PHYSICS 231

INTRODUCTORY PHYSICS I

Lecture 19

• First Law of Thermodynamics

• Work done by/on a gas

Last Lecture

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Q = ΔU +Wby the gas

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Wby the gas = PΔV = −Won the gas

Some Vocabulary

• Isobaric• P = constant

• Isovolumetric• V = constant• W = 0

• Isothermal• T = constant• U = 0 (ideal gas)

• Adiabatic• Q = 0

V

V

V

V

P

P

P

P

P-V DiagramsP

V

Path moves to right:

Wby the gas = Area under curve

P

V

Path moves to left:

Wby the gas = - Area under curve

(Won the gas = - Wby the gas)

Work from closed cycles

WA->B->A= Area

(work done by gas)

WA->B->A= -Area

Clockwise cycle:

Counterclockwise cycle:

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U = 0 in closed cycles

Example 12.8a

Consider an ideal gas undergoing the trajectory through the PV diagram.In going from A to B to C, the work done BY the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

C

Example 12.8b

In going from A to B to C, the change of the internal energy of the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

C

Example 12.8c

In going from A to B to C, the amount of heat added to the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

CD

Example 12.8d

In going from A to B to C to D to A, the work done BY the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

CD

Example 12.8e

In going from A to B to C to D to A, the change of the internal energy of the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

CD

Example 12.8f

In going from A to B to C to D to A, the heat added to the gas is _______ 0.

P

V

AB

a) >

b) <

c) =

CD

Example 12.7Consider a monotonic ideal gas.

a) What work was done by the gas from A to B?

b) What heat was added to the gas between A and B?

c) What work was done by the gas from B to C?

d) What heat was added to the gas between B and C?

e) What work was done by the gas from C to A?

f) What heat was added to the gas from C to A?

V (m3)

P (kPa)

25

50

75

0.2 0.4 0.6

A

BC

20,000 J

20,000 J

-10,000 J

-25,000 J

0

15,000 J

Example 12.7 (Continued)

g) What was total work done by gas in cycle?

h) What was total heat added to gas in cycle?

V (m3)

P (kPa)A

BC

WAB + WBC + WCA = 10,000 J

QAB + QBC + QCA = 10,000 JThis does NOT mean that the engine is 100% efficient!|Qin| = QAB + QCA= 35,000 J

|Qout| = |QBC| = 25,000 JWeng = |Qin|-|Qout|

|Qin|

|Qout|Exhaust!!!

Heat Engines

• Described by a cycle with:

Qhot= heat that flows into engine from source at Thot

Qcold= heat exhausted from engine at lower temperature, Tcold

W= work done by engine

• Efficiency is defined:

Qhot

engine

Qcold

W

=Qhot −QcoldQhot

=1−QcoldQhot

engine:

e=WQhot

€

W =Qhot −Qcoldusing

2nd Law of Thermodynamics(version 1)

The most efficient engine is the Carnot Engine (an idealized engine), for which:

No heat engine can be 100% efficient

€

⇒

€

eCarnot = WQhot

=1− QcoldQhot

=1− TcoldThot

In practice, we always have

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e < eCarnot

€

QcoldQhot

= TcoldThot

(T in Kelvin)

Carnot Cycle

Example 12.9

An ideal engine (Carnot) is rated at 50% efficiency when it is able to exhaust heat at a temperature of 20 ºC. If the exhaust temperature is lowered to -30 ºC, what is the new efficiency.

e = 0.585

Refrigerators

Qhot

fridge

Qcold

W

Just a heat engine run in reverse!• Pull Qcold from fridge• Exhaust Qhot to outside

Most efficient is Carnot refrigerator:

Note: Highest COP for small T differences

Coefficient of Performance:

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COP(cooling) = QcoldW

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COP(cooling) < COPCarnot = TcoldThot −Tcold

Heat Pumps

Qhot

heatpump

Qcold

W

Like Refrigerator: Best performance for small T

Same as refrigerator, except• Pull Qcold from environment• Exhaust Qhot to inside of house

Again, most efficient is Carnot:

Coefficient of Performance:

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COP(heating) = QhotW

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COP(heat) < COPCarnot = ThotThot −Tcold

Example 12.10

A modern gas furnace can work at practically 100% efficiency, i.e., 100% of the heat from burning the gas is converted into heat for the home. Assume that a heat pump works at 50% of the efficiency of an ideal heat pump.

If electricity costs 3 times as much per kw-hr as gas, for what range of outside temperatures is it advantageous to use a heat pump?Assume Tinside = 295 ºK.

T =29556=245.8°K = -27 °C

Entropy• Measure of Disorder of the system

(randomness, ignorance)• S = kBlog(N)

N = # of possible arrangements for fixed E and Q

0

100

200

300

400

500

600

700

800

900

1000

(0,12) (1,11) (2,10) (3,9) (4,8) (5,7) (6,6) (7,5) (8,4) (9,3) (10,2) (11,1) (12,0)

Relative probabilities for 12 molecules to arrange on two halves of container.

On a macroscopic level, one finds that adding heat raises entropy:

Defines temperature in Kelvin!

S =Q /T

2nd Law of Thermodynamics(version 2)

The Total Entropy of the Universe can never decrease.

Why does Q flow from hot to cold?

• Consider two systems, one with TA and one with TB

• Allow Q > 0 to flow from TA to TB

• Entropy changes by:

S = Q/TB - Q/TA

• This can only occur if S > 0, requiring TA > TB.

• System will achieve more randomness by exchanging heat until TB = TA

Carnot Engine

Carnot cycle is most efficient possible, because the total entropy change is zero. It is a “reversible process”.

For real engines:

€

S = ΔSenvironment = QcoldTcold

− QhotThot

> 0 ⇒

€

e = WQhot

=1− QcoldQhot

<1− TcoldThot

= eCarnot

Example 12.11a

An engine does an amount of work W, and exhausts heat at a temperature of 50 degrees C. The chemical energy contained in the fuel must be greater than, and not equal to, W.

a) Trueb) False

Example 12.11b

A locomotive is powered by a large engine that exhausts heat into a large heat exchanger that stays close to the temperature of the atmosphere. The engine should be more efficient on a very cold day than on a warm day.

a) Trueb) False

Example 12.11c

An air conditioner uses an amount of electrical energy U to cool a home. The amount of heat removed from the home must be less than or equal to U.

a) Trueb) False

Example 12.11d

A heat pump uses an amount of electrical energy U to heat a home. The amount of heat added to a home must be less than or equal to U.

a) Trueb) False