11/12/2013phy 113 c fall 2013 -- lecture 211 phy 113 c general physics i 11 am - 12:15 pm mwf olin...

PHY 113 C Fall 2013 -- Lecture 21 111/12/2013

PHY 113 C General Physics I11 AM - 12:15 pM MWF Olin 101

Plan for Lecture 21:

Chapter 20: Thermodynamics

1. Heat and internal energy

2. Specific heat; latent heat

3. Thermodynamic work

4. First “law” of thermodynamics


In this chapter T ≡ temperature in Kelvin or Celsius units


Webassign question from Assignment #18

A rigid tank contains 1.50 moles of an ideal gas. Determine the number of moles of gas that must be withdrawn from the tank to lower the pressure of the gas from 28.5 atm to 5.10 atm. Assume the volume of the tank and the temperature of the gas remain constant during this operation.

removed moles 2316.1

moles2684.028.5

5.1moles5.1

21

1

212

2

1

2

1

22

11

22

11

22221111

nn

P

Pnn

n

n

P

P

RTn

RTn

VP

VP

RTnVPRTnVP


T1T2 Q

Heat – Q -- the energy that is transferred between a “system” and its “environment” because of a temperature difference that exists between them.

Sign convention:

Q<0 if T1 < T2

Q>0 if T1 > T2


Heat withdrawn from system

Thermal equilibrium – no heat transfer

Heat added to system


Units of heat: Joule

Other units: calorie = 4.186 J

Kilocalorie = 4186 J = Calorie

Note: in popular usage, 1 Calorie is a measure of the heat produced in food when oxidized in the body


iclicker question:According to the first law of thermodynamics, heat and work are related through the “internal energy” of a system and generally cannot be interconverted. However, we can ask the question: How many times does a person need to lift a 500 N barbell a height of 2 m to correspond to 2000 Calories (1 Calorie = 4186 J) of work?

A. 4186B. 8372C. 41860D. 83720E. None of these


Heat capacity: C = amount of heat which must be added to the “system” to raise its temperature by 1K (or 1o C).

Q = C DT

Heat capacity per mass: C=mc

Heat capacity per mole (for ideal gas): C=nCv

C=nCp

f

i

T

Tfi dTTCQ )(

:generally More


Some typical specific heats

Material J/(kg·oC) cal/(g·oC)

Water (15oC) 4186 1.00

Ice (-10oC) 2220 0.53

Steam (100oC) 2010 0.48

Wood 1700 0.41

Aluminum 900 0.22

Iron 448 0.11

Gold 129 0.03


iclicker question:Suppose you have 0.3 kg of hot coffee (at 100 oC). How much heat do you need to remove so that the temperature is reduced to 40 oC? (Note: for water, c=1000 kilocalories/(kg oC))

A. 300 kilocalories (1.256 x 106J)B. 12000 kilocalories (5.023 x 107J)C. 18000 kilocalories (7.535 x 107J)D. 30000 kilocalories (1.256 x 108J)E. None of these

Q=m c DT = (0.3)(1000)(100-40) = 18000 kilocalories


Q=

333J

Heat and changes in phase of materials

Example: A plot of temperature versus Q added to

1g = 0.001 kg of ice (initially at T=-30oC)

Q=2260J


Ti Tf C/L Q

A -30 oC 0 oC 2.09 J/oC 62.7 J

B 0 oC 0 oC 333 J 333 J

C 0 oC 100 oC 4.186 J/oC 418.6 J

D 100 oC 100 oC 2260 J 2260 J

E 100 oC 120 oC 2.01 J/oC 40.2 J

m=0.001 kg


Some typical latent heats

Material J/kg

Ice ÞWater (0oC) 333000

Water Þ Steam (100oC) 2260000

Solid N Þ Liquid N (63 K) 25500

Liquid N Þ Gaseous N2 (77 K) 201000

Solid Al Þ Liquid Al (660oC) 397000

Liquid Al Þ Gaseous Al (2450oC) 11400000


iclicker question:Suppose you have 0.3 kg of hot coffee (at 100 oC) which you would like to convert to ice coffee (at 0 oC). What is the minimum amount of ice (at 0 oC) you must add? (Note: for water, c=4186 J/(kg oC) and for ice, the latent heat of melting is 333000 J/kg. )

A. 0.2 kgB. 0.4 kgC. 0.6 kgD. 1 kgE. more

kg 0.377

333000/)100)(4186)(3(. 0)(

iceiceifwater mLmTTcm


Review

Suppose you have a well-insulated cup of hot coffee (m=0.3kg, T=100oC) to which you add 0.3 kg of ice (at 0oC). When your cup comes to equilibrium, what will be the temperature of the coffee?

C10.22

J/kg 333000 C) J/(kg 4186

3.0 )(

100

100)(

0)0()100(

o

o

f

icewater

icewaterwatericewater

iceicewaterwaterf

iceicewaterwaterfwatericewater

fwatericeiceicefwaterwater

T

Lc

kgmmcmm

LmcmT

LmcmTcmm

TcmLmTcmQ


Work defined for thermodynamic process

: workof definition General

f

i

dW fi

r

r

rF

the “system”

In most text books, thermodynamic work is work done on the “system”


f

i

f

i

f

i

V

V

y

y

y

y

PdVAdyA

FFdyW


Thermodynamic work

f

i

V

V

PdVW

Sign convention: W > 0 for compression W < 0 for expansion


Work done on system --

“Isobaric” (constant pressure process)P

(1.

013

x 10

5)

Pa

Po

Vi Vf

W= -Po(Vf - Vi)


Work done on system --

“Isovolumetric” (constant volume process)P

(1.

013

x 10

5)

Pa

Vi

Pi

Pf

W=0


Work done on system which is an ideal gas

“Isothermal” (constant temperature process)P

(1.

013

x 10

5)

Pa

Pi

Vi Vf

i

fii

i

fV

V

V

V

fi

V

VVP

V

VnRTdV

V

nRTPdVW

f

i

f

i

ln

ln

For ideal gas:

PV = nRT


Work done on a system which is an ideal gas:

“Adiabatic” (no heat flow in the process process)P

(1.

013

x 10

5)

Pa

Pi

Vi Vf

Qi®f = 0

11

1

i

fii

V

V

ii

V

V

fi V

VVPdV

V

VPPdVW

f

i

f

i

For ideal gas:

PVg = PiVig

1

11

f

iii

V

VVP

PHY 113 C Fall 2013 -- Lecture 2111/12/201324


Thermodynamic processes: Q: heat added to system (Q>0 if TE>TS) W: work done on system (W<0 if

system expands

f

i

f

i

V

V

T

T

PdVW

dTTCQ )(


iclicker question:What happens to the “system” when Q and W are applied?

A. Its energy increasesB. Its energy decreaseC. Its energy remains the sameD. Insufficient information to answer question


Internal energy of a system

Eint(T,V,P….)

The internal energy is a “state” property of the system, depending on the instantaneous parameters (such as T, P, V, etc.). By contrast, Q and W describe path-dependent processes.

),,(),,( intintint iiifff PVTEPVTEE


First law of thermodynamics:

WQE int

Ei Ef

Q W


Applications of first law of thermodyamics

System ideal gas

nRTPV

pressure in Pascals

volume in m3 # of moles

temperature in K

8.314 J/(mol K)


Ideal gas -- continued

..............................

diatomicfor

monoatomicfor

gas ideal of on type dependingparameter

1

1

1

1 :energy Internal

:state ofEquation

57

35

int

PVnRTE

nRTPV


iclicker question:We are about to see several P-V diagrams. Why is this helpful?

A. It will help us analyze the thermodynamic work

B. Physicists like nice graphsC. It is not actually helpful


Constant volume process on an ideal gas

PVnRTE

nRTPV

1

1

1

1 :energy Internal

:state ofEquation

int

P

V

Pf Vf=Vi

Pi Vi

1

0

int

iif VPP

QE

W


Constant pressure process on an ideal gas

PVnRTE

nRTPV

1

1

1

1 :energy Internal

:state ofEquation

int

P

V

Pf = Pi VfPi Vi

1

1int

ifi

ifi

ifi

VVPQ

VVPE

VVPW


Constant temperature process on an ideal gas

PVnRTE

nRTPV

1

1

1

1 :energy Internal

:state ofEquation

int

f

iii

i

f

V

V

P

PVP

V

VnRT

PdVWQ

Ef

i

ln

ln

0int


Consider the process described by ABCA

iclicker exercise:What is the net change in total energy?

A. 0B. 12000 JC. Who knows?



iclicker exercise:What is the net heat added to the system?

A. 0B. 12000 JC. Who knows?



iclicker exercise:What is the net work done on the system?

A. 0B. -12000 JC. Who knows?


Webassign problem from Assignment #19

11/12/2013phy 113 c fall 2013 -- lecture 211 phy 113 c general physics i 11 am - 12:15 pm mwf olin...

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