The problems of the first law PDF

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Summary

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 ...

Description

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 CP of lead may be taken as 29.3J/(mol K)

Solution:

Qabsorb  Qincrease  Qmelt  n(C p T  H melting ) 1 2 1 mv  nMv 2 2 2 Qabsorb  W

W

n[29.3  (327  25)  4.8  103 ] 

1 n 207.2  10 3 v 2 2

V  363(m / s ) 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

QBurning  2500  10 3  4.1868  10467000( J ) 10467000  121( J / S ) 24  60  60 h 20  mg  75  9.8   245( J / S ) t 60

P  W / t  QBurning / t  Solution

Pincreasin g

1.3 One cubic decimeter (1 dm3) 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/cm2) is also used for surface tension dyn/cm) Solution

Stotal  nS Single 

(1  101 )3

 4    (0.5  10 6 ) 2  6  103 (m 2 )

4    (0.5  10 6 )3 3 W  S  72.75  10 5  2  (3  103  6  10 2 )  436.6( J )

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:

(a) Adiabatic T P  ( ) R / CP T0 P0 T  298  (

1 0.4 )  118( K ) 10

(b) W 1  (500  5)  101325  10 3   118( K ) nC p 10  50  101325  10 3  2.5R R  298 T  T0  T  298  118  180( K ) T 

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 T0. The valve on the tank is oðened 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 tèe gaS in the tank? The heat cap!city mf the gas, Cp and Cv each íay be(assumed to be c/nsuant over thå temperature rang!spanNed by the døperiment. You answer may be meft in terms of Cp and SvMhint: one way to approach the xroblem is to define the system as the gas ends up in the tank. hint: one way to approach the xroblem is to define the system as the gas ends up in the tank.

Adiabatic T P  ( ) R / CP T0 P0

solution

T  T0 (

P0  0 R / CP )  T0 P0

1.6 Calculate the heat of reaction of methane with oxygen at 298K, assuming that the products of reaction are CO2 and CH4 (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 FOR

DATA:

0 H 298 [ Kcal / g  mol ]

CO2 ( g )

 17.89  94.05

H 2 O( g )

 57.80

CH 4 ( g )

solution

CH 4  2O2  CO2  2 H 2O H 298  (H CO2  2H H 2 O  H CH 4 )  (94.05  2  57.80  17.89) H 298  191.76( Kcal / g  mol) 

191.76  103 1   26.9( Btu / 1000SCF ) 10 3 0.30483  103  252  103

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% O2 and 79% N2) (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?

(c) The furnace processes 2000kg of glass hourly, and its heat losses to the surroundings average 400000 kJ/h. calculate the fuel consumption at STP (in m3/h) assuming that for gas H1600-H298=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 for CH 4 CO 2 H 2O N2 O2

C P (cal / mol   C ) 16 13.7 11.9 8.2 8.2 (a) CH 4  2O2  CO2  2 H 2 O 1  8.71% 79 3  2  1.1   2  (1.1  1) 21 H 2 O%  2CO2 %  17.43%

CO2 % 

Solution

N 2 %  72.12% O2 %  0.87% (b) C p , p   C p ,i X i  0.01[13.7  8.71  11.9  17.43  8.2  (72.12  0.87)]  9.25(cal / mol   C ) T  T0  T  298 

191.76  1000  2104( K ) 9.25  11.48

(C ) P  1200  2000  400000  2800000( KJ / h) VConsumin g 

2800000  10 3  3214(m 3 / h) 191.76  1000  9.25  11.48(1600  298) 1 (  4.1868  ) 11.48 0.0224

(d ) C p ,r   C p ,i X i  [16  8.2  2.2 

100 ] / 11.48  8.87(cal / mol   C ) 21

H  H 298   ( C p , p ,i ni   C p ,r ,i ni )dT  191.76  1000  [(9.25  11.48)  8.87  11.48)](800  298)  189570(cal / g  mol ) VConsumin g 

2800000  10 3  1644(m 3 / h) 189570  9.25  11.48(1600  800) 1 (  4.1868  ) 11.48 0.0224

1.8 In an investigation of the thermodynamic properties of a-manganese, the following heat contents were determined: H700-H298=12113 J/(g atom)

H1000-H298=22803 J/(g atom)

Find a suitable equation for HT-H298 and also for CP as a function of temperature in the form (a+bT) Assume that no structure transformation takes place in the given tempeture rang.

Solution

b H   C P dT   a  bTdT  [aT  T 2 ]T298 2 b a(700  298)  (700 2  298 2 )  12113 2 b a(1000  298)  (1000 2  298 2 )  22803 2 a  35.62 b  0.011 C P  35.62  0.011T T H 298  35.62(T  298)  0.0055(T 2  298 2 )

1.9 A fuel gas containing 40% CO, 10% CO2, and the rest N2 (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

 H f , 298,CO  110458 J / mol 0

 H f , 298,CO2  393296 J / mol 0

C P ,CO  28.45  3.97 10 3 T  0.42 105 T 2 ( J / molK ) C P ,CO2  44.35  9.20 10 3 T  8.37 105 T  2 ( J / molK ) C P ,O2  19.92  4.10 10 3 T  1.67 105 T  2 ( J / molK ) C P , N 2  29.03  4.184 10 3 T ( J / molK ) 2CO  O2  2CO2 when 1mole fuel need 1mole air ( N 2 / O2  4) then reation CO  20% CO2  5% N 2  65% O2  10% production CO2  27.8% N 2  72.2%  H i , 298    H

0 f , 298, r

ni   H

0 f , 298, p

ni  393296  110458  282838( J / mol )

(a) C p , r   C p , p , i ni  0.2C P , CO  0.05C P , CO2  0.65C P , N 2  0.1C P , O2  28.77  4.38  10  3 T  0.67  105 T  2 ( J / molK ) C p , r   C p , p , i ni  0.278C P , CO2  0.722C P , N 2  33.28  5.58  10  3 T  0.19  105 T  2 ( J / molK ) H 



 H i , 298ni   ( C p , p , i ni   C p , r , i ni ) dT

 282838  0.2  

T

298

T



 28.77  4.38  10

(33.28  5.58  10  3 T  0.19  105 T  2 )  0.9dT 3

T  0.67  105 T  2 dT

733

 28283.8  2  ( 28.77T  2.19  10  3 T 2  0.67  105 T 1 ) 733 298

Solution

 (1.18T  0.321T 2  0.499T 1 ) T298  0 T ?

 H    H i , 298ni   ( C p , p ,i ni   C p , r ,i ni )dT  282838  0.2  

T

298

T



 28.77  4.38  10

(33.28  5.58  10 3 T  0.19  105 T  2 )  0.9dT 3

T  0.67  105 T  2 dT

1250

 28283.8  2  (28.77T  2.19  10 3 T 2  0.67  105 T 1 ) 1250 298  (1.18T  0.321T 2  0.499T 1 ) T298  0 T (b) 1250

1250 Q298  282838  0.2  

298 1250

1250 Q298  282838  0.2  

298

(33.28  5.58  10 3 T  0.19  105 T  2 )  0.9dT (33.28  5.58  10 3 T  0.19  105 T  2 )  0.9dT

1 2

1.10 (a) for the reaction CO  O2  CO2 ,what is the enthalpy of

reaction ( H 0 ) at 298 K ? (b) a fuel gas, with composition 50% CO, 50% N2 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 temperature the flame may attain (adiabatic flame temperature)? DATA :standard heats of formation

H f

at 298 K

CO  110000( J / mol) CO2  393000( J / mol)

Heat capacities [J/(mol K)] to be used for this problem N2=33, O2=33, CO=34, CO2=57 (a)H 0    H f , 298, r ni   H f , 298, P ni  110000  393000  283000( J / mol ) 0

0

0.5  22.2%, N 2 %  66.6%, O2 %  11.1% 1  0.25 / 0.2 22.2 product CO2 %   25%, N 2 %  75%, 100  11.1 (C )C p , P   Ci , p , P X i  33  0.666  33  0.222  34  0.111  33( J / mol  K ) (b) fuel CO % 

Solution

Cr , P   Ci , r , P X i  57  0.25  33  0.75  39( J / mol  K ) H  H 0   n pC p , P dT  0 283000  0.222  (0.889  39)(T  298)  0 T  2110( K )

1.11 a particular blast furnace gas has the following composition by (volume): N2=60%, H2=4, CO=12%, CO2=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? Air : n(O2 )  0.104 n( N 2 )  0.416

DaTA(k J?mol) FOR

H f (kJ / mol )

CO CO2

 110.523

CP [ J / m o l K ]

??

FOR CO

 393.513

CO2 H 2O ( g )

33 57 50 34

N 2 , O2

Solution 1 (a)CO  O2  CO2 2 1 H 2  O2  H 2O 2

Fuel : n(CO )  0.12 n(CO2 )  0.24 n( H 2 )  0.04

Air : n(O2 )  0.08 n( N 2 )  0.32

n( N 2 )  0.6

Flue : n(CO2 )  0.36 n( H 2O )  0.04 n( N 2 )  0.92

 H   H COCO2   H H 2 H 2O  0.12  (393.51  110.523)  0.04  (241.8  0)  43.6308KJ

C

n  0.36CCO2  0.04CH 2O  0.92C N2  0.36  57  0.04  50  0.92  34  53.8( J / K )

P ,r ,i i

43.6308 103  810( K ) 53.8 T  1108.98( K ) T 

(b)repeat the calculation for 30% excess0combustion air at 298K  H   H CO  CO2   H H 2  H 2 O  0.12  (393.51  110.523)  0.04  (241.8  0)  43.6308KJ

C

n  0.36CCO2  0.04CH 2 O  1.016C N 2  0.024CO2

P , r ,i i

 0.36  57  0.04  50  1.016  34  0.024  34  57.88( J / K ) 43.6308  103  753.8( K ) 57.88 T  1051.8( K ) T 

Fuel : n(CO )  0.12 n(CO2 )  0.24 n( H 2 )  0.04

Flue :

Air : n(O2 )  0.104

n(CO2 )  0.36

n( N 2 )  0.416

n( N 2 )  1.016

n( H 2O)  0.04

Maret 8, 2013

n(O2 )  0.024

n( N 2 )  0.6

(C)what is the adiabatic flame temperature when the blasp furnace gas is preheated to 700K (the dry air is at 298K) Fuel : n(CO )  0.12 n(CO2 )  0.24 n( H 2 )  0.04

Air : n(O2 )  0.08 n( N 2 )  0.32

n( N 2 )  0.6

Flue : n(CO2 )  0.36 n( H 2O )  0.04 n( N 2 )  0.92

 H   H COCO2   H H 2 H 2O  H fuel 700298  0.12  (393.51  110.523)  0.04  (241.8  0)  (700  298)  (0.12  33  0.24  57  0.04  28  0.6  34)  59.373KJ

C

n  0.36CCO2  0.04CH 2O  0.96C N2  0.36  57  0.04  50  0.92  34  53.8( J / K )

P ,r ,i i

59.373 103  1103.6( K ) 53.8 T  1401.6( K ) T 

(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? Fuel :

Air :

Flue :

n(CO )  0.12

n(O2 )  0.08

n(CO2 )  0.36

n(CO2 )  0.24

n( N 2 )  0.32

n( H 2 )  0.04

n( H 2 O)  0.048

15 n ( H 2O )  0.4  0.008 760  15

n( N 2 )  0.92

n( N 2 )  0.6

 H   H COCO2   H H 2 H 2O  0.12  (393.51  110.523)  0.04  (241.8  0)  43.6308KJ

C

n  0.36CCO2  0.048CH 2O  0.92C N2

P ,r ,i i

 0.36  57  0.048  50  0.92  34  54.2( J / K ) 43.6308 103  805( K ) 54.2 T  1103( K ) T 

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) CP,L=7.5(cal/mol℃), CP,S=5.41+(1.5*10-3T )(cal/mol℃) Solution 1803

1798

1798

1803

H SL,1803  

CP , S dT  

CP , L dT  H LS,1798  0

H SL,1798  3100  5.41  5  1.5  10 3 (18032  17982 )  0.5  7.5  5  3103(cal / mol)

1.13 Cuprous oxide (Cu2O) is being reduced by hydrogen in a furnace at 1000K, (a)write the chemical reaction for the reduced one mole of Cu2O (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 Cu2O=-41900 H2O=-59210 solution

Cu2O  H 2  Cu  H 2O H  59210  41900  17310(cal / mol)

,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, Cp,s=26(J/molK), Cp,L=29(J/molK), Melting point=932K, Heat of fusion=10700J/mol 932

1000

H l ,1000   CP , S dT  

CP , L dT  H LS

Solution  26  (932  298)  29  (1000  932)  10700  27184( J / mol) 298

P

932

27184 1  1000   279.7(W )  0.28(kW ) 27 3600

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, Cp,s,Al=26(J/molK),

Cp,s,

Al2O3=104J/mol,

heat

formation

of

Al2O3=-1676000J/mol Solution;

1 1676000   27  2 T  302( K )

1

16  1.5  27  99 27  104  T 102

T  600( K )

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: Cp=4.184J/g k, Density=1g/cm3; for copper: molecular weight=63.54g/mol Cp=7cal/mol k, heat of fusion=3120 cal/mol QCopper / min  [3120  7  (1356  400)] 

Solution: Q

Water

1000 63.54  min

/ min  1(80  20)V

V  2.573  10 3 (m3 / min)

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. DATA; For water Cp=4.184J/g k, Density=1g/cm3 Solution: W  4.184  (60  20)  5  1000  13947(W ) 60

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/m3 what is the internal energy change for the evaporation of water?

PV  101325 

Solution:

373  0.0224  3101( J / mol ) 273

3101  7.6% 2261  18 U  W  Q  3101  2261  18  37597( J / mol )

%

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 m(2261  4.18 100)  1000  334, m  125( g ) 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: CP,L=4.18J/(g k),

CP,v=2.00J/(g k),

Solution:

△HV=2261J/g, △Hm=334 J/g

(2261  2  80  4.18  80) x  4.18  80  (1000  x), x  138( g ) irreversible

The problems of the second law 2.1 The solar energy flux is about 4J cm2/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

T H  TL (90  25) 104 W  QH  4 St TH 90  273 60 W  746(W ) t S  6.25(m 2 ) P 

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 ma...