Vi-Vf)/2

•  A thermal path which returns to its initial condition is called a cycle.

•  The work done by the gas on a clockwise cycle is the area contained in the path.

•  The work done by the gas on a counterclockwise cycle is the negative of the area in the path.

•  The work done on the gas is the negative of the work done by the gas.

Example

•  The work done on the gas to compress one mole of a monatomic ideal gas is 6200 J. The temperature of the gas changes from 350 K to 550 K. How much heat flows between the gas and its surroundings? Determine whether the heat flows into or out of the gas.

Work done by the gas is : Wgas = −6200J

Internal energy change in the gas is: ΔU = 3

2nR ΔT( )

Q = ΔU +Wgas

= 32

nR ΔT( )− 6200J

= 32

8.31J / K( ) 550− 350K( )− 6200J

Q = −3700J

Heat flows out of the gas

Example •  The drawing refers to one mole of monatomic

ideal gas and shows a process that has four steps, two isobaric (A to B and C to D) and two isovolumetric (B to C and D to A). Complete the following table by calculating ΔU, ΔW and ΔQ (including the algebraic signs) for each of the four steps. Note that the gas has returned to its initial state at the end of the process, so that the value for the total ΔU can be predicted in advance without any calculation.

Path ΔU ΔW ΔQ

A-B B-C C-D D-A

( ) ( )

WUQ 1662 3324

)(32 )(

32

1 02493 4986 9972 4986

23

23

23

23

ABA

A

Δ+Δ=Δ−==

−=−=

−=−=−======

====

====

−−

−−

−

−−

JWJW

UUWUUW

TTRVPVPVVPWnPPPPWW

JUJUJUJU

RTURTURTURTU

DCBA

CDDCABBA

ABABABBA

DCBCBAD

DCBA

DDCCBBAAPath ΔU ΔW ΔQ

A-B 5.0kJ 3.3kJ 8.3kJ B-C - 5.kJ 0 - 5.kJ C-D -2.5kJ -1.7kJ -4.2kJ D-A 2.5kJ 0 2.5kJ

Example •  A person takes in a breath of 0°C air and holds it until it warms

to 37.0°C. The air has an initial volume of 0.600 L and a mass of 7.70 x 10–4 kg. Determine (a) the work done by the air on the lungs if the pressure remains constant at 1 atm, (b) the change in internal energy of the air, and (c) the energy added to the air by heat. Model the air as if it were a monatomic gas.

( )

( )( )

( )

( ) JTTnRWUQ

JWTT

nRTTTnRU

JmPaxW

TT

VPVPVP

VPVVPW

PaxPPP

ifgas

gasi

fiif

gas

i

fiatm

iatm

fatmiatmifatmgas

atmif

5.2025

3.12231

23

23

:gas monotonic Assume

2.812733100006.1001.1

11

1001.1

35

5

=−=+Δ=Δ

==⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=Δ

=⎟⎠⎞⎜

⎝⎛ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛−=⎟⎟⎠

⎞⎜⎜⎝

⎛−=−=

===

Conceptual quiz

•  A gas is taken from an initial state of pressure and volume, Pi Vi, to a final, different, state, Pf Vf. As it changes work is done on the gas. –  a) The amount of work depends on the path taken from i to f –  b) The amount of work does not depend on the path taken from i to f

Pf(Vf-Vi) Pi(Vf-Vi)

When I do work on a gas in an adiabatic process, compressing it, I add energy to the gas. Where does this energy go?

A.  The energy is transferred as heat to the environment.

B.  The energy is converted to thermal energy of the gas.

C.  The energy converts the phase of the gas.

Additional Questions

Slide 12-56

Engines •  In a heat engine, thermal energy Qh is used to do

work, Weng. Some of the original thermal energy Qc escapes and ends up heating something else

•  A heat engine involves some working substance in a cyclical process

•  Thermal efficiency is defined as the ratio of the work done by the engine to the energy absorbed at the higher temperature. For simplicity both can be computed over one cycle:

h

c

h

ch

h

eng

QQ

QQQ

QW

−=−

== 1 e

•  e = 1 (100% efficiency) only if Qc = 0 –  No energy expelled to cold reservoir,

which is theoretically possible for Tc=0, but practically impossible.

Example

•  The energy absorbed by an engine is three times as large as the work it performs. (a) What is its thermal efficiency? (b) What fraction of the energy absorbed is expelled to the cold reservoir?

a) Qh = 3W

e = W

Qh

= W3W

= 13

b) Qh =W +Qc ⇒ Qc = Qh −W = 2W = 2

3Qh

⇒

Qc

Qh

= 23

Quiz

•  A heat engine performs 200 J of work in each cycle and has an efficiency of 30%. For each cycle of operation, (1) how much energy is taken in from the hot reservoir (Qh) and (2) how much energy is expelled to the cold reservoir (QC)?

•  Answers below are in the form Qh, Qc. –  a) 667J, 467J –  b) 467J, 667J –  c) 60J, 140J –  d) 140J, 60J

a) e = W

Qh

⇒ Qh =We= 200J

0.3= 667J

b) QH =W +Qc ⇒ Qc = Qh −W = 667J − 200J = 467J

Reversible and Irreversible Processes •  reversible process is one in which every state along some path is an

equilibrium state. –  And one for which the system can be returned to its initial state along

the same path. –  Volume and pressure changes are “slow”. –  When objects are brought into thermal contact, they are at the same

temperatures. –  Carnot cycle is an example of a reversible process.

•  An irreversible process does not meet these requirements –  Most natural processes are irreversible

•  Burning fueling in an automobile engine •  Dropping ice into warm water •  Heating water on a range

•  Reversible process are an idealization, but some real processes are good approximations.

•  If a process is reversible, the changes in entropy ΔSi=ΔQ/Ti of the system add up to zero and the process can be reversed without a gain in entropy.

•  Irreversible processes always increase the entropy and thus cannot be reversed because it is improbably that you can do so.

Carnot Engine

•  A Carnot engine is the most efficient possible engine that takes heat from a hot reservoir at temperature Th and expels heat into a cold reservoir at temperature Tc . –  It uses gas as the working substance. –  It absorbs heat Qh during an

isothermal expansion while in contact with the hot reservoir.

–  It expands adiabatically a little further. Q and the entropy change are zero here.

–  It compresses isothermally while in contact with cold reservoir. Here the entropy change cancels that during the isotheral expansion.

–  It compresses adiabatically a little further.

Fig. 12.13, p. 374

Slide 22

Carnot Cycle, A to B •  A to B is an isothermal expansion •  The gas is placed in contact with the high

temperature reservoir •  The gas absorbs heat Qh •  The gas does work WAB in raising the piston •  What happens to the internal energy of the gas?

–  a) It increases –  b) It decreases –  c) It stays the same

•  What happens to the pressure? –  a) It increases –  b) It decreases –  c) It stays the same

•  What happens to the entropy of the gas? –  a) It increases –  b) It decreases –  c) It stays the same

•  What happens the entropy of the hot reservoir? –  a) It is the same as for the gas –  b) It is minus that of the gas. –  c) It does not change.

WAB = nRTh ln

VB

VA

⎛⎝⎜

⎞⎠⎟

P =nRTh

V

Q = ΔU +WQ = ΔU + PΔV

ΔU = f2

NRΔT

ΔS = ΔQ / T

Fig. 12.13, p. 374

Slide 22

Carnot Cycle, B to C

•  B to C is an adiabatic expansion •  The base of the cylinder a thermally insulating wall •  No heat enters or leaves the system •  The temperature falls from Th to Tc •  The gas does work WBC

•  What happens to the pressure? –  a) increases –  b) decreases –  c) stays the same

•  What happens to the internal energy? –  a) increases –  b) decreases –  c) stays the same.

•  What happens to the entropy of the gas? –  a) increases –  b) decreases –  c) stays the same.

•  What happens the entropy of the hot reservoir? –  a) It increases –  b) It decreases. –  c) It does not change

P ∝V −γ

Q = ΔU +WQ = ΔU + PΔV

ΔU = f2

NRΔT

ΔS = ΔQ / T

Fig. 12.13, p. 374

Slide 22

Carnot Cycle, C to D

•  The gas is placed in contact with the cold temperature reservoir

•  C to D is an isothermal compression •  The gas expels energy QC •  Work WCD is done by the gas •  What happens to the internal energy of the gas?

–  a) It increases –  b) It decreases –  c) It stays the same

•  What happens to the pressure? –  a) It increases –  b) It decreases –  c) It stays the same

•  What happens to the entropy of the gas? –  a) It increases –  b) It decreases –  c) It stays the same

•  What happens the entropy of the hot reservoir? –  a) It is the same as for the gas –  b) It is minus that of the gas. –  c) It does not change

WCD = nRTC ln

VD

VC

⎛⎝⎜

⎞⎠⎟

P =nRTc

V

Q = ΔU +WQ = ΔU + PΔV

ΔU = f2

NRΔT

ΔS = ΔQ / T

Fig. 12.13, p. 374

Slide 22

Carnot Cycle, D to A

•  D to A is an adiabatic compression. •  The gas is again placed against a thermally

nonconducting wall –  So no heat is exchanged with the surroundings

•  The temperature of the gas increases from Tc to Th •  The work done on the gas is WDA •  What happens to the pressure?

–  a) increases –  b) decreases –  c) stays the same

•  What happens to the internal energy? –  a) increases –  b) decreases –  c) stays the same

•  What happens to the entropy of the gas? –  a) increases –  b) decreases –  c) stays the same

•  What happens to the entropy of the reservoir? –  a) increases –  b) decreases. –  c) it does not change

P ∝V −γ

Q = ΔU +WQ = ΔU + PΔV

ΔU = f2

NRΔT

ΔS = ΔQ / T

Carnot cycle efficiency

•  The efficiency of the Carnot cycle depends only on the temperatures of the hot and cold reservoirs:

•  A heat engine operates between two reservoirs at temperatures of 20°C and 300°C. What is the maximum efficiency possible for this engine?

h

c

h

ch

h

eng

TT

TTT

QW

−=−== 1 e

–  No engine operating between these two temperatures is more efficient than the Carnot engine

a) emax = ecarnot = 1−

TC

TH = 1− 293

573= 0.49

Example

•  An engine does 20900 J of work and rejects 7330 J of heat into a cold reservoir at 298K. What is the smallest possible temperature of the hot reservoir?

a) Qh =W +Qc = 20900J + 7330J = 28230J

e = W

Qh

= 20900J28230J

= 0.74 ≤ ecarnot = 1−

Tc

Th

0.74 ≤1−

Tc

Th ⇒

Tc

Th

≤1− .74 = .26

⇒

Tc

Th

≤ .26 ⇒

Tc

.26≤ Th ⇒

298K.26

≤ Th

⇒1146K ≤ Th

Heat pumps and refrigerators •  Heat engines can run in reverse

–  Send in energy –  Energy is extracted from the cold reservoir –  Energy is transferred to the hot reservoir

•  This process means the heat engine is running as a heat pump –  A refrigerator is a common type of heat pump –  An air conditioner is another example of a heat

pump –  In the south or in Asia, people often use heat

pumps to heat homes •  One usually rates refrigerators in terms of their

coeficient of performance COP

COP =

Qc

W=

Qc

Qh −Qc

= 1Qh / Qc −1

COPrev =

Qc

W=

Tc

Th −Tc

= 1Th / Tc −1

•  A reversible engine run as refrigerator has the highest

possible COP

Quiz •  A heat engine operating between a hot reservoir at 500 K and a cold

reservoir at 200 K has an efficiency that is 70% of its maximum possible value. If it receives lx106 J heat energy from the hot reservoir in 25 minutes, it can do a quantity of work equal to –  a)6.3x105J. –  b)4.2x105J. –  c)3.lxl05J. –  d)2.5x105J. –  e)1.7x105J.

a) emax = ecarnot = 1−

Tc

Th

= 1− 200500

= 0.6

COPcarnot =1

Th

Tc

−1= 1

500200

−1= 1

2.5−1= 2 / 3

•  If one replaced this engine with a Carnot energy running at the same temperatures and ran it in reverse, it would be a reversible heat pump (refrigerator) that would extract heat from the cold reservoir and put it the heat into the hot reservoir. For that reversible heat pump, what would be its COP? –  a) .25 –  b) .67 –  c) 1.0 –  d) 1.5

⇒ e = 0.7emax = 0.7(0.6) = .42

e = W

Qh

⇒W = e ⋅Qh = .42( ) 1x106 J( ) = 4.2x105 J

Example

•  A Carnot refrigerator maintains the food inside it at 276 K while the temperature of the kitchen is 298 K. The refrigerator removes 3.00x 104 J of heat from the food. How much heat is delivered to the kitchen?

As a Carnot heat engine, we know that

ecarnot = 1−Qc

Qh

= 1−Tc

Th

⇒Qc

Qh

=Tc

Th

The Carnot engine run in reverse takes mechanical energy W to move Qc from the inside of the refrigerator

and deposit Qh in the kitchen.

⇒

Qh

Qc

=Th

Tc

⇒ Qh = Qc

Th

Tc

= 3x104 J298276

= 3.23x104 J

Is it smart to try to cool your kitchen by opening the refrigerator door? a) yes b)no

•  Entropy can only be calculated from a reversible path, and must be done that that way even if the system actually follows an irreversible path –  To calculate the entropy for an irreversible process, model it as a

reversible process •  When energy is absorbed, Q is positive and entropy increases •  When energy is expelled, Q is negative and entropy decreases •  S ≈ ln(probability). •  A disordered state with energy and matter spread out everywhere is more

probable than having all of the energy stored in an organized way that can be used to do work.

TQS reversible =Δ

Example

•  The surface of the Sun is approximately at 5700 K, and the temperature of Earth’s surface is approximately 290 K. What entropy change occurs when 1000 J of energy is transferred by heat from the Sun to Earth?

The sun loses Q of heat and therefore decreases its entropy by the amount

ΔSsun = −QTsun

The earth gains Q of heat and therefore increases its entropy by the amount

ΔSearth = QTearth

The total entropy change is:

ΔS = ΔSsun + ΔSearth = Q 1Tearth

− 1Tsun

⎛⎝⎜

⎞⎠⎟= 1000 J

1290 K

− 15700 K

⎛⎝⎜

⎞⎠⎟= 3.27J / K

Example

•  A power plant has been proposed that would make use of the temperature gradient in the ocean. The system is to operate between 20.0°C (surface-water temperature) and 5.00°C (water temperature at a depth of about 1 km). (a) What is the maximum efficiency of such a system? (b) If the useful power output of the plant is 75.0 MW, how much energy is absorbed per hour? (c) In view of your answer to (a), do you think such a system is worthwhile (considering that there is no charge for fuel)?

a) emax = ecarnot = 1−

Tc

Th

= 1− 278293

= 0.051

b) e = W

Qh

⇒ Qh =We=

75MW( ) 3600s( )0.051

= 5.3x1012 J

c) What is the energy required to pump the water?

Story of Hawaiian deep water project

•  Keahole sits at a point where underwater land slopes sharply down into the sea, it was a place where warm water can be piped from the surface of the sea and cold water can be piped from depths of about a half-mile.

•  A process called ocean thermal energy conversion, or OTEC, used the temperature difference between hot and cold sea water to produce 50 KW of electricity at Keahole in 1993. The process worked but it was uneconomical.

•  KAILUA, HAWAI'I — Koyo USA Corp., a company selling deep-sea water from Keahole Hawai'i, is expanding its plant and has applied to sell the water in the United States

•  The company is producing more than 200,000 bottles a day and says it can't keep up with demand in Japan, where it sells 1.5 liter bottles of its MaHaLo brand for $4 to $6 each.