lecture 26 - iowa state universitycourse.physastro.iastate.edu/phys111/lectures/lec... · two...

Lecture 26Lecture 26

Second law of thermodynamics.Heat engines and refrigerators.

The Second Law ofThermodynamics – Introduction

The absence of the process illustrated above indicates that conservation of energy is not the whole story. If it were, movies run backwards would look perfectly normal to us!

The Second Law ofThermodynamics – Introduction

The second law of thermodynamics is a statement about which processes occur and which do not. There are many ways to state the second law; here is one:

Heat can flow spontaneously from a hot object to a cold object; it will not flow spontaneously from a cold object to a hot object.

Heat engine

= device with a working substance (eg. gas) that operates in a thermodynamic cycle. In each cycle, the net result is that the system absorbs heat (Q > 0) and does work (W > 0). Examples:

- Car engine: burns fuel, heats air inside piston. Piston expands, does mechanical work to move car

- Animal: burns “food” to be able to move

Steam engine Car engine (internal compusion engine)

Hot and cold reservoirs

Stages of the cycle–Absorb heat from hot reservoir (QH) –Perform mechanical work (W )–Dump excess heat into cold reservoir (QL < 0)

Reservoir = large body whose temperature does not change when it absorbs or releases heat.

L

L

Energy flow of Heat engines

Working substance in engine completes a cycle, so ΔU = 0:

This relation follows naturally from the diagram (QH “splits”). Draw it every time!

(QH+ QL )−W = 0

W = QH+ QL =∣QH∣−∣QL∣L

L

Energy flow diagrams in heat engines

L

L

Limitations

We are not saying that you can absorb 10 J of heat from a hot source (a burning fuel) and produce 10 J of mechanical work...

You can absorb 10 J of heat from a hot source (a burning fuel) and produce 7 J of mechanical work and release 3 J into a cold source (cooling system).

… so at the end you absorbed 10 J but used (= converted to work) only 7 J.

(We’ll see later that it is impossible to make QH = W, or QL = 0)

Efficiency

what you useEfficiency what you pay for

For a heat engine: H

WeQ

Example: A heat engine does 30 J of work and exhausts 70 J by heat transfer. What is the efficiency of the engine?

0 1e

W = 30 J|QL|= 70 J ⇒ QL=−70 J

e =WQH

= 0.3 (or30%)

QH = W−QL = 100 J

L

L

e =

QH−QL

QH

= 1−

QL

QH

ACT: Two engines

Two engines 1 and 2 with efficiencies e1 and e2 work in series as shown. Let e be the efficiency of the combination. Which of the following is true?

A. e > e1 + e2

B. e = e1 + e2

C. e < e1 + e2

TC

TH

Q3

Q1

Q2

W2

W1e1

e2

Carnot engine

e =WQH

=QH − QL

QH

=TH − TL

TH

e =QH − QL

QH

= 1−QL

QH

= 1 −TL

TH

QH

QL

Another way to write it:

Carnot engine is ideal (reversible)

L

L

Refrigerators

• Absorb heat from cold reservoir (QL > 0) • Work done on engine (W < 0)• Dump heat into hot reservoir (QH < 0)

(We want as much QL while paying for the smallest possible W .)

Energy balance:

W = QH + QL

|W |=|QH|− QL

Coefficient of performance (refrigerator)

0 < COP < ∞COP =QL

|W |=

QL

|QH|−QL

L

L

Carnot Refrigerator 0 < COP < ∞COP =T L

TH−TL

Outside of house TC

Inside of house TH

Heat pump

L

L

Heat pumps

A very efficient way to warm a house: bring heat from the colder outside.

Same energy diagram as refrigerator

Coefficient of performance (heat pump)

This time we are interested in QH :

1 < COP < ∞

COP =|QH||W|

=|QH|

|QH|−QL

Carnot pump 1 < COP < ∞COP =TH

TH−TL

ACT: Leaving the fridge open

If you leave the door of your fridge open, you will get a heart-stopping electricity bill, but you will also:

A. Freeze the kitchen

B. Warm up the kitchen

Another way of stating theSecond law of thermodynamics

It is impossible for any process to have as its sole result the transfer of heat from a cooler to a hotter body (“refrigerator” or Clausius statement)

i.e.,

It is impossible to build a 100%-efficient heat engine (e = 1)

It is impossible for any system to undergo a process in which it absorbs heat form a reservoir at a single temperature and convert the heat completely into mechanical work, with the system ending in the same state as it began (“engine” or Kelvin-Plank statement)

Or

i.e.,

It is impossible to build a workless refrigerator (COP ∞)

Clausius Kelvin-Plank

100%-efficient engine

Workless refrigerator

ClausiusKelvin-Plank

Kelvin-Plank Clausius

100%-efficient engine

Workless refrigerator

ClausiusKelvin-Plank

New state variable: entropy

C H 0Q Q (QC > 0), (QH < 0) : same magnitude Q

HC

C H

0QQ

T T

TH TC

QHC

C H

QQT T

since TC < TH

positive large

negative small

This is positive (> 0) for all types of interacting systems.

QST

Define new state variable entropy (S ):

The change of entropy gives the direction of the process (direction of energy dispersion): entropy must increase

C H 0S S S

ACT: Glass of cold water

TW TR

QA glass of cold water is placed in a hot room. Consider a small heat transfer Q. Which of the following statements is a true?

A.SW > 0 , SR > 0B.SW < 0 , SR > 0C.SW > 0 , SR < 0

Terminology

TC TH

Q Glass of cold water (system) in hot room (environment)

Interacting components naturally exchange energy: real, spontaneous processes.Often there are two interacting parts labeled as system and environment

Second law of thermodynamics in terms of entropy

Δ Sinteracting parts = (Δ Ssystem + Δ Senvirenment) ⩾ 0

Reversibility

The processes we have analyzed are irreversible:• spontaneous dispersion of energy (in one direction)• the entropy of the universe increases

If a system and its environment are almost at the same temperature:• a small change in any temperature can reverse the direction of the heat flow reversible process• the entropy of the universe (not of each part) remains constant

Tsys Tenv=Tsys+ Q

Tsys Tenv=Tsys− Q

Change of entropy in an adiabatic process

That’s the easiest case: since Q = 0, ΔS = 0

We can now write the list of thermodynamics processes:

• Isobaric: Constant p

• Isochoric: Constant V

• Isothermal: Constant T• Adiabatic: Constant S (∆S = 0 )

Overall, ΔSall = ΔSgas + ΔS'environment > 0

this cycle is an irreversible process

Any process involving exchange of heat between bodies at different temperature is irreversible.

Only ideal isothermal and adiabatic processes can be reversible.

And this is why the Carnot cycle is reversible…

Exercise: Prove that ΔSall = 0 for a Carnot cycle.

In-class example: Second law

Which of the following is NOT an accurate statement of the second law of thermodynamics?

A. The change of entropy in an object can never be less than zero.

B. A perfect heat engine (e = 1) cannot exist.

C. A perfect refrigerator (COP = ∞) cannot exist.

D. ΔStotal (isolated system) > 0

E. All of the above are accurate statements of the second law of thermodynamics.

In-class example: Second law

Which of the following is NOT an accurate statement of the second law of thermodynamics?

A. The change of entropy in an object can never be less than zero.

B. A perfect heat engine (e = 1) cannot exist.

C. A perfect refrigerator (COP = ∞) cannot exist.

D. ΔStotal (isolated system) > 0

E. All of the above are accurate statements of the second law of thermodynamics.

Entropy of a part can decrease. Entropy of the total (universe) cannot decrease.