Physics II Reviewer for Prelims

PHYSICS 2 PRELIMS REVIEWER (2nd

Semester 2011-2012) TEMPERATURE FORMULAS

degF = 95 degC + 32 degR =

825 degC + 460

degF = 458 degR ndash

51112 degC =

59 (degF ndash 32)

degC = 258 (degR ndash 460) degR =

845 [degF +

51112 ]

K = 273 + C

Cdeg = 95 Fdeg = K

THERMOMETER SCALES

REFERENCE POINTS

CELSIUS FAHRENHE

IT KELVIN RANKINE

STEAM PT 100 oC 212

oF 373 K 492

oR

FREEZING PT

0 oC 32

oF 273K 460

oR

ABSOLUTE ZERO

-273 oC 0 K

SAMPLE PROBLEMS

1 Convert a) -15

oF to KELVIN

answer 24689 K b) 303 K to CELSIUS and to FAHRENHEIT answer 30

oC and 86

oF

c) -459 oF to CELSIUS

answer -27278 oC

2 A thermometer has 0

oA freezing point and 80

oA the

boiling point of water at normal pressure Give the equation to change A to

oC A to

oF and A to K

CELSIUS FAHRENHEIT KELVIN 100

OC 212

OF 373 K

C F K 0

OC 32

OF 273 K

TEMPERATURE SCALE OF ldquoArdquo

80 OA

A 0

OA

A to C A to K 80minus 0

Aminus 0=

100minus 0

Cminus 0

80minus 0

Aminus 0=

373minus 273

Kminus 273

A to F

80minus 0

Aminus 0=

212minus 32

Fminus 32

Gamitin nyo lang yung law of proportionality between SCALE A and STANDARD SCALE para makuha nyo ang equivalent formula ng A to the

other standard temperature scale

3 The change in temperature of a body is 45 Co How

much is it in Fo

In K

45Cdeg x

95 Fdeg

1Cdeg = 81Fdeg

45Cdeg x 1K1Cdeg = 45K

Fo and Co is different from oF and oC Fo and Co is a notation for the change in temperature

ADDITIONAL PROBLEMS

1 While vacationing in Italy you see on local TV one summer morning that the temperature will rise from the current 18

oC to a high of 39

oC What is the

corresponding increase in the Fahrenheit temperature

degF = 95 degC + 32

degF = 95 (18) + 32 degF =

95 (39) + 32

degF = 644degF degF = 1022degF 1022 ndash 644 = 378 F deg

2 Two beakers of water A and B initially are at the same temperature The temperature of the water in beaker A is increased 10

oF and the temperature of the water in

beaker B is increased 10 K After those temperature changes which beaker of water has the higher temperature Explain

degCA = 59 (degF ndash 32) degCB = K ndash 273

degCA = 59 (10 ndash 32) degCB = 10 ndash 273

degCA = -1222degC degCB = -263deg

Beaker A increased -1222 oC while beaker B

increased -263 oC which means beaker A has a

higher temperature than beaker B

3 You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 10 K What is its temperature change in F

o and C

o

-10K = -10degC

-10degC x

95 Fdeg

1Cdeg = -18Fdeg

THERMAL EXPANSION Linear Lf = Lo + ΔL Lf = Lo (1 + α ΔT) ΔL = α Lo ΔT where α = coefficient of thermal expansion Area x Δx ΔA = Af ndash Ao

ΔA = 2xΔx ΔA = 2x (x α ΔT) x ΔA = 2α Ao ΔT Δx Volume ΔV = 3α Vo ΔT ΔV = β Vo ΔT x x where βsolid lt βliquid

Over Flow = (βliquid ndash βsolid) Vo ΔT x

THERMAL STRESS (Tensile or Compressive)

S = FA = Y

∆LLO

where Y = Youngrsquos Modulus of Elasticity ∆LLo

= fractional change due to stress

∆LLo

= α∆T = fractional change due to change in length

COMPRESSION Tf gt To

TENSION Tf lt To

UNITS F N dynes lb A m

2 cm

2 in

2

S Nm

2 dynescm

2 lbin

2 or lbft

2

Y Nm

2 dynescm

2 lbin

2 or lbft

2

EXAMPLE PROBLEMS

1 The ends of a steel rod exactly 02 in2 in cross-sectional

area are rigidly held between two fixed points at a temperature of 30

oC When the temperature drops to

20 oC find

a) the stress developed at the ends of the rod b) the pull in the rod

Ysteel = 33 x 10

6 lbin

2 αsteel = 11 x 10

-6 C

o

a) S = FA = Yα∆T

S = (33 x 106)(11 x 10

-6)(-10) = -3630

lbin

2

b) F = SA F = (-3630)(02) = -726lb

2 An aluminum baseball bat has a length of 86 cm at a

temperature of 17 oC When the temperature of the bat is

raised the bat lengthens by 16 x 10

-4 m Determine the

final temperature of the bat ( αAl = 23 x 10

-6 C

o)

Lo = 86 cm or 086 m To = 17

oC

ΔL = 16 x 10-4

m

∆T = ∆LLoα

Tf ndash 17 = 16 x 10

-4

(086)(23 x 10-6

)

Tf = 25089degC

3 The brass bar and the aluminum bar as shown are each attached to an immovable wall At 28 degC the air gap between the rod is 13 x 10

-3 m At what temperature will

the gap be closed BRASS ALUMINUM 2m 1m αBr = 19 X 10

-6 Cdeg

αAl = 23 X 10

-6 Cdeg

∆LBr + ∆LAl = 13 x 10-3

m

(α Lo ΔT)Br + (α Lo ΔT)Al = 13 x 10-3

m [(19 X 10

-6)(2)∆T + (23 X 10

-6)(1)∆T = 13 x 10

-3 m

∆T = 2131Cdeg Tf = 28 + 2131 = 4931Cdeg

4 A copper cylinder is initially at 200

oC At what

temperature will its volume be 0150 larger than it is at 200

oC (βCu = 51 x 10

-5)

Let Xo = the initial volume Let Xo + 00015 Xo = the final volume Let ∆V = 0150Vo ΔV = βVo ΔT 15 x 10

-3Vo = (51 x 10

-5)(Vo)(Tf ndash 20)

Tf = 4941degC

5 A steel tank is completely filled with 28 m3 of ethanol

when both the tank and the ethanol are at temperature of 320degC When the tank and its contents have cooled to 18degC what additional volume of ethanol can be put into the tank

(βe = 75 x 10-5

Co

βs = 36 x 10-5

Co)

ΔVs = Vo βs ΔT = (28) (36 x 10-5

) (-14) ΔVs = - 000141 ΔVs = - 141 x 10

-3 m

3 ~ - 141 L

ΔVe = Vo βe ΔT = (28) (75 x 10

-5) (-14)

ΔVe = - 00294 ΔVe = -294 x 10

-2 m

3 ~ -294 L

6 A machinist bores a hole of diameter 135 cm in a steel

plate at a temperature of 25degC What is the cross sectional area of the hole a) at 25degC b) when the temperature of the plate is increased to 175degC Assume that the coefficient of linear expansion remains constant over this temperature range

a A0 = 120587r0

2 = 120587 (0675)

2 =1431 cm

2

b ∆A = (2prop)Ao∆T = 2(12 x 10

-6)(1431)(175-25)

= 515 x 10-5

cm2

A = Ao + ∆A

= 1431 + (515 x 10-6

) = 1436 cm2

7 The outer diameter of a glass jar and the inner diameter of

an iron lid are both 725 mm at room temperature (20degC) What will be the size of the mismatch between the lid and the jar if the lid is briefly held under hot water until its temperature rises to 50degC without changing the temperature of the glass

Find the change ΔL in the diameter of the lid

ΔL = αLΔT = (1210minus (Cdeg)minus)(725 mm)(300 Cdeg) = 026 mm

8 An iron rod and a zinc rod have lengths of 2555 cm and

2550 cm respectively at 32degF At what temperature will the rods have the same lengths The coefficients of linear expansion of iron and zinc are 0000010 per Cdeg and 0000030 per Cdeg respectively

L0 iron = 2555 cm L0 zinc = 2550 cm T0 = 32degF = 0degC Tf= ∆T = Tf - To Liron = Lzinc Lo iron (1 + prop ∆T) = Lo zinc (1 + prop ∆T) 2555 [1 + (0000010)(∆T)] = 2550 [1 + (0000030)(∆T) 2555 + 00002555 ∆T = 255 + 0000765 ∆T 00002555∆119879 ndash 0000765 ∆119879 = 255 ndash 255 - 00005095 ∆119879 = 005 ∆119879 = 9814 119862deg Tf = 9814 + 0degC Tf = 9814 degC

9 a) A wire that is 150 m long at 20degC is found to increase in

length by 190 cm when warmed to 420degC Compute its average coefficient of linear expansion for this temperature range (b) The wire is stretched just tant (zero tension) at 420degC Find the stress in the wire if it is cooled to 20degC without being allowed to contract Youngrsquos modulus for the wire is 20 x 10

11 Pa

(a) ΔL=αLΔT

prop = ∆ L

Lo∆T=

0019 m

150m (420degCminus20degC)=32 x 10

-5(Cdeg)

-1

(b) Stress FA= minusYαΔT

ΔT=20degCminus420degC=minus400 Cdeg

ΔT always means final temperature minus initial temperature

FA= minus(2010 11 Pa)(3210 -5 (Cdeg) -1)(minus400 Cdeg) = +2610 9 Pa

10 A glass flask whose volume is 1000 m

3 at 0degC is completely

filled with mercury at this temperature When flask and mercury are warmed to 55degC 895 cm

3 of mercury

overflow If the coefficient of volume expansion of mercury is 18 x 10

-5K compute the coefficient of volume expansion

of the glass

∆VHg - ∆Vglass = 895 cm3

∆VHg = Vo120573Hg∆119879

= (1000) (18 x 10-5

) (55) = 99cm3

∆Vglass = ∆VHg - 895 = 99 ndash 895 = 095

120573glass= ∆Vglass

119881119900∆119879 =

095

1000 (55) = 17 x 10

-5K

HEAT Q Q = m c ΔT where c = specific heat capacity UNITS Q calorie Joule (kcal) Btu m gram Kg lb ΔT C

o C

o F

o

c calgC

o JkgC

o or kcalkgC

BtulbF

o

CONVERSION FACTOR 1 Btu = 252 cal 1 kcal = 1000 cal 1 cal = 4186 J

cwater= 1cal

gCo= 1

kcal

kgCo= 1

Btu

lbFo

CALORIMETRY (Principle of conservation energy) HEAT LOST by warmer bodies is equal to the HEAT GAINED by the cooler bodies

QLOSS = QGAINED

SAMPLE PROBLEMS

1 The Calorimeter cup 015kg in mass is made of aluminum and contains 02kg of water Initially the water and the cup have a common temperature of 18

oC A 40g mass of solid material is heated to a

temperature of the water the cup and the solid material was observed at 22

oC after thermal

equilibrium was established

QLOSS = QGAINED

40 c (97 ndash 22)Co = [Cal mAl + cw mw] (22 ndash 18)

c =

[ (200022) + (1501) ](4)(40)(75)

c = 031calgCdeg

2 A substance boils at 120 oC and freezes at -20

oC Its

specific heat capacity as a solid liquid and gas are 1 15 and 04 calgC

o respectively When 15g of this

substance at 180 oC is brought to -40

oC 6274 Btu of

heat is liberated If the Lf of the substance is 200 calg determine its Lv

6274Btu x 252cal1Btu = 1581048 cal

Q = mc∆T + mLf + mc∆T + mLv + mc∆T 1581048 = 15[-20-(-40)] + (15)(200) + (15)(15)[120-(-20)] + 15Lv + (15)(04)(180-120) 1581048 = 6810 _ 15Lv

Lv = 600 calg

3 How much water could be cooled from 22degC to 5degC by a 10-g ice cube placed into a glass of water The glass container is initially at 5deg C and its water is equivalent is 10-g

water equivalent means yung mass ng water is equal to mass of container Equal din ang value of C

Qlost = Qgained miceCwater(Tf-Ti) + miceLf = mwaterCwater(TfmdashTi) +

mglassCwater(Tf-Ti)

(10g)(1)(22degC - 0degC) + (10g)(80 calg) = mw (1)(22degC - 5degC) +

(10g)(1)(22degC - 5degC) 220 + 800 = 17 mw + 170

17 mw = 850

mw = 50 g

4 When 200g of copper at 100degC is dropped into an aluminum container of mass 120g and containing 200g of an unknown liquid at 15degC the mixture reaches a temperature of 45degC What is the specific heat of the unknown liquid C of copper is 0093 and that of aluminum is 021

Qlost = Qgained

mcopperCcopper(Tf-Ti) = malCal(Tf-Ti) + miquidCliquid(Tf-Ti) 200g(0093)(100degC-45degC) = 120g(021)(45degC-15degC) +

200g(Cliquid)(45deg-15degC) 1023 = 756 + 6000 Cliquid

Cliquid = 00445

5 A worker needs to know the temperature inside an

oven He removes a 2-lb iron bar from the oven and places it in a 1-lb aluminum container partially filled inside an oven with 2-lb of water The system is immediately insulated and the temperature of the water and container rises from 70degF to the equilibrium temperature of 120degF What was the oven temperature C of iron = 0113 and C of Al = 022

Qlost = Qgained

mironCiron∆T = mwateCwate∆T + malCal∆T (2)(0113)(T - 120) = [(2)(1)(120 ndash 70) + (1)(022)(120 ndash 70)]

-2712 + 0226 T = 100 + 11 0226 T = 13812

T = 61115 degF

6 A substance boils at at 120degC and freezes at -20degC Its specific heat as a solid a liquid and a gas are 1 15 and 040 calg-Cdeg respectively When 15g of this substance at 180degC is brought to -40degC 62738 of heat is liberated If the heat of fusion of the substance is 200 calg what is the heat of vaporization

may tatlong119898119888∆T kasi 3 beses siya maguundergo ng temperature change First from -40deg119862 to -20degCSecond from -20degC to 120degCThird from 120degC to 180degCmaymLfkasimaguundergosiyang phase change from solid to liquid Lv from liquid to steam

Q = mc∆T + mLf + mc∆T + mLv +mc∆T = 15(1)(20) + 15(200) + 15(15)(140) + 15Lv + 15(04)(60) = 5010 + Lv Lv= 600 calg

HEAT TRANSFER

I CONDUCTION

H = KA∆T

L

where H = rate of heat flow t = time of heat flow K = thermal conductivity A = cross-sectional area L = thickness of material

∆TL = temperature gradient

II CONVECTION

H = Qt =

KL A∆T = h A∆T =

AT1hf

+LK +

1hf

where h = film coefficient

III RADIATION P = eσAT

4

where P = radiant power (Heat Rate)

P = Qt

e = emissivity of material A = surface area T = temperature of the material net radiant power P = eσA(T

4 ndash Ts

4)

where Ts = surrounding temperature σ = Stefan ndash Boltzmann Constant

σ =57 x 10-8 W

m2K

4

or

57 x 10-12 W

cm2K

4

Note for BLACKBODY e = 1 SAMPLE PROBLEMS

1 The temperature at the ends of a bar are 85 oC and 27

oC The bar has a length of 68 cm What is the

temperature at a point 22 com from the cooler end of the bar

H2T = HT1

KA∆T2T

L2T =

KA∆TT1

LT1

T2 - T68 - 22 =

T - T1

22

T = 4576degc

2 Compute the heat conducted per hour through a plate-glass window of area 15 ft

2 and thickness frac14 in when

the inside temperature is 70 oF and the outside

temperature is 20 oF The thermal conductivity of the

glass is 58 Btu[h ft2

(Foin)] and the film coefficient for

the fluid is about 2 Btu(h ft2

Fo)

Qt =

A∆T1h +

Lk +

1h

Qt =

15(70 - 20)

12 +

14

58 + 12

Qt = 71901

Btuh

in one day find the amount of heat flow

H = Qt

Q = Ht

Q = 71901 Btuh x 24h

Q = 172562 Btu

3 A spherical body emits 1 x 108 watts at 1000K surface

temperature Assuming the body is a perfect emitter find its radius

P = eσAT4

A = P

eσT4 A = 4πr

2

4πr2 =

PeσT

4

r = P

4πeσT4

r = 1 x 10

8

4π(1)(56 x 10-8

)(1000)4

r = 1185m

4 (1766page 604) One end of an insulated metal rod is maintained at 1000˚C and the other end is maintained at 000˚C by an ice-water mixture The rod is 600 cm long and has a cross-sectional area of 125 cm

2 The heat conducted by

the rod melts 850 g of ice in 100min Find the thermal conductivity k of the metal Given TH = 1000˚C TC = 000˚C L = 600 cm = 0600 m A = 125 cm

2 = 125 x 10

-4 m

2

m = 85 g = 00085 kg

Lf = 80 kcalkg times4186 119869

1 119896119888119886119897= 334 times 105 119869

119896119892

Solution

Q = mLf = (00085 kg)( 334 times 105 119869

119896119892)

Q= 284 times 103119869 119876

119905=

284times 103119869

10 119898119894119899 times

1 119898119894119899

60 119904= 473 119882

119867 =119876

119905=

119896119860 (119879119867minus 119879119862)

119871

119896 = 119867 119871

119860 (119879119867minus 119879119862)=

473 119882 ( 0600 119898)

125 times10minus4 1198982 (100119862deg)= 227

119882

119898∙119896

5 (1767page 604) A carpenter builds an exterior house wall with a layer of wood 30 cm thick on the outside and layer of Styrofoam insulation 22 cm thick on the inside wall surface The wood has k = 0080 WmK and the Styrofoam has k= 0010 WmK the interior surface

temperature is 190 the exterior temperature is -100 (a) What is the temperature at the plane where the wood meets the Styrofoam (b) What is the rate of heat flow per square meter through the wall Given T1= -10˚C = 263 K T2=19˚C=292 K 119871119908 = 30 cm=003 m 119871119904 = 22cm = 0022 m 119896119908 = 0080 WmK 119896119904= 0010 WmK Solutions (a)

119867 =119896119860 (119879119867 minus 119879119862)

119871

119867119908119900119900119889 =119896119908119860 (119879 minus 1198791)

119871119908

119867119904119905119910119903119900 =119896119904119860 (1198792 minus 119879)

119871119904

119867119908 = 119867119904

119896119908119860 119879minus 1198791

119871119908=

119896119904119860 1198792minus 119879

119871119904

119896119908119871119904 119879 minus 1198791 = 119896119904119871119908 1198792 minus 119879 (0080 WmK) (0022m) (T ndash 263)K = (0010 WmK)

(003m)( 292 ndash T)K

T 000176 WK ndash 046288 W = 00876 - T00003 WK

T000206 WK = 055048 W T = 26722 K

T = - 58 ˚C (b)

119867 =119896119860 (119879119867 minus 119879119862)

119871

119867

119860= 119896(

119879119867 minus 119879119888

119871)

Wood 119867119908

119860= 119896119908 (

119879minus 1198791

119871119908)

= (0080 WmK) 26722 ndash 263 K

003m

= 113 119882 1198982

Styrofoam 119867119904

119860= 119896119904

1198792minus 119879

119871119904

= (0010 WmK) 292minus26722 K

0022m

= 113 119882 1198982

6 Heat Flow and Cooling your Coffee

The author likes to bring a thermos bottle of coffee in his office each morning The walls and the lid of a thermos bottle have low thermal conductivities so the rate of heat flow of the bottle is small Suppose the coffee inside has a mass m= 030 kg and an initial temperature of 65˚C Coffee is mostly water so its specific heat is about the same as that of water If the average rate of heat flow through the walls of the thermos bottle is Qt = 30 W approximately how long will it take the coffee to cool to a final temperature of 55˚C Take a room temperature to be 25˚C

Given 119898 = 030 119896119892 119879119900 = 65 119867 = 30 119882 119900119903 119869119904 119879119891 = 55

119888 = 1119896119888119886119897

119896119892∙ 119862deg = 4186

119869

119896119892∙ 119870

Solution

119867 = 119876

119905

119876 = 119898119888∆119879

119876 = 030 119896119892 (4186119869

119896119892∙ 119870) 55 minus 65 119870

= minus13 times 104 119869 The negative sign indicates that the energy flows out

of the coffee

119905 = 119876

119867=

13 times 104 119869

119869119904= 4300 119904

7 A small dog has thick fur with a thermal conductivity of

0040 W(m _ K) The dogrsquos metabolism produces heat at a rate of 40 W and its internal (body) temperature is 38degC If all of this heat flows out through the dogrsquos fur what is the outside temperature Assume the dog has a surface area of 050 m2 and that the length of the dogrsquos hair is 10 cm

Given 119867 = 40 119882

119896 = 004119882

119898 ∙ 119870

119879119894 = 38 = 311 119870 119871 = 10 119888119898 = 01 119898 119860 = 050 1198982 Solution

119867 =119870119860∆119879

119871

40 119882 = 0040

119882119898 ∙ 119870

050 1198982 311 minus 119879119900

001 119898

119879119900 = 291119870 119879119900 = 291119870 minus 273119870 = 18

8 A long rod insulated to prevent heat loss along its sides is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other The rod consists of a 100 m section of copper (one end in boiling water) joined end to end to a length L2 of steel (one end in the ice-water mixture) Both reactions of the rod have cross-sectional areas of 4 cm

2 The temperature of the copper-steel

junction is 65degC after a steady state has been set up (a) How much heat per second flows from the boiling water to the ice-water mixture (b) What is the length L2 of the steel section For copper c k=385 WmsdotK For steel s k = 502 WmsdotK

Qt is the same for both sections of the rod For the copper

119876

119905 =

385119882

119898119870 00004 (100minus65)

1 119898 = 539 Js

For steel

L = 119896119860∆119879

119876119905 =

502119882

119898119870 00004 (65minus0)

539 119869119904= 120782 120784120786120784 119950

9 Rods of copper brass aluminum and steel are welded

together to form an X-shaped figure Assume that the cross-sectional area of each rod is the same The end of the copper rod is maintained at 120degC The end of the aluminum rod is at 100degC and the ends of the brass and steel rods are at 0degC The lengths and thermal conductivities of the rods are copper 50cm and 092 aluminum 60cm and 049 brass 25cm and 026 steel 15cm and 012 Thermal conductivities are expressed in the CGS units Calculate the temperature of the junction

(119876

119905)Cu + (

119876

119905)Al = (

119876

119905)Br + (

119876

119905)St

(092 (120 minus 119879))50 + (049 (100 minus 119879))60= (026 (119879 minus 0))25 + (012 (119879 minus 0))15

T = 6726 degC

10 The emissivity of tungsten is 0350 A tungsten sphere with radius 150 cm suspended within a large evacuated enclosure whose walls are at 290K What power input is required to maintain the sphere at a temperature of 3000 K if heat conduction along the supports is neglected

H=Aeσ(Tf

4minusTi

4)

H=4π (00150 m)2(035)(56710minus8 Wm

2sdotK

4 )([3000 K]

4 minus290 K+

4

) = 454 x 104 W

11 A square chip of width L = 15 mm on a side are

mounted to a substrate that is installed in an enclosure whose walls and air are maintained at a temperature of Tsurr = 25degC The chips have an emissivity of 060 and a maximum allowable temperature of T s = 85degC

(a) If heat is neglected from the chips by radiation and natural convection what is the maximum operating power of each chip The convection coefficient depends on the chip-to-air temperature difference and may be approximated as h = C (Ts ndash Tsurr)

025 where C =

42 Wm2-K

54

(b)If a fan is used to maintain air flow through the enclosure and heat transfer is by forced convection with h = 250 WmK what is the maximum operating power

SOLUTION (a) The maximum operating chip power is the summation

of heat transfer due to convection and radiations

Pmax = Qtot = Qconv + Qrad = hA (Ts ndash Tsurr) + 120590efA (Ts

4 ndash Tsurr

4)

= (42)(0015)2(85 ndash 25)

125 + 567 x 10

-8 (06)(1)(0015)

2

(3584-298

4)

= 02232 W (b) Maximum operating power if a fan used is Pmax = Qtot = Qconv + Qrad = hA (Ts ndash Tsurr) + 120590efA (Ts

4 ndash Tsurr

4)

= 250 (0015)2 (85 ndash 25) + 567x10

-8 (06)(1)(0015

2)

(3584 ndash 298

4)

= 344 W

12 What is rate of energy radiation per unit area of a blackbody at a temperature of (a) 273 K (b) 2730 K

Remember that the e of blackbody is 1

119867

119860 = e120575T

4

(a) 119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(273 K)

4 = 315 Wm

2

(b)119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(2730 K)

4 = 315 x 10

6

Wm2

13 One end of a solid cylindrical copper rod 0200 m long

is maintained at a temperature of 20 K The other end is blackbody and exposed to thermal radiation from surrounding walls at 500 K The sides of the rod are insulated so no energy is lost or gained except at the ends of the rod When equilibrium is reached what is the temperature of the blackened end

Dahil blackened yung end e = 1 so T1 = 20 K and T2yunghahanapin Ts = 500 K

T2 = T1 + 120575119871

119896 (Ts

2 ndash T1

4)

= T1 + (679 x 10-12

K-3

) (Ts4 ndash T1

4)

= T1 + 0424 K = 2042 K

EXERCISES 1 The melting point of tin is 232degC What is this temperature

on the Fahrenheit and Kelvin scales 2 What is the value of the ldquoabsolute zerordquo of temperature on

the Fahrenheit scale 3 How much heat is required to raise the temperature of a

30-kg block of steel by 50 K 4 While working in a research lab you find that it takes 400 J

of heat to increase the temperature of 012 kg of a material by 30 K What is the specific heat of the material

5 You are a blacksmith and have been working with 12 kg of

steel When you are finished shaping it the steel is at a temperature of 400degC To cool it off you drop it into a bucket containing 50 kg of water at 0degC How much of this water is converted to steam Assume the steel the water and the steam all have the same final temperature

6 One section of a steel railroad track is 25 m long If its

temperature increases by 25 K during the day how much does the track expand

Cu Brass

Al Steel

7 A steel building is 120 m tall when the outside temperature is 0degC How tall is the building on a hot summer day (30degC) The change in height is small so express your answer with five significant figures

8 A rectangular aluminum plate has an area of 040 m

2 at

15degC If it is heated until its area has increased by 40 x 106

m2 what is the final temperature of the plate

9 An aluminum beam 80 m long and with a cross-sectional

area of 010 m2 is used as part of a bridge The beam is

clamped rigidly at both ends (a) If the temperature of the beam is increased by 30degC what is the extra force exerted by the beam on one of its supports (b) What is the force if the temperature of the beam is decreased by 40degC


0040 W(m K) The dogrsquos metabolism produces heat at a rate of 40 W and its internal (body) temperature is 38degC If all of this heat flows out through the dogrsquos fur what is the outside temperature Assume the dog has a surface area of 050 m

2 and that the length of the dogrsquos hair is 10 cm

11 A jogger generates heat energy at a rate of 800 W If all this

energy is removed by sweating how much water must evaporate from the joggerrsquos skin each hour

12 A typical incandescent lightbulb has a filament temperature

of approximately 3000 K At what wavelength is the intensity of the emitted light highest

13 The intensity of sunlight at the Earthrsquos surface is about 1000

Wm2 What is the total power emitted by the Sun

14 An aluminum block slides along a horizontal surface The

block has an initial speed of 10 ms and an initial temperature of 10degC The block eventually slides to rest due to friction Assuming all the initial kinetic energy is converted to heat energy and all this energy stays in the block what is the final temperature of the block

15 The Rankine scale Named after William John Macquorn

Rankine (a Scottish engineer and physicist who proposed it in 1859) the Rankine scale is similar to the Kelvin scale in that the zero point is placed at absolute zero but the size of temperature differences are the same as that of the Fahrenheit scale (eg 1degF = 1degR) (a) Determine the conversion formula to go from the Fahrenheit to the Rankine scale (b) Find the formula to convert from Kelvin to Rankine (c) What is the temperature of the freezing point of water on the Rankine scale What is room temperature on this scale







19 A long rod insulated to prevent heat losses has one end immersed in boiling water and the other end in water-ice mixture all at atmospheric pressure The rod consists of 100 cm of copper (one end in steam) and length L2 of steel

(one end in ice) Both rods have cross-sectional area of 5

cm2 The temperature of the Cu-steel junction is 60degC KCu

= 092 and Ksteel = 012 cals-cm-Cdeg (a) How many calories

per second flow from the steam bath to the ice water mixture (b) How long is L2

20 One end of an insulated metal rod is maintained at 100degC

and the other end at 0degC by an ice-water mixture The rod

has a length of 40cm and a cross-section of 0750 cm2 The heat conducted by the rod melts 300 g of ice in 500 min Calculate the thermal conductivity of the metal

21 What is the final result when 400g of water and 100g of ice

at 0degC are in a calorimeter whose water equivalent is 50g into which is passed 10 g of steam at 100degC

22 A cylindrical hot plate is maintained at a temperature of

120degC by an electrical heating coil placed beneath it Heat is being transferred the surrounding air through natural convection The room temperatue is 30degC Determine the diameter of the hot plate if the heating coil is penetrating heat at a rate of 800 W Assume that the heat generated by the coil is totally effective in heating the plate h for a

horizontal place facing upward is 0595 x 10-4 ()14

23 A long rod insulated to prevent heat losses has one end

immersed in boiling water and the other end in water-ice mixture all at atmospheric pressure The rod consists of 100 cm of copper (one end in steam) and length L2 of steel (one end in ice) Both rods have cross-sectional area of 5 cm

2 The temperature of the Cu-steel junction is 60degC KCu =

092 and Ksteel = 012 cals-cm-Cdeg (a) How many calories per second flow from the steam bath to the ice water mixture (b) How long is L2


and the other end at 0degC by an ice-water mixture The rod has a length of 40cm and a cross-section of 0750 cm

2 The

heat conducted by the rod melts 300 g of ice in 500 min Calculate the thermal conductivity of the metal




120degC by an electrical heating coil placed beneath it Heat is being transferred the surrounding air through natural convection The room temperatue is 30degC Determine the diameter of the hot plate if the heating coil is penetrating heat at a rate of 800 W Assume that the heat generated by the coil is totally effective in heating the plate h for a horizontal place facing upward is 0595 x 10

-4 (∆119879)

14

27 A tungsten filament reaches a temperature of 2000 K when

its power consumption is 16 watts On the assumption that the filament radiates heat as a blackbody what power must be supplied to maintain a filament temperature of 3000 K

ANSWERS 1 TF = 450degF and TK = 505 K 2 460degF 3 68 X 10

4 J

4 110 J(kg K) 5 026 kg 6 75 X 10

3 m

7 12004 m 8 155degC 9 (a) 47 X 10

6 N (compression)

(b) 63 X 106 N (tension)

10 18degC 11 13 kgh 12 96 X 10

7 m

13 28 X 1026

W 14 1006degC 15 (a) TR = TF + 459

(b) TR = (95)TK

(c) Freezing point of water = 491degR room temperature = 530degR

16 A0 = 1431 cm2

A = 1436 cm2

17 ΔL = 026 mm 18 Tf = 9814 degC

19 184 cals 1957 cm 20 0427 21 0degC 490 g H2O 20g ice

22 12146 cm 23 184 cals 1957 cm 24 0427 25 0degC 490 g H2O 20g ice 26 12146 cm 27 81 Js

THERMAL STRESS (Tensile or Compressive)

S = FA = Y

∆LLO

where Y = Youngrsquos Modulus of Elasticity ∆LLo

= fractional change due to stress

∆LLo

= α∆T = fractional change due to change in length

COMPRESSION Tf gt To

TENSION Tf lt To

UNITS F N dynes lb A m

2 cm

2 in

2

S Nm

2 dynescm

2 lbin

2 or lbft

2

Y Nm

2 dynescm

2 lbin

2 or lbft

2

EXAMPLE PROBLEMS

1 The ends of a steel rod exactly 02 in2 in cross-sectional

area are rigidly held between two fixed points at a temperature of 30

oC When the temperature drops to

20 oC find

a) the stress developed at the ends of the rod b) the pull in the rod

Ysteel = 33 x 10

6 lbin

2 αsteel = 11 x 10

-6 C

o

a) S = FA = Yα∆T

S = (33 x 106)(11 x 10

-6)(-10) = -3630

lbin

2

b) F = SA F = (-3630)(02) = -726lb

2 An aluminum baseball bat has a length of 86 cm at a

temperature of 17 oC When the temperature of the bat is

raised the bat lengthens by 16 x 10

-4 m Determine the

final temperature of the bat ( αAl = 23 x 10

-6 C

o)

Lo = 86 cm or 086 m To = 17

oC

ΔL = 16 x 10-4

m

∆T = ∆LLoα

Tf ndash 17 = 16 x 10

-4

(086)(23 x 10-6

)

Tf = 25089degC

3 The brass bar and the aluminum bar as shown are each attached to an immovable wall At 28 degC the air gap between the rod is 13 x 10

-3 m At what temperature will

the gap be closed BRASS ALUMINUM 2m 1m αBr = 19 X 10

-6 Cdeg

αAl = 23 X 10

-6 Cdeg

∆LBr + ∆LAl = 13 x 10-3

m

(α Lo ΔT)Br + (α Lo ΔT)Al = 13 x 10-3

m [(19 X 10

-6)(2)∆T + (23 X 10

-6)(1)∆T = 13 x 10

-3 m

∆T = 2131Cdeg Tf = 28 + 2131 = 4931Cdeg

4 A copper cylinder is initially at 200

oC At what

temperature will its volume be 0150 larger than it is at 200

oC (βCu = 51 x 10

-5)

Let Xo = the initial volume Let Xo + 00015 Xo = the final volume Let ∆V = 0150Vo ΔV = βVo ΔT 15 x 10

-3Vo = (51 x 10

-5)(Vo)(Tf ndash 20)

Tf = 4941degC

5 A steel tank is completely filled with 28 m3 of ethanol

when both the tank and the ethanol are at temperature of 320degC When the tank and its contents have cooled to 18degC what additional volume of ethanol can be put into the tank

(βe = 75 x 10-5

Co

βs = 36 x 10-5

Co)

ΔVs = Vo βs ΔT = (28) (36 x 10-5

) (-14) ΔVs = - 000141 ΔVs = - 141 x 10

-3 m

3 ~ - 141 L

ΔVe = Vo βe ΔT = (28) (75 x 10

-5) (-14)

ΔVe = - 00294 ΔVe = -294 x 10

-2 m

3 ~ -294 L



a A0 = 120587r0

2 = 120587 (0675)

2 =1431 cm

2

b ∆A = (2prop)Ao∆T = 2(12 x 10

-6)(1431)(175-25)

= 515 x 10-5

cm2

A = Ao + ∆A

= 1431 + (515 x 10-6

) = 1436 cm2



Find the change ΔL in the diameter of the lid

ΔL = αLΔT = (1210minus (Cdeg)minus)(725 mm)(300 Cdeg) = 026 mm



L0 iron = 2555 cm L0 zinc = 2550 cm T0 = 32degF = 0degC Tf= ∆T = Tf - To Liron = Lzinc Lo iron (1 + prop ∆T) = Lo zinc (1 + prop ∆T) 2555 [1 + (0000010)(∆T)] = 2550 [1 + (0000030)(∆T) 2555 + 00002555 ∆T = 255 + 0000765 ∆T 00002555∆119879 ndash 0000765 ∆119879 = 255 ndash 255 - 00005095 ∆119879 = 005 ∆119879 = 9814 119862deg Tf = 9814 + 0degC Tf = 9814 degC



11 Pa

(a) ΔL=αLΔT

prop = ∆ L

Lo∆T=

0019 m


-5(Cdeg)

-1








3 of mercury



of the glass


∆VHg = Vo120573Hg∆119879

= (1000) (18 x 10-5

) (55) = 99cm3

∆Vglass = ∆VHg - 895 = 99 ndash 895 = 095


119881119900∆119879 =

095

1000 (55) = 17 x 10

-5K


o C

o F

o

c calgC

o JkgC

o or kcalkgC

BtulbF

o


cwater= 1cal

gCo= 1

kcal

kgCo= 1

Btu

lbFo


QLOSS = QGAINED

SAMPLE PROBLEMS




oC after thermal


QLOSS = QGAINED


c =

[ (200022) + (1501) ](4)(40)(75)

c = 031calgCdeg


oC Its




oC 6274 Btu of


6274Btu x 252cal1Btu = 1581048 cal


Lv = 600 calg




mglassCwater(Tf-Ti)


(10g)(1)(22degC - 5degC) 220 + 800 = 17 mw + 170

17 mw = 850

mw = 50 g


Qlost = Qgained



Cliquid = 00445



Qlost = Qgained


-2712 + 0226 T = 100 + 11 0226 T = 13812

T = 61115 degF




HEAT TRANSFER

I CONDUCTION

H = KA∆T

L



II CONVECTION

H = Qt =


AT1hf

+LK +

1hf



4


P = Qt


4 ndash Ts

4)


σ =57 x 10-8 W

m2K

4

or

57 x 10-12 W

cm2K

4





H2T = HT1

KA∆T2T

L2T =

KA∆TT1

LT1

T2 - T68 - 22 =

T - T1

22

T = 4576degc








Fo)

Qt =

A∆T1h +

Lk +

1h

Qt =

15(70 - 20)

12 +

14

58 + 12

Qt = 71901

Btuh


H = Qt

Q = Ht

Q = 71901 Btuh x 24h

Q = 172562 Btu



P = eσAT4

A = P

eσT4 A = 4πr

2

4πr2 =

PeσT

4

r = P

4πeσT4

r = 1 x 10

8

4π(1)(56 x 10-8

)(1000)4

r = 1185m




2 = 125 x 10

-4 m

2

m = 85 g = 00085 kg


1 119896119888119886119897= 334 times 105 119869

119896119892

Solution

Q = mLf = (00085 kg)( 334 times 105 119869

119896119892)

Q= 284 times 103119869 119876

119905=

284times 103119869

10 119898119894119899 times

1 119898119894119899

60 119904= 473 119882

119867 =119876

119905=

119896119860 (119879119867minus 119879119862)

119871

119896 = 119867 119871

119860 (119879119867minus 119879119862)=

473 119882 ( 0600 119898)

125 times10minus4 1198982 (100119862deg)= 227

119882

119898∙119896



119867 =119896119860 (119879119867 minus 119879119862)

119871

119867119908119900119900119889 =119896119908119860 (119879 minus 1198791)

119871119908

119867119904119905119910119903119900 =119896119904119860 (1198792 minus 119879)

119871119904

119867119908 = 119867119904

119896119908119860 119879minus 1198791

119871119908=

119896119904119860 1198792minus 119879

119871119904


(003m)( 292 ndash T)K

T 000176 WK ndash 046288 W = 00876 - T00003 WK

T000206 WK = 055048 W T = 26722 K

T = - 58 ˚C (b)

119867 =119896119860 (119879119867 minus 119879119862)

119871

119867

119860= 119896(

119879119867 minus 119879119888

119871)

Wood 119867119908

119860= 119896119908 (

119879minus 1198791

119871119908)

= (0080 WmK) 26722 ndash 263 K

003m

= 113 119882 1198982


119860= 119896119904

1198792minus 119879

119871119904

= (0010 WmK) 292minus26722 K

0022m

= 113 119882 1198982



Given 119898 = 030 119896119892 119879119900 = 65 119867 = 30 119882 119900119903 119869119904 119879119891 = 55

119888 = 1119896119888119886119897

119896119892∙ 119862deg = 4186

119869

119896119892∙ 119870

Solution

119867 = 119876

119905

119876 = 119898119888∆119879

119876 = 030 119896119892 (4186119869

119896119892∙ 119870) 55 minus 65 119870


of the coffee

119905 = 119876

119867=

13 times 104 119869

119869119904= 4300 119904



Given 119867 = 40 119882

119896 = 004119882

119898 ∙ 119870

119879119894 = 38 = 311 119870 119871 = 10 119888119898 = 01 119898 119860 = 050 1198982 Solution

119867 =119870119860∆119879

119871

40 119882 = 0040

119882119898 ∙ 119870

050 1198982 311 minus 119879119900

001 119898

119879119900 = 291119870 119879119900 = 291119870 minus 273119870 = 18





119876

119905 =

385119882

119898119870 00004 (100minus65)

1 119898 = 539 Js

For steel

L = 119896119860∆119879

119876119905 =

502119882

119898119870 00004 (65minus0)

539 119869119904= 120782 120784120786120784 119950



(119876

119905)Cu + (

119876

119905)Al = (

119876

119905)Br + (

119876

119905)St


T = 6726 degC


H=Aeσ(Tf

4minusTi

4)

H=4π (00150 m)2(035)(56710minus8 Wm

2sdotK

4 )([3000 K]

4 minus290 K+

4

) = 454 x 104 W




025 where C =

42 Wm2-K

54





4 ndash Tsurr

4)

= (42)(0015)2(85 ndash 25)

125 + 567 x 10

-8 (06)(1)(0015)

2

(3584-298

4)


4 ndash Tsurr

4)

= 250 (0015)2 (85 ndash 25) + 567x10

-8 (06)(1)(0015

2)

(3584 ndash 298

4)

= 344 W



119867

119860 = e120575T

4

(a) 119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(273 K)

4 = 315 Wm

2

(b)119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(2730 K)

4 = 315 x 10

6

Wm2




T2 = T1 + 120575119871

119896 (Ts

2 ndash T1

4)

= T1 + (679 x 10-12

K-3

) (Ts4 ndash T1

4)

= T1 + 0424 K = 2042 K










Cu Brass

Al Steel



2 at












































2 The






-4 (∆119879)

14




4 J

4 110 J(kg K) 5 026 kg 6 75 X 10

3 m

7 12004 m 8 155degC 9 (a) 47 X 10

6 N (compression)


10 18degC 11 13 kgh 12 96 X 10

7 m

13 28 X 1026

W 14 1006degC 15 (a) TR = TF + 459

(b) TR = (95)TK


16 A0 = 1431 cm2

A = 1436 cm2

17 ΔL = 026 mm 18 Tf = 9814 degC





11 Pa

(a) ΔL=αLΔT

prop = ∆ L

Lo∆T=

0019 m


-5(Cdeg)

-1








3 of mercury



of the glass


∆VHg = Vo120573Hg∆119879

= (1000) (18 x 10-5

) (55) = 99cm3

∆Vglass = ∆VHg - 895 = 99 ndash 895 = 095


119881119900∆119879 =

095

1000 (55) = 17 x 10

-5K


o C

o F

o

c calgC

o JkgC

o or kcalkgC

BtulbF

o


cwater= 1cal

gCo= 1

kcal

kgCo= 1

Btu

lbFo


QLOSS = QGAINED

SAMPLE PROBLEMS




oC after thermal


QLOSS = QGAINED


c =

[ (200022) + (1501) ](4)(40)(75)

c = 031calgCdeg


oC Its




oC 6274 Btu of


6274Btu x 252cal1Btu = 1581048 cal


Lv = 600 calg




mglassCwater(Tf-Ti)


(10g)(1)(22degC - 5degC) 220 + 800 = 17 mw + 170

17 mw = 850

mw = 50 g


Qlost = Qgained



Cliquid = 00445



Qlost = Qgained


-2712 + 0226 T = 100 + 11 0226 T = 13812

T = 61115 degF




HEAT TRANSFER

I CONDUCTION

H = KA∆T

L



II CONVECTION

H = Qt =


AT1hf

+LK +

1hf



4


P = Qt


4 ndash Ts

4)


σ =57 x 10-8 W

m2K

4

or

57 x 10-12 W

cm2K

4





H2T = HT1

KA∆T2T

L2T =

KA∆TT1

LT1

T2 - T68 - 22 =

T - T1

22

T = 4576degc








Fo)

Qt =

A∆T1h +

Lk +

1h

Qt =

15(70 - 20)

12 +

14

58 + 12

Qt = 71901

Btuh


H = Qt

Q = Ht

Q = 71901 Btuh x 24h

Q = 172562 Btu



P = eσAT4

A = P

eσT4 A = 4πr

2

4πr2 =

PeσT

4

r = P

4πeσT4

r = 1 x 10

8

4π(1)(56 x 10-8

)(1000)4

r = 1185m




2 = 125 x 10

-4 m

2

m = 85 g = 00085 kg


1 119896119888119886119897= 334 times 105 119869

119896119892

Solution

Q = mLf = (00085 kg)( 334 times 105 119869

119896119892)

Q= 284 times 103119869 119876

119905=

284times 103119869

10 119898119894119899 times

1 119898119894119899

60 119904= 473 119882

119867 =119876

119905=

119896119860 (119879119867minus 119879119862)

119871

119896 = 119867 119871

119860 (119879119867minus 119879119862)=

473 119882 ( 0600 119898)

125 times10minus4 1198982 (100119862deg)= 227

119882

119898∙119896



119867 =119896119860 (119879119867 minus 119879119862)

119871

119867119908119900119900119889 =119896119908119860 (119879 minus 1198791)

119871119908

119867119904119905119910119903119900 =119896119904119860 (1198792 minus 119879)

119871119904

119867119908 = 119867119904

119896119908119860 119879minus 1198791

119871119908=

119896119904119860 1198792minus 119879

119871119904


(003m)( 292 ndash T)K

T 000176 WK ndash 046288 W = 00876 - T00003 WK

T000206 WK = 055048 W T = 26722 K

T = - 58 ˚C (b)

119867 =119896119860 (119879119867 minus 119879119862)

119871

119867

119860= 119896(

119879119867 minus 119879119888

119871)

Wood 119867119908

119860= 119896119908 (

119879minus 1198791

119871119908)

= (0080 WmK) 26722 ndash 263 K

003m

= 113 119882 1198982


119860= 119896119904

1198792minus 119879

119871119904

= (0010 WmK) 292minus26722 K

0022m

= 113 119882 1198982



Given 119898 = 030 119896119892 119879119900 = 65 119867 = 30 119882 119900119903 119869119904 119879119891 = 55

119888 = 1119896119888119886119897

119896119892∙ 119862deg = 4186

119869

119896119892∙ 119870

Solution

119867 = 119876

119905

119876 = 119898119888∆119879

119876 = 030 119896119892 (4186119869

119896119892∙ 119870) 55 minus 65 119870


of the coffee

119905 = 119876

119867=

13 times 104 119869

119869119904= 4300 119904



Given 119867 = 40 119882

119896 = 004119882

119898 ∙ 119870

119879119894 = 38 = 311 119870 119871 = 10 119888119898 = 01 119898 119860 = 050 1198982 Solution

119867 =119870119860∆119879

119871

40 119882 = 0040

119882119898 ∙ 119870

050 1198982 311 minus 119879119900

001 119898

119879119900 = 291119870 119879119900 = 291119870 minus 273119870 = 18





119876

119905 =

385119882

119898119870 00004 (100minus65)

1 119898 = 539 Js

For steel

L = 119896119860∆119879

119876119905 =

502119882

119898119870 00004 (65minus0)

539 119869119904= 120782 120784120786120784 119950



(119876

119905)Cu + (

119876

119905)Al = (

119876

119905)Br + (

119876

119905)St


T = 6726 degC


H=Aeσ(Tf

4minusTi

4)

H=4π (00150 m)2(035)(56710minus8 Wm

2sdotK

4 )([3000 K]

4 minus290 K+

4

) = 454 x 104 W




025 where C =

42 Wm2-K

54





4 ndash Tsurr

4)

= (42)(0015)2(85 ndash 25)

125 + 567 x 10

-8 (06)(1)(0015)

2

(3584-298

4)


4 ndash Tsurr

4)

= 250 (0015)2 (85 ndash 25) + 567x10

-8 (06)(1)(0015

2)

(3584 ndash 298

4)

= 344 W



119867

119860 = e120575T

4

(a) 119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(273 K)

4 = 315 Wm

2

(b)119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(2730 K)

4 = 315 x 10

6

Wm2




T2 = T1 + 120575119871

119896 (Ts

2 ndash T1

4)

= T1 + (679 x 10-12

K-3

) (Ts4 ndash T1

4)

= T1 + 0424 K = 2042 K










Cu Brass

Al Steel



2 at












































2 The






-4 (∆119879)

14




4 J

4 110 J(kg K) 5 026 kg 6 75 X 10

3 m

7 12004 m 8 155degC 9 (a) 47 X 10

6 N (compression)


10 18degC 11 13 kgh 12 96 X 10

7 m

13 28 X 1026

W 14 1006degC 15 (a) TR = TF + 459

(b) TR = (95)TK


16 A0 = 1431 cm2

A = 1436 cm2

17 ΔL = 026 mm 18 Tf = 9814 degC





Qlost = Qgained


-2712 + 0226 T = 100 + 11 0226 T = 13812

T = 61115 degF




HEAT TRANSFER

I CONDUCTION

H = KA∆T

L



II CONVECTION

H = Qt =


AT1hf

+LK +

1hf



4


P = Qt


4 ndash Ts

4)


σ =57 x 10-8 W

m2K

4

or

57 x 10-12 W

cm2K

4





H2T = HT1

KA∆T2T

L2T =

KA∆TT1

LT1

T2 - T68 - 22 =

T - T1

22

T = 4576degc








Fo)

Qt =

A∆T1h +

Lk +

1h

Qt =

15(70 - 20)

12 +

14

58 + 12

Qt = 71901

Btuh


H = Qt

Q = Ht

Q = 71901 Btuh x 24h

Q = 172562 Btu



P = eσAT4

A = P

eσT4 A = 4πr

2

4πr2 =

PeσT

4

r = P

4πeσT4

r = 1 x 10

8

4π(1)(56 x 10-8

)(1000)4

r = 1185m




2 = 125 x 10

-4 m

2

m = 85 g = 00085 kg


1 119896119888119886119897= 334 times 105 119869

119896119892

Solution

Q = mLf = (00085 kg)( 334 times 105 119869

119896119892)

Q= 284 times 103119869 119876

119905=

284times 103119869

10 119898119894119899 times

1 119898119894119899

60 119904= 473 119882

119867 =119876

119905=

119896119860 (119879119867minus 119879119862)

119871

119896 = 119867 119871

119860 (119879119867minus 119879119862)=

473 119882 ( 0600 119898)

125 times10minus4 1198982 (100119862deg)= 227

119882

119898∙119896



119867 =119896119860 (119879119867 minus 119879119862)

119871

119867119908119900119900119889 =119896119908119860 (119879 minus 1198791)

119871119908

119867119904119905119910119903119900 =119896119904119860 (1198792 minus 119879)

119871119904

119867119908 = 119867119904

119896119908119860 119879minus 1198791

119871119908=

119896119904119860 1198792minus 119879

119871119904


(003m)( 292 ndash T)K

T 000176 WK ndash 046288 W = 00876 - T00003 WK

T000206 WK = 055048 W T = 26722 K

T = - 58 ˚C (b)

119867 =119896119860 (119879119867 minus 119879119862)

119871

119867

119860= 119896(

119879119867 minus 119879119888

119871)

Wood 119867119908

119860= 119896119908 (

119879minus 1198791

119871119908)

= (0080 WmK) 26722 ndash 263 K

003m

= 113 119882 1198982


119860= 119896119904

1198792minus 119879

119871119904

= (0010 WmK) 292minus26722 K

0022m

= 113 119882 1198982



Given 119898 = 030 119896119892 119879119900 = 65 119867 = 30 119882 119900119903 119869119904 119879119891 = 55

119888 = 1119896119888119886119897

119896119892∙ 119862deg = 4186

119869

119896119892∙ 119870

Solution

119867 = 119876

119905

119876 = 119898119888∆119879

119876 = 030 119896119892 (4186119869

119896119892∙ 119870) 55 minus 65 119870


of the coffee

119905 = 119876

119867=

13 times 104 119869

119869119904= 4300 119904



Given 119867 = 40 119882

119896 = 004119882

119898 ∙ 119870

119879119894 = 38 = 311 119870 119871 = 10 119888119898 = 01 119898 119860 = 050 1198982 Solution

119867 =119870119860∆119879

119871

40 119882 = 0040

119882119898 ∙ 119870

050 1198982 311 minus 119879119900

001 119898

119879119900 = 291119870 119879119900 = 291119870 minus 273119870 = 18





119876

119905 =

385119882

119898119870 00004 (100minus65)

1 119898 = 539 Js

For steel

L = 119896119860∆119879

119876119905 =

502119882

119898119870 00004 (65minus0)

539 119869119904= 120782 120784120786120784 119950



(119876

119905)Cu + (

119876

119905)Al = (

119876

119905)Br + (

119876

119905)St


T = 6726 degC


H=Aeσ(Tf

4minusTi

4)

H=4π (00150 m)2(035)(56710minus8 Wm

2sdotK

4 )([3000 K]

4 minus290 K+

4

) = 454 x 104 W




025 where C =

42 Wm2-K

54





4 ndash Tsurr

4)

= (42)(0015)2(85 ndash 25)

125 + 567 x 10

-8 (06)(1)(0015)

2

(3584-298

4)


4 ndash Tsurr

4)

= 250 (0015)2 (85 ndash 25) + 567x10

-8 (06)(1)(0015

2)

(3584 ndash 298

4)

= 344 W



119867

119860 = e120575T

4

(a) 119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(273 K)

4 = 315 Wm

2

(b)119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(2730 K)

4 = 315 x 10

6

Wm2




T2 = T1 + 120575119871

119896 (Ts

2 ndash T1

4)

= T1 + (679 x 10-12

K-3

) (Ts4 ndash T1

4)

= T1 + 0424 K = 2042 K










Cu Brass

Al Steel



2 at












































2 The






-4 (∆119879)

14




4 J

4 110 J(kg K) 5 026 kg 6 75 X 10

3 m

7 12004 m 8 155degC 9 (a) 47 X 10

6 N (compression)


10 18degC 11 13 kgh 12 96 X 10

7 m

13 28 X 1026

W 14 1006degC 15 (a) TR = TF + 459

(b) TR = (95)TK


16 A0 = 1431 cm2

A = 1436 cm2

17 ΔL = 026 mm 18 Tf = 9814 degC



Solution

Q = mLf = (00085 kg)( 334 times 105 119869

119896119892)

Q= 284 times 103119869 119876

119905=

284times 103119869

10 119898119894119899 times

1 119898119894119899

60 119904= 473 119882

119867 =119876

119905=

119896119860 (119879119867minus 119879119862)

119871

119896 = 119867 119871

119860 (119879119867minus 119879119862)=

473 119882 ( 0600 119898)

125 times10minus4 1198982 (100119862deg)= 227

119882

119898∙119896



119867 =119896119860 (119879119867 minus 119879119862)

119871

119867119908119900119900119889 =119896119908119860 (119879 minus 1198791)

119871119908

119867119904119905119910119903119900 =119896119904119860 (1198792 minus 119879)

119871119904

119867119908 = 119867119904

119896119908119860 119879minus 1198791

119871119908=

119896119904119860 1198792minus 119879

119871119904


(003m)( 292 ndash T)K

T 000176 WK ndash 046288 W = 00876 - T00003 WK

T000206 WK = 055048 W T = 26722 K

T = - 58 ˚C (b)

119867 =119896119860 (119879119867 minus 119879119862)

119871

119867

119860= 119896(

119879119867 minus 119879119888

119871)

Wood 119867119908

119860= 119896119908 (

119879minus 1198791

119871119908)

= (0080 WmK) 26722 ndash 263 K

003m

= 113 119882 1198982


119860= 119896119904

1198792minus 119879

119871119904

= (0010 WmK) 292minus26722 K

0022m

= 113 119882 1198982



Given 119898 = 030 119896119892 119879119900 = 65 119867 = 30 119882 119900119903 119869119904 119879119891 = 55

119888 = 1119896119888119886119897

119896119892∙ 119862deg = 4186

119869

119896119892∙ 119870

Solution

119867 = 119876

119905

119876 = 119898119888∆119879

119876 = 030 119896119892 (4186119869

119896119892∙ 119870) 55 minus 65 119870


of the coffee

119905 = 119876

119867=

13 times 104 119869

119869119904= 4300 119904



Given 119867 = 40 119882

119896 = 004119882

119898 ∙ 119870

119879119894 = 38 = 311 119870 119871 = 10 119888119898 = 01 119898 119860 = 050 1198982 Solution

119867 =119870119860∆119879

119871

40 119882 = 0040

119882119898 ∙ 119870

050 1198982 311 minus 119879119900

001 119898

119879119900 = 291119870 119879119900 = 291119870 minus 273119870 = 18





119876

119905 =

385119882

119898119870 00004 (100minus65)

1 119898 = 539 Js

For steel

L = 119896119860∆119879

119876119905 =

502119882

119898119870 00004 (65minus0)

539 119869119904= 120782 120784120786120784 119950



(119876

119905)Cu + (

119876

119905)Al = (

119876

119905)Br + (

119876

119905)St


T = 6726 degC


H=Aeσ(Tf

4minusTi

4)

H=4π (00150 m)2(035)(56710minus8 Wm

2sdotK

4 )([3000 K]

4 minus290 K+

4

) = 454 x 104 W




025 where C =

42 Wm2-K

54





4 ndash Tsurr

4)

= (42)(0015)2(85 ndash 25)

125 + 567 x 10

-8 (06)(1)(0015)

2

(3584-298

4)


4 ndash Tsurr

4)

= 250 (0015)2 (85 ndash 25) + 567x10

-8 (06)(1)(0015

2)

(3584 ndash 298

4)

= 344 W



119867

119860 = e120575T

4

(a) 119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(273 K)

4 = 315 Wm

2

(b)119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(2730 K)

4 = 315 x 10

6

Wm2




T2 = T1 + 120575119871

119896 (Ts

2 ndash T1

4)

= T1 + (679 x 10-12

K-3

) (Ts4 ndash T1

4)

= T1 + 0424 K = 2042 K










Cu Brass

Al Steel



2 at












































2 The






-4 (∆119879)

14




4 J

4 110 J(kg K) 5 026 kg 6 75 X 10

3 m

7 12004 m 8 155degC 9 (a) 47 X 10

6 N (compression)


10 18degC 11 13 kgh 12 96 X 10

7 m

13 28 X 1026

W 14 1006degC 15 (a) TR = TF + 459

(b) TR = (95)TK


16 A0 = 1431 cm2

A = 1436 cm2

17 ΔL = 026 mm 18 Tf = 9814 degC







119876

119905 =

385119882

119898119870 00004 (100minus65)

1 119898 = 539 Js

For steel

L = 119896119860∆119879

119876119905 =

502119882

119898119870 00004 (65minus0)

539 119869119904= 120782 120784120786120784 119950



(119876

119905)Cu + (

119876

119905)Al = (

119876

119905)Br + (

119876

119905)St


T = 6726 degC


H=Aeσ(Tf

4minusTi

4)

H=4π (00150 m)2(035)(56710minus8 Wm

2sdotK

4 )([3000 K]

4 minus290 K+

4

) = 454 x 104 W




025 where C =

42 Wm2-K

54





4 ndash Tsurr

4)

= (42)(0015)2(85 ndash 25)

125 + 567 x 10

-8 (06)(1)(0015)

2

(3584-298

4)


4 ndash Tsurr

4)

= 250 (0015)2 (85 ndash 25) + 567x10

-8 (06)(1)(0015

2)

(3584 ndash 298

4)

= 344 W



119867

119860 = e120575T

4

(a) 119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(273 K)

4 = 315 Wm

2

(b)119867

119860 = (1)(567 x 10

-8 Wm

2 K

4)(2730 K)

4 = 315 x 10

6

Wm2




T2 = T1 + 120575119871

119896 (Ts

2 ndash T1

4)

= T1 + (679 x 10-12

K-3

) (Ts4 ndash T1

4)

= T1 + 0424 K = 2042 K










Cu Brass

Al Steel



2 at












































2 The






-4 (∆119879)

14




4 J

4 110 J(kg K) 5 026 kg 6 75 X 10

3 m

7 12004 m 8 155degC 9 (a) 47 X 10

6 N (compression)


10 18degC 11 13 kgh 12 96 X 10

7 m

13 28 X 1026

W 14 1006degC 15 (a) TR = TF + 459

(b) TR = (95)TK


16 A0 = 1431 cm2

A = 1436 cm2

17 ΔL = 026 mm 18 Tf = 9814 degC





2 at












































2 The






-4 (∆119879)

14




4 J

4 110 J(kg K) 5 026 kg 6 75 X 10

3 m

7 12004 m 8 155degC 9 (a) 47 X 10

6 N (compression)


10 18degC 11 13 kgh 12 96 X 10

7 m

13 28 X 1026

W 14 1006degC 15 (a) TR = TF + 459

(b) TR = (95)TK


16 A0 = 1431 cm2

A = 1436 cm2

17 ΔL = 026 mm 18 Tf = 9814 degC




4 J

4 110 J(kg K) 5 026 kg 6 75 X 10

3 m

7 12004 m 8 155degC 9 (a) 47 X 10

6 N (compression)


10 18degC 11 13 kgh 12 96 X 10

7 m

13 28 X 1026

W 14 1006degC 15 (a) TR = TF + 459

(b) TR = (95)TK


16 A0 = 1431 cm2

A = 1436 cm2

17 ΔL = 026 mm 18 Tf = 9814 degC



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Physics II Reviewer for Prelims