The problems of the first law1. 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) (1.1)Solution: )/(5.112.20721]108.4)25327(3.29[2121)(2322s m V v n n WQ nMv mv W H T C n Q Q Q absorb melting p melt increase absorb ==⨯+-⨯===∆+∆=+=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 eh rate of 20 m/min (1.2)Solution )/(24560208.975)/(12160602410467000//)(104670001868.4102500sin 3S J t h mg P S J t Q t W P J Q gincrea Burning Burning =⨯⨯=∆==⨯⨯====⨯⨯=3 One cubic decimeter (1 dm 3) of water is broken into droplets having a diameter of onemicrometer (1 um) at 20℃. (1.3)(a) what is the total area of the droplets?(b) Calculate the minimum work required to produce the droplets. Assume that the dropletsare 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)(25.218)106103(1075.72)(103)101(4)101(34)101(232523263631J S W m nS S Single total =⨯-⨯⨯⨯=∆=⨯=⨯⨯⨯⨯⨯⨯⨯⨯==-+----σππ4.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 thehelium entering the quench chamber when the pressure in the tank has fallen to 1 atm? (1.4)Solution: )(180118298)(1185.229810101325501010101325)5500(1)()(118)101(298)()(0334.0/00K T T T K RR nC W T b K T P PT T Adiabatic a p C R P=-=∆-==⨯⨯⨯⨯⨯⨯⨯-⨯==∆=⨯==--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 opened and the surrounding gas is allowed to flow quickly into the tank until the pressure inside the tank is equals the pressure outside. Assume that no heat flow takes place. What is the final tempeture of the gas in the tank? The heat capacity of the gas, C p and C v each may be assumed to be constant over the temperature rang spanned by the experiment. You answer may be left in terms of C p and C vhint: one way to approach the problem is to define the system as the gas ends up in the tank. (1.5)solution 0/000/00)()(T P P T T P PT T Adiabatic PPC R C R ≈-==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 1atmNOTE: this value is a good approximation for the low calorific powder of natural gas (1.6)DA TA:)()()(224g O H g CO g CH FOR80.5705.9489.17]/[0298---∙∆mol g Kcal Hsolution)1000/(9.2610252103048.01101076.191)/(76.191)89.1780.57205.94()2(22333332982982224422SCF Btu mol g Kcal H H H H H OH CO O CH CH O H CO =⨯⨯⨯⨯⨯=∙=∆+⨯---=∆-∆+∆-=∆+=+-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) (1.7)(a) Assuming complete combustion, what is the composition of the flue gas (the gasfollowing combustion)?(b) What is the temperature of the gas, assuming no heat loss?(c) The furnace processes 2000kg of glass hourly, and its heat losses to the surroundingsaverage 400000 kJ/h. 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 thecombustion air. Calculate the decrease in fuel consumption if the combustion air is heated to 800KDA TA STP means T=298K, P=1atm22224O N O H CO CH for 2.82.89.117.1316)/(C mol cal C P ∙Solution)(210448.1125.9100076.191298)/(25.9)]87.012.72(2.843.179.1171.87.13[01.0)(%87.0%%12.72%%43.17%2%%71.8)11.1(221791.1231%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/]211002.22.816[)()/(3214)0224.011868.448.11)2981600(48.1125.9100076.191(102800000)/(280000040000020001200)(33min ,,,,298,,33min h m V mol g cal dTn C n C H H C mol cal X C C d h m V h KJ P C gConsu i i r p i i p p i i p r p g Consu =⨯⨯-⨯-⨯=∙=-⨯-⨯-⨯=--∆=∆∙=⨯⨯+===⨯⨯-⨯-⨯⨯==+⨯=⎰∑∑∑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. (1.8)Solution )298(0055.0)298(62.35011.062.35011.062.3522803)2981000(2)2981000(12113)298700(2)298700(]2[2229822222982---=∆-=-===-+-=-+-+=+==∆⎰⎰T T H TC b a ba ba T baT bTdT a dT C H TP T P9.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 (1.9)molJ Hmol 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(28.282831067.01038.477.289.0)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/(1122298127332981523733253253298,,,,298,253,,,,,253,,,,,,,0,298,0,298,298,22222222222222==+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆⨯-⨯+=+==⨯-⨯+=+++===-=∆-∆=∆========+-----------⎰⎰⎰∑∑⎰∑∑∑∑T T T T T T T dT T T dTT 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 Hn H H N CO production O N CO CO reation then O N air mole need fuel mole when CO O CO T TT 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 pf i rf idTT T Q dT T T Q b T T T T T T T dT T T dTT T dT n C n C n H H T TT 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(28.282831067.01038.477.289.0)1019.01058.528.33(2.0282838)(253125029812502982531250298125029829812125029815231250253253298,,,,298,⨯⨯-⨯++⨯-=⨯⨯-⨯++⨯-===+--⨯+⨯++⨯=⨯-⨯++⨯⨯-⨯+-⨯=--∆=∆-----------⎰⎰⎰⎰⎰∑∑⎰10. (a) for the reaction 2221CO 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?(c) If the fuel gas and the air enter there burner at 298 K, what is the highest temperaturethe flame may attain (adiabatic flame temperature)? DA TA :standard heats of formation f H ∆ at 298 K (1.10))/(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 ==-⨯-⨯=-∆=∆∙=⨯+⨯==∙=⨯+⨯+⨯====-====+==+-=∆-∆=∆⎰∑∑∑∑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 flame temperature be affected? DA TA(k J/mol) （1.11）2CO CO FOR513.393523.110)/(--∆m o lkJ H f 2222,)(O N g O H CO CO FOR34505733]/[K mol J C P ∙Solution)(1052)(75438286370])295.03450(241604[026.0])335.03457(110523393513[079.0])([%8.66%%,8.6%%,6.2%%,8.15%%,9.72.0/83.110012%)()(1122)(82538313430])295.03450(241604[029.0])335.03457(110523393513[086.0])([%7.65%%,7.5%%,9.2%%,1.17%%,6.82.0/810012%2121)(,,,,,,,02222,,,,,,,0222222222K T K T T n C T T X C dT n C n C H x H N O H CO CO b K T K T T n C T T X C dT n C n C H x H N O H CO CO OH O H CO O CO a i i r P ii P i i r P i i p P i i i i r P ii P i i r P i i p P i i ===∆=∆-∆-⨯--+∆-⨯---=+--∆=∆=====⨯+====∆=∆-∆-⨯--+∆-⨯---=+--∆=∆=====+=→+→+∑∑∑⎰∑∑∑∑∑⎰∑∑)(1419),(11213842594034286.0)402(2.39714.0])295.03450(241604[029.0])335.03457(110523393513[086.0)3(K T K T T T T T H ===∆=∆⨯--∆⨯-∆-⨯--+∆-⨯---=∆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 ℃) （1.12） Solution)/(310355.75.0)17981803(105.1541.5310002231798,1798,17981803,18031798,1803,mol cal H H dT C dT C HL S SL L P S P LS =⨯-⨯-⨯+⨯+==+++-⎰⎰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 （1.13） solution)/(173104190059210222mol cal H OH Cu H O Cu =-=∆+=+,exothermic reaction14. (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 (1.14)Solution )(28.0)(7.2793600110002727184)/(2718410700)9321000(29)298932(261000932,932298,1000,kW W P mol J H dT C dT C H SLL P S P l ==⨯⨯==+-⨯+-⨯=++=⎰⎰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(1.15)Solution;)(600)(3021041029927275.116122711676000K T K T T ==∆∆⨯⨯++⨯⨯=⨯⨯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 (1.16)Solution:min)/(10573.2)2080(1min /min54.631000)]4001356(73120[min /33m V VQ Q Water Copper -⨯=-=⨯⨯-⨯+=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. DATA; For water C p =4.184J/g k, Density=1g/cm 3 (1.17) Solution: )(139476010005)2060(184.4W W =⨯⨯-⨯=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? (1.18)Solution: )/(375971822613101%6.71822613101%)/(31010224.0273373101325mol J Q W U mol J V P =⨯+-=+=∆=⨯==⨯⨯=∆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 (1.19) Solution )(125,3341000)10018.42261(g m m =⨯=⨯+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 (1.20) Solution:leirreversib g x x x )(138),1000(8018.4)8018.48022261(=-⨯⨯=⨯-⨯+The problems of the second law1 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. (2.1)Solution )(664.0)(74660104273900)25900(24m S W tWP StQ T T T W H H L H ===⨯⨯+-=-=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. (2.2)Solution:)(114474625.02035202733475.0%75W P P T T T P Q T T T W L LLLH HHLH =⨯⨯+-⨯=-=-=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 runningcontinuously? Assume that the refrigerator is perfectly insulated and that the efficiencies involved have their largest possible value. (2.3)Solution: )(4576033474625.020273g m M m P P T T T P L LLLH ===⨯⨯=-=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. (2.4)Solution: )(52.0)(393'60284216.4216.4300'5.0%50hp W P P T T T P P Q T T T W L L L H LLLH ==⨯⨯-=-==-= 5 if a fossil fuel power plant operating between 540 and 50℃ provides the electrical powerto 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 ofperformance of the heat pump is 10% of maximum(c) if a furnace can use 80% of the energy in fossil foe to heat the house would it be moreeconomical in terms of overall fissile fuel consumption to use a heat pump or a furnace ? do the calculations for cases a and b (2.5)solution:1,2,2,1,212,2,2,2,21,1,1,1,198.82527352527354050540)(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)(1,2,not is b ok is a c P P b H H =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 adiabaticenclosure. Assume that C p is 77J /(mol K) from 273K to 373K (2.6) Solution:)/(933.0)273323ln(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 =+=+=∆=∆ 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 heatprovided to the boiler?(b) How many metric tons (1000kg) of coal per hour is required if the plant out put is to be500MW (megawatts). Assume the maximum efficiency for the plant. The heat of combustion of coal is 29.0 MJ/k g(c) Electricity is used to heat a home at 25℃ when the out door temperature is 10℃ bypassing a current through resistors. What is the maximum amount of heat that can be added to the home per kilowatt-hour of electrical energy supplied? (2.7)Solution:)(3.69)(6937136005000.29)()(89.013054030540)(ton kg m T T T mb J Q T T T W a LH LH H L H ==⨯=-=+-=-=)(9.191102525273)(J Q Q T T T W c H HHLH =-+=-=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 containeris 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 therate 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 (2.8)solution:)(23.2,2510027310010004000018)()(45.0,10004000018)(g m m b g m ma =-+===9 some aluminum parts are being quenched (cooled rapidly ) from 480℃ to -20℃ byimmersing 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 minuspower rating in kilowatts, of motor required to operate the refrigerator? Data for aluminum heat capacity is 28J/mol K; Molecular weight 27g/mol (2.9)Solution:)(5.102)(102474202732030)20480(28271000kW W P P T T T P P L L L L H W L ==---=-=--⨯=10 an electric power generating plant has a rated output of 100MW. The boiler of the plantoperates 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 morethan 10℃. What volume of cooling water in cubic meters per hour, is require to operate the plant?(c) The boiler tempeture is to be raised to 540℃,but the condensed temperature and electricoutput will remain the same. Will the cooling water requirement be increased, decreased, or remain the same?Data heat capacity 4.184, density 1g/cm 3 (2.10)Solution: )(109.7)(102.21040300273300)(1188J 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 ⨯==⨯⨯⨯⨯=noW P T T T P c L H H H )(10626.11040540273540)(88⨯=-+=-=11 (a) Heat engines convert heat that is available at different temperature to work. Theyhave 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? (2.11)Solution:)/(1006.136001000)()(055.0127320420)(6h kW hmg P b J Q T T T W a H H L H ⨯=⨯∆==+-=-=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? (2.12)Solution:)(1014.1101527320273)()(77.33600/10152731520)(555kJ Q b kW P T T T P a H L L L H ⨯=-+==-+=-=13solution:)/(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 AlNN Al a -=+-⨯-⨯=∆=∆-=-⨯-⨯=∆=∆=+14solution:)/(2257412000)27340273ln 184.4273336263273ln1.2()(40,010,K J dT T C T H dT T C m S WATER P m mICE P =+++=+∆+=∆⎰⎰- 15)(70428)(2896100077773002J W J Q T T T W L L L H ==-=-=16)(4.3719))2.4300(314.85.13.83(3002.4300)(7.58663.832.42.4300J Q T T T W J Q T T T W H H L H L L L H =-⨯+-=-==-=-=17yesd Q c K J PPnR S b J pdV n W Q OU T a )(0)()/(1.1910ln 314.81ln )()(570410ln 298314.810)(0==⨯⨯==∆=⨯⨯=-=-==∆=∆⎰18)(122233527302033560500g m m m T T T L L H =-=-=⨯教材各章习题参考答案 (魏)3.2 ΔG = -108.9 J/mol; ΔS = -21.42 J/(mol.K)3.6 (a ) 22.09/(.)S J mol K ∆=；(b) At 0︒C, ∆G =0; (c) ∆H = 5841.9 J;(d) ∆S =21.39J /(mol.K)，∆G = 109.38 J/mol4.1 (a ) 2898.28J/mol; ( b ) No; ( c ) 345 J/mol; ( d ) 14939 atm; ( e )4921 J/mol4.2 ( a ) 272.8K; ( b ) Pa P 610345⨯≈∆ ; ( c ) 249.46K 4.3 1202K4.4 P=5.73⨯10-6 atm 4.5 0.16P4.7 08.10430685ln +-=TP 4.8 ( a ) 1180K; ( b ) 695.3K; ( c ) 114.4kJ/mol; ( d ) 7123 J/mol; ( e )4.2J/mol4.9 In the initial state: 4.06 mol %; in the final state:5.3 mol% 4.10 ( a )348 kJ; ( b ) 2.3×10-3Pa ;( c ) “ solution not possible ”; (d ) “solution not possible ”5.1 atm p H 0005.0= 5.2、atmp o 1221007.1-⨯=If the error in enthalpy is 500cal, the uncertainty in the pressure calculated is 28.6%, and if the error in enthalpy is -500cal, the uncertainty is -22.1%5.3、(a) T =462K; (b) T = 420K5.4 (a) atm P O 2621014.1-⨯=, (b) P O2 =2.28⨯10-10 atm., (c) The equilibriumoxygen pressure remains the same when the total pressure increases, which means a higher purity level of N 2 .5.5 (a) 略； (b) Pa atm P H 8.181013056.1800019.0)('2=⨯==; (c) 21.5L Ar isneeded to be bubbled into the melt.5.6（a ）l n K a1/T, 10-31/K=∆-=∆o o G kJ H 1000;50- 66.6kJ(b) Ja = 3 < Ka, the reaction will proceed from left to right, and theatmosphere will not oxidize Ni. 5.7 略5.8. (a) P SiO = 8.1⨯10-8 (atm) (b) ∆H o = 639500J; ∆So =334.9J/K (c ) PO2 =10-30 atm 5.9 5.10．J H o72250=∆，the reaction is an endothermic one.5.11. (a),166528J H o =∆ the reaction is an endothermic one.; (b) At 1168K, the equilibrium pressure of CO2 equals one atmosphere.)(106.08)(atm Pg u -⨯=5.12 (a) 略 ， (b) Mg CO P P =; (c) T = 2037 K 5.13 (a) 略； (b) 13109.2⨯=K ; (c) ppm 186.0 5.14 (a) 略； (b) kJ H 52.267=∆; (c) K T 1592= 5.15 (a) )(106.13atm -⨯≈; (b) )(1028.210)(2atm P g O H -⨯=5.16 (a) 97.9=K ; (b) atm x 14.4=; (c) if the temperature is increased, the fraction of water reacted will increase since the equilibria constant increases with increasing temperature.6.2 （a ）1.287V;(b) When the water impure, the voltage will go higher; (c) 1.219V 6.4 (a) 145.3kJ;(b) The maximum work that could be derived is 702.36kJ; (c) In this case, the maximum work that could be derived is696.56kJ.6.5 (a) -6252J/mol; (b) 370.0)(=II Cd a ; (c) )(42.3mmHg P Cd =; 6.67.87⨯10-4 V 6.7 (a))(22g Cl Mg MgCl +=(b) Pa P Cl 21'1086.82-⨯=;(c) 2.485V6.8 (a) Pa P O 11'2105.5-⨯=;(b) Anode: e Ni Ni 2+→Cathode: -→+2222/1O e O ;(c) 0.757V; (d) 0.261V6.10 (a) )(509.3V E o=;(b) 0.074kJ;(c) 4.1⨯106J;(d) Yes. In this case, the open circuit voltage is 3.648V;(e) In this case, to keep the temperature constant, 3.92⨯106J heatshould be removed from the battery per hour. 6.11(a) TG CO Al C O Al o 26.3211008.12/322/36232-⨯=+=+Δ(b) The minimum voltage at which the electrolysis may be carriedout at 1250K is 1.172V .7.1 0.117 atm 7.5 ( a ) ,82.5 2.5 2.5B A BA BB T PV V V x x x x x ⎛⎫∂=+=--⎪∂⎝⎭ ,102.5 2.5 2.5A B A A B A T PV V V x x x x x ⎛⎫∂=+=-- ⎪∂⎝⎭( b) B A M x x V 5.2=7.7 2)1(736.0ln Sn Sn x --=γ7.8 The maximum solubility of MgF2 in liquid MgCl at 900︒C is 19。
