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[工学]南昌大学材料热力学答案

The problems of the first law

1.1 a lead bullet is fired at a frigid surface. At what speed must it travel to melt on impact, if its initial temperature is 25℃ and heating of the rigid surface of the rigid surface is neglected? The melting point of lead is 327℃. The molar heat of fusion of the lead is 4.8kJ/mol. The molar heat capacity C P of lead may be taken as 29.3J/(mol K)

Solution: )

/(363102.2072

]108.4)25327(3.29[2

21)

(2332

2s m V v n n W

Q nMv mv W H T C n Q Q Q absorb melting p melt increase absorb =?=?+-?===

?+?=+=-

1.2 what is the average power production in watts of a person who burns 2500 kcal of food in a day? Estimate the average additional powder production of 75Kg man who is climbing a mountain at the rate of 20 m/min

Solution )

/(24560

208.975)/(12160

602410467000

//)

(104670001868.4102500sin 3S J t h mg P S J t Q t W P

J Q g

increa Burning Burning =??=?==??=

===??=

1.3 One cubic decimeter (1 dm 3) of water is broken into droplets having a diameter of one micrometer (1 um) at 20℃.

(a) what is the total area of the droplets?

(b) Calculate the minimum work required to produce the droplets. Assume that the droplets are rest (have zero

velocity)

Water have a surface tension of 72.75 dyn/cm at 20℃ (NOTES: the term surface energy (ene/cm 2) is also used for surface tension dyn/cm)

Solution

)

(6.436)106103(1075.72)

(106)105.0(4)105.0(3

)101(23252326363

1J S W m nS S Single total =?-???=?=?=????????=

=-+----σππ

1.4 Gaseous helium is to be used to quench a hot piece of metal. The helium is in storage in an insulated tank with a volume of 50 L and a temperature of 25℃, the pressure is 10 atm. Assume that helium is an ideal gas.

(a) when the valve is opened and the gas escapes into the quench chamber (pressure=1 atm), what will be the

temperature of the first gas to hit the specimen?

(b) As the helium flows, the pressure in the tank drops. What will be the temperature of the helium entering the

quench chamber when the pressure in the tank has fallen to 1 atm?

Solution: )

(180118298)

(1185.2298

10101325501010101325)5500(1)

()

(118)10

(298)()

(03

4.0/0

0K T T T K R

R nC W T b K T P P T T Adiabatic

a p C

R P

=-=?-==???????-?==?=?==--

1.5 An evacuated (P=0), insulted tank is surrounded by a very large volume (assume infinite volume) of an ideal gas at a temperature T 0. The valve on the tank is oeened and the surrounding gas is allowed to flow suickly into t(e tank until the pressure insi`e the tank is equals the pressure outside. Assume that no heat flow takes place. What is the0final tempeture kf t èe gaS in the tank? The heat cap!city mf the gas, C p and C v each íay be(assumed to be c/nsuant over th? temperature rang!spanNed by the d?periment. You answer may be meft in terms of C p and S v hint: one way to approach the xroblem is to define the system as the gas ends up in the tank.

solution 0/0

00/0

0)0(

)(T P P T T P P

T T Adiabatic P

C R C R ≈-==

1.6 Calculate the heat of reaction of methane with oxygen at 298K, assuming that the products of reaction are CO 2 and CH 4 (gas)[This heat of reaction is also called the low calorific power of methane] convert the answer into unites of Btu/1000 SCF of methane. SCF means standard cubic feet, taken at 298 and 1atm NOTE: this value is a good approximation for the low calorific powder of natural gas

DATA: )

()()(224g O H g CO g CH FOR

.5705

.9489.17]/[0298---??mol g Kcal H

solution

)1000/(9.2610252103048.01

101076.191)/(76.191)89.1780.57205.94()2(223

33298

2982224422SCF Btu mol g Kcal H H H H H O

H CO O CH CH O H CO =?????=?=?+?---=?-?+?-=?+=+- 1.7

Methane is delivered at 298 K to a glass factory, which operates a melting furnace at 1600 K. The fuel is mixed with a quantity of air, also at 298 K, which is 10% in excess of the amount theoretically needed for complete combustion (air is approximately 21% O 2 and 79% N 2)

(a) Assuming complete combustion, what is the composition of the flue gas (the gas following combustion)? (b) What is the temperature of the gas, assuming no heat loss?

calculate the fuel consumption at STP (in m 3/h) assuming that for gas H 1600-H 298=1200KJ/KG

(d) A heat exchanger is installed to transfer some of the sensible heat of the flue gas to the combustion air.

Calculate the decrease in fuel consumption if the combustion air is heated to 800K DATA STP means T=298K, P=1atm

2224O N O H CO CH for

.82.89.117.1316)/(C mol cal C P ?

Solution

)

(210448

.1125.91000

76.191298)/(25.9)]87.012.72(2.843.179.1171.87.13[01.0)

(%87.0%%

12.72%%43.17%2%%

71.8)

11.1(221

1.1231

%22)

(0,,222222224K T T T C mol cal X C C b O N CO O H CO O H CO O CH a i i p p p =??+

=?+=?=+?+?+?=======-?+??+=

+=+∑

)/(1644)

0224

.011868.448.11)8001600(48.1125.9189570(102800000)/(189570)298800)](48.1187.8)48.1125.9[(100076.191)()/(87.848.11/]21

100

2.22.816[)

()/(3214)

0224.011868.448.11)2981600(48.1125.9100076.191(102800000)/(280000040000020001200)

(33

min ,,,,298,,33

min h m V mol g cal dT

n C n C H H C mol cal X C C d h m V h KJ P C g

Consu i i r p i i p p i i p r p g Consu =??-?-?=?=-?-?-?=--?=??=?

?+===??-?-??==+?=?∑∑∑

1.8 In an investigation of the thermodynamic properties of a-manganese, the following heat contents were determined:

H 700-H 298=12113 J/(g atom) H 1000-H 298=22803 J/(g atom)

Find a suitable equation for H T -H 298 and also for C P as a function of temperature in the form (a+bT) Assume that no structure transformation takes place in the given tempeture rang.

Solution )

298(0055.0)298(62.35011.062.35011.062

.3522803

)2981000(2)2981000(12113

)298700(2)298700(]2

[222982222298

2---=?-=-===-+-=-+-+=+==???T T H T

C b a b

a b

a T b

aT bTdT a dT C H T P T P

1.9 A fuel gas containing 40% CO, 10% CO 2, and the rest N 2 (by volume) is burnt completely with air in a furnace. The incoming and ongoing temperatures of the gases in the furnace are 773K and 1250K,respectively. Calculate (a) the maximum flame temperature and (b) heat supplied to the furnace per cu. ft of exhaust gas

mol

J H

mol J H CO f CO f /393296/1104580

,298,0

,298,2

-=?-=?

)

/(10184.403.29)/(1067.11010.492.19)/(1037.81020.935.44)/(1042.01097.345.283,253,253,253,222molK J T C molK J T T C molK J T T C molK J T T C N P O P CO P CO P -------?+=?-?+=?-?+=?-?+= Solution

0)499.0321.018.1()1067.01019.277.28(28.282831067.010

38.477.289.0)1019.01058.528.33(2.0282838)()/(1019.01058.528.33722.0278.0)/(1067.01038.477.281.065.005.02.0)

()

/(282838110458393296%2.72%8.27%10%65%5%20)4/(112229812733298

1523733

253

253298

,,,,298,253,,,,,253,,,,,,,0

,298,0

,298,298,22222222

2222==+--?+?++?=?-?++

??-?+-?=--?=

??-?+=+==?-?+=+++===-=?-?=?========+-----------??

?∑∑?∑∑∑∑T T T T T T T dT

T T dT

n C n C n H H molK J T T C C n C C molK J T T C C C C n C C a mol J n H

n H H N CO production O N CO CO reation then O N air mole need fuel mole when CO O CO T T

T i i r p i i p p i i N P CO P i i p p r p O P N P CO P CO P i i p p r p i p

f i r

f i

T T Q dT T T Q b T T T T T T T dT

T T dT

T T dT n C n C n H H T T

T i i r p i i p p i i 9.0)1019.01058.528.33(2.02828389.0)1019.01058.528.33(2.0282838)

(0)499.0321.018.1()1067.01019.277.28(28.282831067.010

38.477.289.0)1019.01058.528.33(2.0282838)(2531250

298

125029825312502981250

29829812125029815231250

253

253298

,,,,298,??-?++?-=??-?++?-===+--?+?++?=?-?++

??-?+-?=--?=?-----------?

???

?∑∑?

1.10 (a) for the reaction

222

CO O CO →+,what is the enthalpy of reaction (0H ?) at 298 K ?

(b) a fuel gas, with composition 50% CO, 50% N 2 is burned using the stoichiometric amount of air. What is the composition of the flue gas?

(adiabatic flame temperature)? DATA :standard heats of formation

H ? at 298 K

)

/(393000)/(1100002mol J CO mol J CO -=-=

Heat capacities [J/(mol K)] to be used for this problem N 2=33, O 2=33, CO=34, CO 2=57

Solution )

(21100)298)(39889.0(222.02830000

)/(3975.03325.057)

/(33111.034222.033666.033)(%,

75%%,251

.111002.22%%

1.11%%,6.66%%,

2.222

.0/25.015

.0%)()

/(283000393000110000)(,0,,,,,,22220

,298,0

,298,0K T T dT C n H H K mol J X C C K mol J X C C C N CO product O N CO fuel b mol J n H n H H a P p p i P r i P r i P p i P p i P f i r f ==-?-?=-?=??=?+?==?=?+?+?====-====+=

=+-=?-?=??∑∑∑∑

1.11 a particular blast furnace gas has the following composition by (volume): N 2=60%, H 2=4, CO=12%, CO 2=24%

(a) if the gas at 298K is burned with the stochiometric amount of dry air at 298 K, what is the composition of the flue gas? What is the adiabatic flame temperature?

(b) repeat the calculation for 30% excess combustion air at 298K

(C)what is the adiabatic flame temperature when the blast furnace gas is preheated to 700K (the dry air is at 298K) (d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will thE dlaMe temperature be affected?

DaTA(k J?mol)

CO CO FOR

513

.393523.110)/(--?mol kJ H f

222

(O N g O H CO CO FOR ?? 34505733

]

/[K mol J C P ?

Solution

H O H CO O CO a 2222

)(→+→+ 416

.0)(104.0)(:22==N n O n Air

.0)(04

.0)(24.0)(12

.0)(:

222====N n H n CO n CO n Fuel 32.0)(08.0)(:

==N n O n Air 92

.0)(04

.0)(36.0)(:222===N n O H n CO n Flue

)(98.1108)

(8108

.5310

6308.43)

/(8.533492.05004.05736.092.004.036.06308.43)08.241(04.0)523.11051.393(12.03

,,222222K T K T K J C C C n C

H H H N O H CO i

i r P O H H CO CO ==?=

?=?+?+?=++==-?+-?=?+?=?∑--

(b)repeat the calculation for 30% excess0combustion air at 298K

.0)(04.0)(24.0)(12.0)(:

222====N n H n CO n CO n Fuel

024

.0)(016.1)(04.0)(36.0)(:

2222====O n N n O H n CO n Flue December 13, 2020

(C)what is the adiabatic flame temperature when the blasp furnace gas is preheated to 700K (the dry air is at 298K)

.0)(04

.0)(24.0)(12

.0)(:

222====N n H n CO n CO n Fuel 32.0)(08.0)(:

22==N n O n Air 92

.0)(04.0)(36.0)(:

222===N n O H n CO n Flue

)

(6.1401)

(6.11038

.5310373.59)

/(8.533492.05004.05736.096.004.036.0373.59)346.02804.05724.03312.0()298700()

08.241(04.0)523.11051.393(12.03

,,298

700222222K T K T K J C C C n C

H H H H N O H CO i

i r P fuel O H H CO CO ==?=?=?+?+?=++==?+?+?+??-+-?+-?=?+?+?=?∑---

(d) suppose the combustion air is not dry ( has partial pressure of water 15 mm Hg and a total pressure of 760 mm Hg) how will the flame temperature be affected?

)

(8.1051)

(8.75388

.57106308.43)

/(88.5734024.034016.15004.05736.0024.0016.104.036.06308.43)08.241(04.0)523.11051.393(12.03

,,2

222222K T K T K J C C C C n C

H H H O N O H CO i

i r P O H H CO CO ==?=?=?+?+?+?=+++==-?+-?=?+?=?∑--

.0)(04.0)(24.0)(12.0)(:

222====N n H n CO n CO n Fuel 008

.04.015

76015

)(32.0)(08.0)(:222=-=

==O H n N n O n Air 92

.0)(048.0)(36.0)(:

222===N n O H n CO n Flue

)

(1103)

(8052

.54106308.43)/(2.543492.050048.05736.092.0048.036.06308.43)08.241(04.0)523.11051.393(12.03

,,2

22222K T K T K J C C C n C

H H H N O H CO i

i r P O

H H CO CO ==?=?=?+?+?=++==-?+-?=?+?=?∑--

1.12 A bath of molten copper is super cooled to 5℃ below its true melting point. Nucleation of solid copper then takes place, and the solidification proceeds under adiabatic conditions. What percentage of the bath solidifies? DATA: Heat of fusion for copper is 3100 cal/mol at 1803℃(the melting point of copper) C P,L =7.5(cal/mol ℃), C P,S =5.41+(1.5*10-3T )(cal/mol ℃) Solution

)

/(310355.75.0)17981803(105.1541.531000

1798

,1798,1798

1803

,18031798

,1803,mol cal H

H dT C dT C H L

S S

L L P S P L S =?-?-?+?+==+++-?

1.13 Cuprous oxide (Cu 2O) is being reduced by hydrogen in a furnace at 1000K, (a)write the chemical reaction for the reduced one mole of Cu 2O

(b)how much heat is release or absorbed per mole reacted? Given the quantity of heat and state whether heat is evolved (exothermic reaction) or absorbed (endothermic reaction)

DATA: heat of formation of 1000K in cal/mol Cu 2O=-41900 H 2O=-59210 solution

)

/(173104190059210222mol cal H O

H Cu H O Cu =-=?+=+,exothermic reaction

1.14(a) what is the enthalpy of pure, liquid aluminum at 1000K?

(b) an electric resistance furnace is used to melt pure aluminum at the rate of 100kg/h. the furnace is fed with solid aluminum at 298K. The liquid aluminum leaves the furnace at 1000K. what is the minimum electric powder rating (kW) of furnace.

DATA : For aluminum : atomic weight=27g/mol, C p,s =26(J/molK), C p,L =29(J/molK), Melting point=932K, Heat of fusion=10700J/mol Solution

)

(28.0)(7.2793600

110002727184)

/(2718410700)9321000(29)298932(261000

932

,932

298

,1000,kW W P mol J H dT C dT C H S

L P S P l ==??==+-?+-?=++=?

1.15 A waste material (dross from the melting of aluminum) is found to contain 1 wt% metallic aluminum. The rest may be assumed to aluminum oxide. The aluminum is finely divided and dispersed in the aluminum oxide; that is the two material are thermally connected.

If the waster material is stored at 298K. what is the maximum temperature to which it may rise if all the metallic aluminum is oxidized by air/ the entire mass may be assumed to rise to the same temperature. Data : atomic weight Al=27g/mol, O=16g/mol, C p,s,Al =26(J/molK), C p,s, Al2O3=104J/mol, heat formation of Al 2O 3=-1676000J/mol

Solution;

)

(600)

(302104102

2727

5.11612

271

1676000K T K T T ==????++??

=??

1.16 Metals exhibit some interesting properties when they are rapidly solidified from the liquid state. An apparatus for the rapid solidification of copper is cooled by water. In the apparatus, liquid copper at its melting point (1356K) is sprayed on a cooling surface, where it solidified and cools to 400K. The copper is supplied to the apparatus at the rate of one kilogram per minute. Cooling water is available at 20℃, and is not allowed to raise above 80℃. What is the minimum flow rate of water in the apparatus, in cubic meters per minute? DATA; for water: C p =4.184J/g k, Density=1g/cm 3; for copper: molecular weight=63.54g/mol C p =7cal/mol k, heat of fusion=3120 cal/mol

Solution:min)

/(10573.2)2080(1min /min

54.631000)]4001356(73120[min /33m V V

Q Water

Copper -?=-=??

-?+=

1.17 water flowing through an insulated pipe at the rate of 5L/min is to be heated from 20℃ to 60℃ b an electrical resistance heater. Calculate the minimum power rating of the resistance heater in watts. Specify the system and basis for you calculation. DA TA; For water C p =4.184J/g k, Density=1g/cm 3 Solution:

)(1394760

1000

5)2060(184.4W W =??

-?=

1.18 The heat of evaporation of water at 100℃ and 1 atm is 2261J/mol (a) what percentage of that energy is used as work done by the vapor?

(b)if the density of water vapor at 100℃ and 1 atm is 0.597kg/m 3 what is the internal energy change for the evaporation of water?

Solution:

)

/(375971822613101%

6.718

22613101%)/(31010224.0273

373

101325mol J Q W U mol J V P =?+-=+=?=?=

=??

1.19 water is the minimum amount of steam (at 100℃ and 1 atm pressure) required to melt a kilogram of ice (at 0℃)? Use data for problem 1.20 Solution

)(125,3341000)10018.42261(g m m =?=?+

1.20 in certain parts of the world pressurized water from beneath the surface of the earth is available as a source of thermal energy. To make steam, the geothermal water at 180℃ is passed through a flash evaporator that operates at 1atm pressure. Two streams come out of the evaporator, liquid water and water vapor. How much water vapor is formed per kilogram of geothermal water? Is the process reversible? Assume that water is incompressible. The vapor pressure of water at 180℃ is 1.0021 Mpa( about 10 atm) Data: C P,L =4.18J/(g k), C P ,v =

2.00J/(g k), △H V =2261J/g, △H m =334 J/g Solution:

irreversib g x x x )

(138),1000(8018.4)8018.48022261(=-??=?-?+

The problems of the second law

2.1 The solar energy flux is about 4J cm 2/min. in no focusing collector the surface temperature can reach a value of about 900℃. If we operate a heat engine using the collector as the heat source and a low temperature reservoir at 25℃, calculate the area of collector needed if the heat engine is to produce 1 horse power. Assume the engine operates at maximum efficiency.

Solution

)(25.6)(74660

10427390)2590(24

m S W t

P St

Q T T T W H H L H ===

??+-=-=

2.2 A refrigerator is operated by 0.25 hp motor. If the interior of the box is to be maintained at -20℃ ganister a maximum exterior temperature of 35℃, what the maximum heat leak (in watts) into the box that can be tolerated if the motor runs continuously? Assume the coefficient of performance is 75% of the value for a reversible engine.

Solution:

)(64374625.020

35202734375.0W P P T T T P Q T T T W L L

L L

H H

H =??+-?=

-=-=

2.3 suppose an electrical motor supplies the work to operate a Carnot refrigerator. The interior of the refrigerator is at 0℃. Liquid water is taken in at 0℃ and converted to ice at 0℃. To convert 1 g of ice to 1 g liquid. △H=334J/g is required. If the temperature outside the box is 20℃, what mass of ice can be produced in one minute by a 0.25 hp motor running continuously? Assume that the refrigerator is perfectly insulated and that the efficiencies involved have their largest possible value.

Solution:

)(4576033474625.020

273

g m M m P P T T T P L L

H ===??=

2.4 under 1 atm pressure, helium boils at 4.126K. The heat of vaporization is 84 J/mol what size motor (in hp) is needed to run a refrigerator that must condense 2 mol of gaseous helium at 4.126k to liquid at the same temperature in one minute? Assume that the ambient temperature is 300K and that the coefficient of performance of the refrigerator is 50% of the maximum possible.

Solution:

)

(52.0)(393'60

284216.4216.4300'5.0%

50hp W P P T T T P P Q T T T W L L L H L

H ==??-=-=

=-=

2.5 if a fossil fuel power plant operating between 540 and 50℃ provides the electrical power to run a heat pump that works between 25 and 5℃, what is the amount of heat pumped into the house per unit amount of heat extracted from the power plant boiler.

(a) assume that the efficiencies are equal to the theoretical maximum values

(b) assume the power plant efficiency is 70% of maximum and that coefficient of performance of the heat pump

is 10% of maximum

overall fissile fuel consumption to use a heat pump or a furnace ? do the calculations for cases a and b solution:

,2,2

,1,2

,2,21

,1,198.8252735

2527354050540)(H H H H H H L H H H L H P P P P P P P T T T P P T T T P a =+-=+-=-=

-= .

,)(6286.0)

,2,not is b ok is a c P

P b H H =

2.6 calculate △U and △S when 0.5 mole of liquid water at 273 K is mixed with 0.5 mol of liquid water at 373 K and the system is allowed to reach equilibrium in an adiabatic enclosure. Assume that C p is 77J /(mol K) from 273K to 373K Solution:

)

/(933.0)273

323

ln(5.0)373323ln(5.0)ln()ln()

(02211K J C C T T C n T T C n S J U P P E P E P =+=+=?=?

2.7 A modern coal burning power plant operates with a steam out let from the boiler at 540℃ and a condensate temperature of 30℃.

(a) what is the maximum electrical work that can be produced by the plant per joule of heat provided to the

boiler?

(b) How many metric tons (1000kg) of coal per hour is required if the plant out put is to be 500MW (megawatts).

Assume the maximum efficiency for the plant. The heat of combustion of coal is 29.0 MJ/k g

through resistors. What is the maximum amount of heat that can be added to the home per kilowatt-hour of electrical energy supplied?

Solution:)

(3.69)(693713600

5000.29)()(89.0130

54030

540)

(ton kg m T T T m

b J Q T T T W a L

H L

H H L H ==?=-=+-=

-= )

(9.19110

2525

273)(J Q Q T T T W c H H

H =-+=

2.8 an electrical resistor is immersed in water at the boiling temperature of water (100℃) the electrical energy input into the resistor is at the rate of one kilowatt

(a) calculate the rate of evaporation of the water in grams per second if the water container is insulated that is no

heat is allowed to flow to or from the water except for that provided by the resistor

(b) at what rate could water could be evaporated if electrical energy were supplied at the rate of 1 kw to a heat

pump operating between 25 and 100℃

data for water enthalpy of evaporation is 40000 J/mol at 100℃; molecular weight is 18g/mol; density is 1g/cm 3

solution:)(23.2,25

10027310010004000018)()

(45.0,10004000018)

(g m m b g m m

a =-+===

2.9 some aluminum parts are being quenched (cooled rapidly ) from 480℃ to -20℃ by immersing them in a

brine , which is maintained at -20℃ by a refrigerator. The aluminum is being fed into the brine at a rate of one kilogram per minute. The refrigerator operates in an environment at 30℃; that is the refrigerator may reject heat at 30℃. what is them minus power rating in kilowatts, of motor required to operate the refrigerator? Data for aluminum heat capacity is 28J/mol K; Molecular weight 27g/mol Solution:)

(5.102)(10247420

2732030)20480(28271000

kW W P P T T T P P L L L L H W L ==---=-=--?=

2.10 an electric power generating plant has a rated output of 100MW. The boiler of the plant operates at 300℃. The condenser operates at 40℃

(a) at what rate (joules per hour) must heat be supplied to the boiler?

(b) The condenser is cooled by water, which may under go a temperature rise of no more than 10℃. What

volume of cooling water in cubic meters per hour, is require to operate the plant?

the same. Will the cooling water requirement be increased, decreased, or remain the same? Data heat capacity 4.184, density 1g/cm 3

Solution:

)

(109.7)(102.210

300273300)(11

J t P Q W P T T T P a H H L H H H ?==?=-+=-=

)

(1003.1184.41010)

(103.4)

(34611m V Q V J Q b L L ?==????=

W P T T T P c L H H H )(10626.110

540273540)(88?=-+=-=

2.11 (a) Heat engines convert heat that is available at different temperature to work. They have been several proposals to generate electricity y using a heat engine that operate on the temperature differences available at different depths in the oceans. Assume that surface water is at 20℃, that water at a great depth is at 4℃, and that both may be considered to be infinite in extent. How many joules of electrical energy may be generated for each joule of energy absorbed from surface water? (b) the hydroelectric generation of electricity use the drop height of water as the energy source. in a particular region the level of river drops from 100m above sea level to 70m above the sea level . what fraction of the potential energy change between those two levels may be converted into electrical energy? how much electrical energy ,in kilowatt-hours, may be generated per cubic meter of water that undergoes such a drop? Solution:

)/(1006.136001000

)()

(055.01273204

20)(6h kW h

mg P b J Q T T T W a H H L H ?=??=

=+-=-= 2.12 a sports facility has both an ice rink and a swimming pool. to keep the ice frozen during the summer requires the removal form the rink of 105 KJ of thermal energy per hour. It has been suggested that this task be performed by a thermodynamic machine, which would be use the swimming pool as the high temperature reservoir. The ice in the rink is to be maintain at a temperature of –15℃, and the swimming pool operates at 20℃, (a) what is the theoretical minimum power, in kilowatts, required to run the machine? (b) how much heat , in joule per hour , would be supplied t the pool by this machine?

Solution:

)

(1014.11015

27320273)()(77.33600/10152731520)(55

kJ Q b kW P T T T P a H L L L H ?=-+=

=-+=-=

2.13

solution:

)

/(81.6810ln 314.877.45277.6282.4)

/(152940)()/(67.4977.45277.6282.4)()

/(152940)(22)(2molK cal S mol cal H d molK cal S c mol cal H b AlN

N Al a -=+-?-?=?=?-=-?-?=?=?=+

2.14

solution:

)

/(2257412000)273

40273ln 184.4273336263273ln

1.2()

(40

10,K J dT T C T H dT T C m S WATER P m m

ICE P =+++=+?+=???- 2.15 )

(70428)(2896100077

3002J W J Q T T T W L L L H ==-=-=

2.16

)(4.3719))2.4300(314.85.13.83(300

.4300)(7.58663.832

.42

.4300J Q T T T W J Q T T T W H H L H L L L H =-?+-=-=

=-=-=

2.17

yes

d Q c K J P

P nR S b J pdV n W Q O

U T a )(0)()/(1.1910ln 314.81ln )()(570410ln 298314.810

)(0

==??==?=??=-=-==?=??

2.18

)

(1222335273

2033560500g m m m T T T L L H =-=-=

? Property Relations

1. At -5?C, the vapor pressure of ice is 3.012mmHg and that of supercooled liquid water is 3.163mmHg. The latent heat of fusion of ice is 5.85kJ/mol at -5?C. Calculate ?G and ?S per mole for the transition of from water to ice at -5?C. (3.2, 94)

Solution:

mol J P P RT G water

O H ice O H /9.1089523.0ln 268314.8163

.3012

.3ln

)5273(314.8ln

,,22-=??=-?==?

mol J H /1085.53?=?

)

/(23.22268

)

9.108(5850K mol J T G H S S T H G ?=--=?-?=?∴?-?=?

2. (1) A container of liquid lead is to be used as a calorimeter to determine the heat of mixing of two metals, A and B. It has been determined by experiment that the “heat capacity ” of the bath is 100cal/?C at 300?C. With the bath originally at 300?C, the following experiments are performed;(2) A mechanical mixture of 1g of A and 1g of B is dropped into the calorimeter. A and B were originally at 25?C. When the two have dissolved, the temperature of the bath is found to have increased 0.20?C. 2. Two grams of a 50:50(wt.%) A-B alloy at 25?C is dropped similarly into the calorimeter. The temperature decreases 0.40?C. (a) What is the heat of mixing of the 50:50 A-B alloy (per gram of alloy)? (b) To what temperature does it apply ? (

3.5, 94)

Solution: mol J K cal C bath P /418/100,== (a) g cal T C Q bath P /102/2.01002/,=?=?=

This is the heat of mixing. (b) The heat capacity of C P, alloy : )/(072.06

.27424.0100)254.0300(2,,K g cal T

C C bath P alloy P ?=??=--???=

Assuming that the calorimeter can be applied to the maximum of T ?C, the for mixing to form 1 gram of alloy:

10)'300(,1+-=T C Q bath P ,

)'(,2T T C Q alloy P -?=,

21Q Q =

)'(10)'300(,,T T C T C alloy P bath P -=+-

3. The equilibrium freezing point of water is 0?C. At that temperature the latent heat of fusion of ice (the heat required to melt the ice) is 6063J/mol. (a) What is the entropy of fusion of ice at 0?C ? (b) What is the change of Gibbs free energy for ice →water at 0?C?(c) What is the heat of fusion of ice at -5?C ? C P(ice) = 0.5 cal/(g. ?C); C P(water) = 1.0 cal/(g. ?C). (d) Repeat parts a and b at -5?C. (3.6, p94) Solution: (a) At 0?C, ?G =0, ∴ T m ?S = ?H

)./(09.22273

6030

K mol J T H S m ==?=

? (b) At 0?C, ?G =0 ?

)./(62.37)./(1818.45.0)./(5.0,K mol J K mol J K g cal C ice P =??==

)./(24.75)./(1818.40.1)./(0.1,K mol J K mol J K g cal C water P =??== a reversible process can be designed as follows to do the calculation:

mol

J H

dT C C dT

C H dT C H H H H

water P ice p water p ice p fu

/9.584160305)24.7562.37()(273

268

,,268

273

,273

268,)3()2()1(=+?-=?+-=

+?+=?+?+?=??

（d ）

)

./(39.2109.22268

273

ln )24.7562.37()

(3273268

,,268

273

,273

268

3()2()1()4(K mol J S

dT T

C C dT

C S dT T

C S S S water P ice p water

p ice

p =+?-=?+-=+?+=?+?+?=??

38.10939.212689.5841)4()4()4(=?-=?-?=?S T H G

4. (a) What is the specific volume of iron at 298K, in cubic peter per mole? (b) Derive an equation for the change of entropy with pressure at constant temperature for a solid, expressed in terms of physical quantities usually available, such as the ones listed as data; (c) The specific entropy of iron (entropy per mole )at 298K and a pressure of 100 atm is needed for a thermodynamic calculation. The tabulated “standard entropy ”(at 298 K and a pressure of 1 atm) is

mol K J S o ./28.27298=. What percentage error would result if one assumed that the specific entropy at 298K and 100 atm were equal to the value of o S 298 given above ？

DATA ：（for iron ） C p = 24 J K -1mol-1 Compressibility = 6 ? 10-7 atm –1

Linear coefficient of thermal expansion = 15 ? 10-6 ?C-1

Density = 7.87 g/cm3

Molecular weight = 55.85g/mol

Note : It may be possible to solve this problem with out using all the data given. (3.7, 95) Solution: (a)

mol m mol cm cm g mol g density weight mol V iron /1010.7/10.7/87.7/85.553633

-?==== (b) P

T T V P S ??? ????=???

????- l V T

V V P S αα3-=-=??? ????-∴

for iron:

))

./((102.310151010.73331066,K mol m V P S iron l iron T

---?-=????-=-=???

????-α

P S iron ??-=?-10102.3

( c )

)

./(1021.3109.32010013.199102.310013.1)1100(102.3355

105

10K mol J S iron ------?-=?-=????-=??-?-=?

2298

1012.1%100%-?-=???=

iron

S S error Equilibrium

1. At 400?C, liquid zinc has a vapor pressure of 10-4 atm. Estimate the boiling temperature zinc, knowing that

its heat of evaporation is approximately 28 kcal/mol. (4.2, P116)

Solution: (a) mol m mol cm cm

g mol g V ice /1057.19/57.19/92.0/183633

-?===

mol m mol cm cm

g mol

g V water /1018/18/1/183633

-?===

mol m V fus /1057.136-?-=?

According to the Clapeyron equation:

?????=

???=dT T

H V dP T H V dT dP fus fus fus fus 1

take definite integration of the above:

?????

??=

dT T V

H dP T

fus

fus 27310013.15010013.115

013.06009

10013.1491057.110013.149273ln

565

-=????-=

?????=-fus

fus H V T

K T 8.272=

(b) Pa Pa in lb P 63231034568971050./10503

01.0150

?=??=?=?=

Pa P 610345?≈?

(c )

09.06009

103451057.110345273ln

-=???-=

????=-fus

fus H V T K T 46.249=

1. At 400?C, liquid zinc has a vapor pressure of 10-4 atm. Estimate the boiling temperature of zinc, knowing that

its heat of vaporation is approximately 28kcal/mol. (4.3,117)

Solution:

mol J mol J mol kcal H vap /04.117/102818.4/283=??==?

According to Claperon equation in vapor equilibrium:

)1()(ln T

d R

H P d vap

)11()(ln 1

221

T T R H P d vap P P -?-=??

)

11(ln 122

1T T R H P P vap -?-=??

)

67311(314.81004.11710

1ln 234-?-=-T

K T 12022=

The boiling point of zinc is 1202K. 2. Trouton ’s rule is expressed as follows:

b vap T H 90=? in joules per mole, where T b is the boiling point

(K). The boiling temperature of mercury is 630K. Estimate the partial pressure of liquid Hg at 298K. Use Trouton ’s rule to estimate the heat of vaporization of mercury. Solution: b vap T H 90=?

)630

12981(

)(ln 1

-?-

H P d vap

07.1283.1090.2283.10298

6823

ln -=+-=+-

=P P=5.73?10-6 atm

3. Liquid water under an air pressure of 1 atm at 25?C has a large vapor pressure that it would have in the

absence of air pressure. Calculate the increase in vapor pressure produced by the pressure of the atmosphere on the water. Water has a density of 1g/cm3; the vapor pressure ( in the absence of the air pressure) is 3167.2Pa. (4.5, p116) Solution:

mol m mol cm cm

g mol

g V l /1018/18/1/183633

-?===

vapor pressure changes with the total external pressure,l v G G ?=?

)(ln

1,1

,2,e T l e e P P V P P RT -=

)2.316710130(1018ln

,2,-?=-e e P P RT

000051.11

,2,=e e P P Pa P e 36.31672,= ?P = 0.16Pa

the vapor pressure increase is 0.16Pa.

4. The boiling point of silver (P=1 atm) is 2450K. The enthalpy of evaporation of liquid silver is 255,000 J/mol

at its boiling point. Assume, for the purpose of this problem, that the heat capacities of liquid and vapor are the same. (a) Write an equation for the vapor pressure of silver, in atmospheres, as a function of kelvin

temperature. (b). The equation should be suitable for use in a tabulation, NOT in differential form. Put numerical values in the equation based on the data given. (4.7, p117) Solution:

)245011()(ln 1

-?-

T R H P d vap P

)2450

11(314.8255000ln ---=T P

08.10430685

ln +-

P 6. Zinc may exist as a solid, a liquid, or a vapor. The equilibrium pressure-temperature relationship between solid zinc and zinc vapor is giben by the vapor pressure equation for the solid. A similar relation exists for liquid zinc. At the triple point all three phases, solid, liquid, and vapor exist in equilibrium. That means that the vapor pressure of the liquid and the solid are the same. The vapor pressure of solid Zn varies with T as:

25.19)ln(755.015755

)(ln +--=

T T

atm P and the vapor pressure of liquid Zn varies with T as:79.21)ln(255.115246

)(ln +--=

T T

atm P . Calculate: (a) The boiling point of Zn under 1 atm; (b) The triple-point temperature; (c) the heat of evaporation of Zn at the normal (1 atm) boiling point; (d) The heat of fusion of Zn at the triple-point temperature; (e)The differences between the heat capacities of solid and liquid Zn. (4.8, p118)

Solution :(a) At boiling point, P=1 atm, that is lnP=0

079.21)ln(255.115246

=+--T T

079.21)ln(255.115246=+--T T T

To solve the equation, let y1 = 21.79T-15246, y2 = 1.255TlnT. Plot y-T of the functions, and the intersection is the answer.

T, K

From the plot, the intersection is 1180 K. So at 1180K, zinc boils. (b) At triple point, vapor pressure of solid Zn equals that of liquid Zn;

25.19)ln(755.01575579.21)ln(255.115246+--=+--T T T T 054.2)ln(5.0509

=+-T T

054.2)ln(5.0509=+-T T T

To solve the equation, assume two functions, y1=2.54T+509; y2 = 0.5TlnT. Plot y1-T and y2-T. The intersection is the answer.

T,K

The intersection is 695.3K, and this is the triple point temperature. (c )

)

()255.115246()1)(255.115246(79

.21255

.115246)(ln 22

d T dT T T T T P d +-=-+-=+--=

)1()(ln T d R H P d vap ?-= T R

H vap

255.115246+-=?-

T T R H vap 43.10126755)255.115246(-=-=?

At T b =1180K, mol kJ mol J T H vap /4.114/104.11443.101267553=?=-=?

(d) For solids,

)

()755.015755()1

)(755.015755()755.015755()(ln 25.19)ln(755.015755

ln 22T d dT

T dT T T

P d T T

P +-=-+-=-=+--=

()(ln T d R H P d fus ?-=

T R

H fus

755.015755+-=?-

)755.015755(T R H

fus

-=?

At triple point, T tr = 695.4K

mol kJ mol J T H fus /6.126/106.126)755.015755(314.83=?=-?=?

(e) dT C H

d P fus

?=?)(

755.0)

(-=?=

?dT

H d C fus

7. A particular material has a latent heat of vaporization of 5000J/mol. This heat of vaporization does not

change with temperature or pressure. One mole of the material exists in a two-phase equilibrium (liquid-vapor) in a container of volume V=1L, a temperature of 300K, and pressure of 1 atm. The container (constant volume ) is heated until the pressure reaches 2 atm. (Note that this is not a small ?P.) The vapor phase can be treated as an

ideal monatomic gas and the molar volume of the liquid can be neglected relative to that of the gas. Find the fraction of material in the vapor phase in the initial and final states. (4.9, P118) Solution: In the initial state, L

V Pa atm P K T 1,10013.11,

3001511=?===

mol RT V P n RT n V P 33

11111111006.4300

314.81010130,

--?=??===

%06.4)%(=vapor mol

In the final state, P 2=2 atm, V 2 = 1L According to Clayperon equation:

T T T T R H P P RT vap 6.458)30011(314.8500012ln 11ln 222212=--=???

??-?-= mol

RT V P n RT n V P 3322222222103.56.458314.810101302,

--?=???===

%3.5)%(=vapor mol

8. The melting point of gold is 1336K, and vapor pressure of liquid gold is given by:

)(ln 222.143522

716.23)(ln K T T

atm P --

=. (a) Calculate the heat of vaporization of gold at its melting point; Answer parts b, c, and d only if the data given in this problem statement are sufficient to support the calculation. If there are not enough data, write “solution not possible.” (b) What is the vapor pressure of solid gold at its melting point? (c) What is the vapor pressure of solid gold at 1200K ? (d) What is the v Solution: (a)

T T R H vap 19.10361841)222.143522(-=+--=?

(a) At 1336K, kJ J H vap 34810348133619.103618413-=?-=?-=? (b) “Solution not possible ”; (c) “Solution not possible ”. 9.

(a) At 298K, what is the Gibbs free energy change for the following reaction?

diamond graphite C C =

(b) Is the diamond thermodynamically stable relative to graphite at 298K?

(c ) What is the change of Gibbs free energy of diamond when it is compressed isothermally from 1atm to 1000 atm?

(d) Assuming that graphite and diamond are incompressible, calculate the pressure at which the two exist at

)

()222.143522()1

)(222.143522()222

.143522(

)(ln 22

T d dT T dT

T T P d +-=-

+-=-=

equilibrium at 298K.

(e) What is the Gibbs free energy of diamond relative to graphite at 900K? to simplify the calculation, assume

that the heat capacities of the two materials are equivalent.

DATA Density of graphite is 2.25g/cm 3 Density of diamond is 3.51g/cm 3

)/()298(mol kJ H o f ? )./()298(K mol J H o

Diamond 1.897 2.38 Graphite 0 5.74

Solution: (a) diamond graphite C C =

mol

kJ H H H o

graphite f o diamond f /1897,,=?-?=? )

./(36.338.274.5,,K mol J S S S o

graphite f o diamond f -=+-=?-?=? mol J S T H G /28.2898)36.3(2981897=-?-=?-?=?

(b) No, diamond is not thermodynamically stable relative to graphite at 298K.

(c ) mol J P V G diamand /29.34101309951

.310126

=???=

?=?-

(c ) Assuming N atm , ?G = 0, reversible processes as following can be designed to realize this,

)

(14939028.2898194.028.28981013051

.3101225.2101228.289810130)1)(V V ()(V 28.2898V 6

3()2()1()4(atm N N N N P P G G G G diamond graphite diamond graphite ==+-=+????

??-?-=+?--=?-++?????--＝＋＋＝

,0,0''

=?=?=??

?dT T

C dT C C T T

p T T p p

mol J H H /1897298900=?=? K mol J S S ./36.3298900-=?=?

mol J S T H G /492136.39001897=?+=?-?=?