Physics 231 Final Review •  Final Exam Thursday Dec. 15, 8-10 pm in E100 Vet. Medical Center

–  The exam will be over the entire material for this term •  Don’t forget the Homework set is due TONIGHT (Fri.) 10 pm.

Practical!

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

coefficient 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

Problem 2

There is no mention of air resistance and the option that they are released at different heights is not mentioned among the answers.

v = v0 + at a is constant

a=v − v0

t= 15m / s− 0

9s= 1.667m / s2

Problem 3

Problem 4

x(t) = 12

at2

x(t = 4.99s)− x(t = 2.77s) = 12

a (4.99s)2 − (2.77s)2( ) = 14.36m

v = v0 + at a is constant

a=v − v0

t= 15m / s− 0

9s= 1.667m / s2

Problem 5

Energy is conserved12

mv02 + mgy0 =

12

mv f2 + mgy f

multiply by 2/m and rearrange ⇒ v f2 = v0

2 + 2g( y0 − y f )

v f = v02 + 2g( y0 − y f ) = 8.822 + 2 ⋅9.81⋅1.86m / s = 10.7m / s

Problem 6

!F1 +!F2 = m!a

a =!F1 +!F2 / m

a = sqrt F1,x + F2,x⎡⎣ ⎤⎦

2+ F1,y + F2,y⎡⎣ ⎤⎦

2⎛⎝

⎞⎠ / m

a = sqrt −9.9+8.6⎡⎣ ⎤⎦2+ 6+ 4.6⎡⎣ ⎤⎦

2( ) / 6.8 m / s 2= 1.57m / s 2

Problem 7

F = (m1 + m2 )a

T = m1a < 75N ⇒ a < 75N

m1

F < 75N

m1

m1 + m2( ) = 75N1.77

1.77 + 6.47( ) = 349N

T

Problem 8

T = m1a < 75N

⇒ a < 75N

m1

= 751.77

m / s2 = 42.4m / s2

Problem 9

P = F ⋅v = F ⋅ d

t = 58N ⋅ 109m

1.85min⋅1min.

60s= 56.96 W

Problem 10

W =Weight ⋅h = PE f − PE0

h = l sinθ = 10.7m ⋅sin 12°( )h = 2.225m

W = 640N ⋅2.225m = 1424J

Problem 11

Conservation of energymass 112

m1vbefore2 = m1gh

vbefore = 2gh

vbefore = 2 ⋅9.81⋅10.2m / s

vbefore = 14.15m / s

Totally inelastic collision

m1vbefore = m1 + m2( )vafter

⇒ vafter =m1vbefore

m1 + m2

vafter =13.7

13.7 + 23.6⋅14.15m / s

vafter = 5.20m / s

Problem 12

Problem 13

Problem14 & 15

τ = 0⇒ Ly = Lx

I = mr 2; ω= vr

; L = mvr

⇒ Lx = mvxrx = Ly = mvyry

force is along r

L = Iω

⇒ vy = vx

rx

ry

= 2vx

KEy =12

mvy2 = 4 ⋅ 1

2mvx

2

KE is increased by work of pulling on the string

Problem 16, 17 &18

I = mr 2; ω= vr

; L = mvr

⇒ Lx = mvxrx = Ly = mvyry

⇒ vy = vx

rx

ry

= 2vx

KEy =12

mvy2 = 4 ⋅ 1

2mvx

2

Work energy theoremW = KEy − KEx = 3KEx

I yω y = Ixω x ⇒ω y =

Ix

I y

ω x =r x

2

ry2 ω x = 4ω x

τ = 0⇒ Ly = Lx

force is along r

L = Iω

Problem 19

Problem 20 & 21

Problem 22

m v2

r=

GM Emr 2

⇒ v =

GM E

r

r = RE +8150km = 14520km

v = 6.67x10−11 ⋅5.97x1024

1.452x107 m / s = 5236m / s

Problem 23

mg = ρVg = 94.5N

Fscale = 23.7N = mg − FB ⇒ FB = 94.5N − 23.7N = 70.8N⇒ FB = 70.8N = ρwaterVg

⇒ mgFB

= ρVgρwaterVg

= ρρwater

⇒ ρρwater

= 94.5N70.8N

= 1.334

⇒ ρ = 1.334 ⋅ ρwater = 1.334 ⋅1000kg / m3 = 1334kg / m3

Problem 24

floatingWice = FB = ρ waterVdisplaced g

Vdisplaced =Wice

ρ water g=

Wmelted _ ice

ρ water g

Melted ice takes up the same volume as the ice diplaced.

Problem 25 & 26

incompressible fluid:vq Aq = vr Ar

vq = vr Ar / Aq = vr / 4

Pu +12ρvu

2 + ρgyu = Pq +12ρvq

2 + ρgyq

⇒ Pu = Pq +12ρ vq

2 − vu2( )

but vu = 4vq ⇒ Pu < Pq

Problem 27 & 28

Pt +12ρvt

2 + ρgyt = Ps +12ρvs

2 + ρgys

At = As ⇒ vt = vs fluid is incompressible

Pt + ρgyt = Ps + ρgys ⇒ Pt = Ps + ρgys − ρgyt

yt > ys ⇒ Pt < Ps

At = Ar ⇒ vt = vr fluid is incompressible

Problem 29

ƒ' = ƒ

v + vo

v − vs

⎛⎝⎜

⎞⎠⎟

; vo = 0 ⇒ v − vs( )ƒ' = ƒv ⇒ v − vs =

ƒƒ'

v

vs = v − ƒ

ƒ'v = v 1− ƒ

ƒ'⎛⎝⎜

⎞⎠⎟= 340m / s 1− 244

261⎛⎝⎜

⎞⎠⎟= 22.1m / s

Problem 30

βA = 10log10

I A

I0

⎛⎝⎜

⎞⎠⎟

; βB = 10log10

IB

I0

⎛⎝⎜

⎞⎠⎟

Can plug in I0=10-12 W/m2

and compute it, but I0 value does not change answer

βA − βB = 10log10

I A

I0

⎛⎝⎜

⎞⎠⎟−10log10

IB

I0

⎛⎝⎜

⎞⎠⎟

βA − βB == 10(log10 I A( )− log10 I0( )− log10 IB( ) + log10 I0( ))βA − βB = 10(log10 I A( )− log10 IB( )) = 10log10

I A

IB

⎛⎝⎜

⎞⎠⎟= 10log10

7.4x10−4

2.6x10−8

⎛⎝⎜

⎞⎠⎟

βA − βB = 44.5dB

H = kALΔT

HLeft

HRight

=

kALeft

LLeft

ΔT

kARight

LRight

ΔT=

ALeft

LLeft

ARight

LRight

⋅LLeft ⋅ LRight

LLeft ⋅ LRight

⇒HLeft

HRight

=ALeft

ARight

⋅LRight

LLeft

= 2 ⋅2 = 4

Problem 31

Problem 32

H = kALΔT

H400

H800

=

kAL

200K

kAL

600= 1

3

Problem 33

P = nRTV

= 1.24 ⋅8.31⋅(273.15+ 27.9)4.66x10−3 Pa ⋅ 1atm

1.013x105 Pa= 6.57atm

Problem 34 & 35

PiViγ = PfVf

γ

Pf = Pi

Vi

V f

⎛

⎝⎜

⎞

⎠⎟

γ

= Pi

Vi

V f

⎛

⎝⎜

⎞

⎠⎟

CP /CV

cp / cV >1

vV f >Vi ⇒ Pf < Pi

Problem 36, 37, 38

adiabatic ⇒ΔQ = ΔS = 0ΔQ = ΔU + pΔV = 0

ΔU = f2

nRΔT = − pΔV < 0

Problem 39

ideal means that it has Carnot efficiency

ε = .173= WQh

=Th −Tc

Th

= 74.6Th

Th

74.6= 1

.173= 5.78

Th = 5.78 ⋅74.6K = 431K