phy 113 c general physics i 11 am - 12:15 p m tr olin 101 plan for lecture 20:

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11/07/2013 PHY 113 C Fall 2013 -- Lecture 20 1 PHY 113 C General Physics I 11 AM - 12:15 PM TR Olin 101 Plan for Lecture 20: Chapter 19: The notion of temperature 1.Review of fluid physics 2.Temperature equilibrium 3.Temperature scales 4.Temperature in ideal gases

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PHY 113 C General Physics I 11 AM - 12:15 P M TR Olin 101 Plan for Lecture 20: Chapter 19: The notion of temperature Review of fluid physics Temperature equilibrium Temperature scales Temperature in ideal gases. Review: . The physics of fluids. - PowerPoint PPT Presentation

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Page 1: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 111/07/2013

PHY 113 C General Physics I11 AM - 12:15 PM TR Olin 101

Plan for Lecture 20:Chapter 19: The notion of temperature

1. Review of fluid physics2. Temperature equilibrium3. Temperature scales4. Temperature in ideal gases

Page 2: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 211/07/2013

Page 3: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 311/07/2013

The physics of fluids.•Fluids include liquids (usually “incompressible) and gases (highly “compressible”).•Fluids obey Newton’s equations of motion, but because they move within their containers, the application of Newton’s laws to fluids introduces some new forms.Pressure: P=force/area

1 (N/m2) = 1 Pascal

Density: r =mass/volume 1 kg/m3 = 0.001 gm/ml

Review:

Page 4: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 411/07/2013

)(ρ (constant) fluid, ibleincompressFor 00 yygPP r

Review of equations describing static fluids in terms of pressure P and density r:

gdydP ρ :surface sEarth'near fluids allFor

Note that for compressible fluids (such as air), the relationship between pressure and density is more complicated.

Buoyant force for fluid acting on a solid – net force due to volume Vdisplaced being displaced in fluid:

FB=rfluidVdisplacedg

Page 5: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 511/07/2013

Bernoulli’s equation:

22222

111

212

1 PgyvPgyv rrrr

Page 6: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 611/07/2013

Webassign questions on fluids (Assignment #17)

A large man sits on a four-legged chair with his feet off the floor. The combined mass of the man and chair is 95.0 kg. If the chair legs are circular and have a radius of 0.500 cm at the bottom, what pressure does each leg exert on the floor?

mgmg/4

P=F/A=(mg/4)/A

Page 7: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 711/07/2013

Webassign questions on fluids (Assignment #17)

A swimming pool has dimensions 32.0 m ✕ 7.0 m and a flat bottom. The pool is filled to a depth of 2.50 m with fresh water. (a) What is the force exerted by the water on the bottom?

(b) What is the force exerted by the water on each end? (The ends are 7.0 m.)

(c) What is the force exerted by the water on each side? (The sides are 32.0 m.)

Page 8: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 811/07/2013

Webassign questions on fluids (Assignment #17)

A swimming pool has dimensions 32.0 m ✕ 7.0 m and a flat bottom. The pool is filled to a depth of 2.50 m with fresh water. (a) What is the force exerted by the water on the bottom?

h=2.5m

Fbottom=PA=rghA

Page 9: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 911/07/2013

Webassign questions on fluids (Assignment #17)

A swimming pool has dimensions 32.0 m ✕ 7.0 m and a flat bottom. The pool is filled to a depth of 2.50 m with fresh water.

(b) What is the force exerted by the water on each end? (The ends are 7.0 m.)

h=2.5m

w=7.0m

22

2

0 21

2 gwhhhgwwdyyhgAPF

i

h

iiside rrr

Page 10: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 1011/07/2013

Webassign questions on fluids (Assignment #17)

Page 11: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 1111/07/2013

21

1

: thatNote

2

2

yAhA

yhgg

zyhg

zggP

HgOH

Hg

HgOH

rr

r

rr

22A

M

OHr

y

z

Page 12: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 1211/07/2013

2

1

2

1

21

1

1

2

2

2

AA

h

AAghg

yAhA

yhgg

Hg

OH

HgOH

HgOH

r

r

rr

rr

m0049.02113600

1000m2.0

:2/ m, 0.2For 21

h

AA

Page 13: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 1311/07/2013

Webassign questions on fluids (Assignment #17)

The gravitational force exerted on a solid object is 5.30 N. When the object is suspended from a spring scale and submerged in water, the scale reads 3.50 N (figure). Find the density of the object.

fluid

object

B

B

Fmg

NmgNmgF

rr

3.55.3

Page 14: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 1411/07/2013

Webassign questions on fluids (Assignment #17)

A light balloon is filled with 373 m3 of helium at atmospheric pressure. (a) At 0°C, the balloon can lift a payload of what mass? Note: rair = 2.9 kg/m3 : rHe = 0.179 kg/m3

VmVgF

VgmmmgF

Heairload

airB

Heload

B

rrr

r

0

Page 15: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 1511/07/2013

Webassign questions on fluids (Assignment #17)

A hypodermic syringe contains a medicine with the density of water (see figure below). The barrel of the syringe has a cross-sectional area A = 2.40  10-5 m2, and the needle has a cross-sectional area a = 1.00  10-8 m2. In the absence of a force on the plunger, the pressure everywhere is 1.00 atm. A force  of magnitude 2.65 N acts on the plunger, making medicine squirt horizontally from the needle. Determine the speed of the medicine as it leaves the needle's tip.  

112

2121

22222

111

212

1

/ ;/ ; :case In this

vvaAvAFPPyy

PgyvPgyv

rrrr

Page 16: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 1611/07/2013

Dictionary definition: temperature – a measure of the the warmth or coldness of an object or substance with reference to some standard value. The temperature of two systems is the same when the systems are in thermal equilibrium.

“Zeroth” law of thermodynamics:

If objects A and B are separately in thermal equilibrium with a third object C, then objects A and B are in thermal equilibrium with each other.

T1 T2 T3

Not equilibrium: Equilibrium:

Page 17: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 1711/07/2013

Constant temperature “bath”

T

T

At equilibrium:

Page 18: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 1811/07/2013

-100 -50 0 50 100 150 200-200

-150

-100

-50

0

50

100

150

200

250

300

350

400

450

500

T F

TC

Temperature scales TF=9/5 TC + 32

Kelvin scale:

T = TC + 273.15o

T 0

Page 19: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 1911/07/2013

iclicker question:Suppose you find yourself in a hotel in Europe or Canada. Which Celsius temperature would you set the thermostat for comfort?

A. -20oCB. +20oCC. +40oCD. +60oCE. +80oC

Page 20: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 2011/07/2013

There is a lowest temperature:

T0 = -273.15o C = 0 K

Kelvin (“absolute temperature”) scale

TC = -273.15 + TK

Example –

Room temperature = 68o F = 20o C = 293.15 K

Page 21: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 2111/07/2013

Page 22: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 2211/07/2013

Effects of temperature on matter

Solids and liquids

Li (equilibrium bond length at Ti)

Model of a solid composed of atoms and bonds

L

Thermal exansion:

L = a Li T

Page 23: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 2311/07/2013

Typical expansion coefficients at TC = 20o C:

Linear expansion: L = a Li T

Steel: a = 11 x 10-6/ oC

Concrete: a = 12 x 10-6/ oC

Volume expansion:

V=L3 V = 3a Vi T = b Vi T

Alcohol: b = 1.12 x 10-4/ oC

Air: b = 3.41 x 10-3/ oC

Page 24: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 2411/07/2013

iclicker question

On the last slide – we suggest that b=3a. Is this result

A. One of those mysteries of physics that has no explanation?

B. A result that we can derive?

L L+L

V=L3

V+V=(L+L)3@V(1+3(L/L))

Page 25: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 2511/07/2013

Page 26: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 2611/07/2013

Brass

Steel

Page 27: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 2711/07/2013

Switch in thermostat

Modern thermostats use electrical circuits to detect temperature

Page 28: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 2811/07/2013

Effects of temperature on materials – continued strange case of water:

Page 29: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 2911/07/2013

Effects of temperature on materials – continued -- ideal gas “law” (thanks to Robert Boyle (1627-1691), Jacques Charles (1746-1823), and Gay-Lussac (1778-1850)

nRTPV

pressure in Pascals

volume in m3 # of moles

temperature in K

8.314 J/(mol K)

1 mole corresponds to 6.022 x 1023 molecules

Page 30: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 3011/07/2013

nRTPV

P0 =12.6 atmT0 =27.5oCn0

P=?T=81.0oCn=n0/3

atm.

. atm. PTT

PP

RTnnRT

VPPV

/n nVVnRTPVRTnVP

9.4653003

15354612 3

3

000000

000000

Page 31: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 3111/07/2013

Assuming that air behaves like an ideal gas, what is the density of air at T=0o C and P=1 atm?

V

Mn

Vm

RTP

Vn

nRTPV

iii

rdensity mass

Page 32: PHY 113 C General Physics I 11 AM - 12:15  P M  TR Olin 101 Plan for Lecture 20:

PHY 113 C Fall 2013 -- Lecture 20 3211/07/2013

Typical composition of air: url: http://www.engineeringtoolbox.com/molecular-mass-air-d_679.html

3kg/m 1.29 1000

density mass

RTPM

V

Mn

Vm avgi

ii

r