Resolução Termodinâmica Çengel capítulo 1 exercícios 85 - 125, Exercícios de Termodinâmica

Baixe Resolução Termodinâmica Çengel capítulo 1 exercícios 85 - 125 e outras Exercícios em PDF para Termodinâmica, somente na Docsity! 1-34 Review Problems 1-85 A hydraulic lift is used to lift a weight. The diameter of the piston on which the weight to be placed is to be determined. F2 F1 D2 10 cm 25 kg Weight 2500 kg Assumptions 1 The cylinders of the lift are vertical. 2 There are no leaks. 3 Atmospheric pressure act on both sides, and thus it can be disregarded. Analysis Noting that pressure is force per unit area, the pressure on the smaller piston is determined from kPa 23.31kN/m 23.31 m/skg 1000 kN 1 4/m) 10.0( )m/s kg)(9.81 25( 4/ 2 22 2 2 1 1 1 1 1 ==       ⋅ = == π πD gm A FP From Pascal’s principle, the pressure on the greater piston is equal to that in the smaller piston. Then, the needed diameter is determined from m 1.0=→        ⋅ =→=== 222 2 2 2 2 2 2 2 2 21 m/skg 1000 kN 1 4/ )m/s kg)(9.81 2500(kN/m 23.31 4/ D DD gm A F PP ππ Discussion Note that large weights can be raised by little effort in hydraulic lift by making use of Pascal’s principle. 1-86 A vertical piston-cylinder device contains a gas. Some weights are to be placed on the piston to increase the gas pressure. The local atmospheric pressure and the mass of the weights that will double the pressure of the gas are to be determined. WEIGTHS GAS Assumptions Friction between the piston and the cylinder is negligible. Analysis The gas pressure in the piston-cylinder device initially depends on the local atmospheric pressure and the weight of the piston. Balancing the vertical forces yield kPa 95.7==        ⋅ −=−= kN/m 66.95 m/skg 1000 kN 1 )/4m 12.0( )m/s kg)(9.81 5(kPa 100 2 22 2 piston atm πA gm PP The force balance when the weights are placed is used to determine the mass of the weights kg 115.3=→        ⋅ + += + += weights22 2 weights weightspiston atm m/skg 1000 kN 1 )/4m 12.0( )m/s )(9.81kg 5( kPa 95.66kPa 200 )( m m A gmm PP π A large mass is needed to double the pressure. 1-35 1-87 An airplane is flying over a city. The local atmospheric pressure in that city is to be determined. Assumptions The gravitational acceleration does not change with altitude. Properties The densities of air and mercury are given to be 1.15 kg/m3 and 13,600 kg/m3. Analysis The local atmospheric pressure is determined from kPa 91.8==        ⋅ += += kN/m 84.91 m/skg 1000 kN 1m) )(3000m/s )(9.81kg/m (1.15kPa 58 2 2 23 planeatm ghPP ρ The atmospheric pressure may be expressed in mmHg as mmHg 688=           == m 1 mm 1000 kPa 1 Pa 1000 )m/s )(9.81kg/m (13,600 kPa 8.91 23 atm Hg g P h ρ 1-88 The gravitational acceleration changes with altitude. Accounting for this variation, the weights of a body at different locations are to be determined. Analysis The weight of an 80-kg man at various locations is obtained by substituting the altitude z (values in m) into the relation         ⋅ ×−== − 2 26 m/skg1 N1 )m/s103.32kg)(9.807(80 zmgW Sea level: (z = 0 m): W = 80×(9.807-3.32x10-6×0) = 80×9.807 = 784.6 N Denver: (z = 1610 m): W = 80×(9.807-3.32x10-6×1610) = 80×9.802 = 784.2 N Mt. Ev.: (z = 8848 m): W = 80×(9.807-3.32x10-6×8848) = 80×9.778 = 782.2 N 1-89 A man is considering buying a 12-oz steak for $3.15, or a 320-g steak for $2.80. The steak that is a better buy is to be determined. Assumptions The steaks are of identical quality. Analysis To make a comparison possible, we need to express the cost of each steak on a common basis. Let us choose 1 kg as the basis for comparison. Using proper conversion factors, the unit cost of each steak is determined to be 12 ounce steak: $9.26/kg=                  kg 0.45359 lbm 1 lbm 1 oz 16 oz 12 $3.15 =Cost Unit 320 gram steak: $8.75/kg=            kg 1 g 1000 g 320 $2.80 =Cost Unit Therefore, the steak at the international market is a better buy. 1-38 1-98E An expression for the equivalent wind chill temperature is given in English units. It is to be converted to SI units. Analysis The required conversion relations are 1 mph = 1.609 km/h and T(°F) = 1.8T(°C) + 32. The first thought that comes to mind is to replace T(°F) in the equation by its equivalent 1.8T(°C) + 32, and V in mph by 1.609 km/h, which is the “regular” way of converting units. However, the equation we have is not a regular dimensionally homogeneous equation, and thus the regular rules do not apply. The V in the equation is a constant whose value is equal to the numerical value of the velocity in mph. Therefore, if V is given in km/h, we should divide it by 1.609 to convert it to the desired unit of mph. That is, T T Vequiv ambientF F( ) . [ . ( )][ . . ( / . ) . / . ]° = − − ° − +914 914 0 475 0 0203 1609 0 304 1609V or T T Vequiv ambientF F( ) . [ . ( )][ . . . ]° = − − ° − +914 914 0 475 0 0126 0 240 V where V is in km/h. Now the problem reduces to converting a temperature in °F to a temperature in °C, using the proper convection relation: 1 8 32 914 914 18 32 0 475 0 0126 0 240. ( ) . [ . ( . ( ) )][ . . . ]T Tequiv ambientC C° + = − − ° + − +V V which simplifies to T T Vequiv ambientC( ) . ( . )( . . .° = − − − +330 330 0 475 0 0126 0 240 V ) where the ambient air temperature is in °C. 1-39 1-99E EES Problem 1-98E is reconsidered. The equivalent wind-chill temperatures in °F as a function of wind velocity in the range of 4 mph to 100 mph for the ambient temperatures of 20, 40, and 60°F are to be plotted, and the results are to be discussed. Analysis The problem is solved using EES, and the solution is given below. "Obtain V and T_ambient from the Diagram Window" {T_ambient=10 V=20} V_use=max(V,4) T_equiv=91.4-(91.4-T_ambient)*(0.475 - 0.0203*V_use + 0.304*sqrt(V_use)) "The parametric table was used to generate the plot, Fill in values for T_ambient and V (use Alter Values under Tables menu) then use F3 to solve table. Plot the first 10 rows and then overlay the second ten, and so on. Place the text on the plot using Add Text under the Plot menu." Tequiv [F] Tambient [F] V [mph] -52 -25 10 -46 -20 10 -40 -15 10 -34 -10 10 -27 -5 10 -21 0 10 -15 5 10 -9 10 10 -3 15 10 3 20 10 -75 -25 20 -68 -20 20 -61 -15 20 -53 -10 20 -46 -5 20 -39 0 20 -32 5 20 -25 10 20 -18 15 20 -11 20 20 -87 -25 30 -79 -20 30 -72 -15 30 -64 -10 30 -56 -5 30 -49 0 30 -41 5 30 -33 10 30 -26 15 30 -18 20 30 -93 -25 40 -85 -20 40 -77 -15 40 -69 -10 40 -61 -5 40 -54 0 40 -46 5 40 -38 10 40 -30 15 40 -22 20 40 -30 -20 -10 0 10 20 -80 -70 -60 -50 -40 -30 -20 -10 0 10 20 Tambient T W in dC hi ll W ind Chill Temperature W ind speed =10 mph 20 mph 30 mph 40 mph 0 20 40 60 80 100 -20 -10 0 10 20 30 40 50 60 V [mph] T e qu iv [F ] Tamb = 20F Tamb = 40F Tamb = 60F 1-40 1-100 One section of the duct of an air-conditioning system is laid underwater. The upward force the water will exert on the duct is to be determined. Assumptions 1 The diameter given is the outer diameter of the duct (or, the thickness of the duct material is negligible). 2 The weight of the duct and the air in is negligible. Properties The density of air is given to be ρ = 1.30 kg/m3. We take the density of water to be 1000 kg/m3. FB L = 20 m D =15 cm Analysis Noting that the weight of the duct and the air in it is negligible, the net upward force acting on the duct is the buoyancy force exerted by water. The volume of the underground section of the duct is m 0.353=m) /4](20m) 15.0([)4/( 322 ππ === LDALV Then the buoyancy force becomes kN 3.46=        ⋅ == 2 323 m/skg 0001 kN 1)m )(0.353m/s )(9.81kg/m (1000gVFB ρ Discussion The upward force exerted by water on the duct is 3.46 kN, which is equivalent to the weight of a mass of 353 kg. Therefore, this force must be treated seriously. 1-101 A helium balloon tied to the ground carries 2 people. The acceleration of the balloon when it is first released is to be determined. Assumptions The weight of the cage and the ropes of the balloon is negligible. Properties The density of air is given to be ρ = 1.16 kg/m3. The density of helium gas is 1/7th of this. Analysis The buoyancy force acting on the balloon is N 5958 m/skg 1 N 1 )m )(523.6m/s)(9.81kg/m (1.16 m 523.6/3m) π(54/3r4π 2 323 balloonair 333 =        ⋅ = = === V V gFB balloon ρ Helium balloon m = 140 kg The total mass is kg 226.870286.8 kg 86.8)m (523.6kg/m 7 1.16 peopleHetotal 33 HeHe =×+=+= =     == mmm m Vρ The total weight is N 2225 m/skg 1 N 1 )m/s kg)(9.81 (226.8 2 2 total =        ⋅ == gmW Thus the net force acting on the balloon is N 373322255958net =−=−= WFF B Then the acceleration becomes 2m/s 16.5=        ⋅ == N 1 m/skg1 kg 226.8 N 3733 2 total net m F a 1-43 1-105 A barometer is used to measure the altitude of a plane relative to the ground. The barometric readings at the ground and in the plane are given. The altitude of the plane is to be determined. Assumptions The variation of air density with altitude is negligible. Properties The densities of air and mercury are given to be ρ = 1.20 kg/m3 and ρ = 13,600 kg/m3. h Analysis Atmospheric pressures at the location of the plane and the ground level are kPa100.46 N/m 1000 kPa1 m/s kg1 N 1 m) )(0.753m/s )(9.81kg/m (13,600 )( kPa 92.06 N/m 1000 kPa1 m/s kg1 N 1 m) )(0.690m/s )(9.81 kg/m(13,600 )( 22 23 groundground 22 23 planeplane =                 ⋅ = = =                 ⋅ = = hgP hgP ρ ρ Taking an air column between the airplane and the ground and writing a force balance per unit base area, we obtain 0 Sea level kPa92.06)(100.46 N/m 1000 kPa1 m/s kg1 N 1 ))(m/s )(9.81 kg/m(1.20 )( / 22 23 planegroundair planegroundair −=                ⋅ −= −= h PPhg PPAW ρ It yields h = 714 m which is also the altitude of the airplane. 1-106 A 10-m high cylindrical container is filled with equal volumes of water and oil. The pressure difference between the top and the bottom of the container is to be determined. Water Oil SG = 0.85 h = 10 m Properties The density of water is given to be ρ = 1000 kg/m3. The specific gravity of oil is given to be 0.85. Analysis The density of the oil is obtained by multiplying its specific gravity by the density of water, 33OH kg/m 850)kg/m 0(0.85)(100SG 2 ==×= ρρ The pressure difference between the top and the bottom of the cylinder is the sum of the pressure differences across the two fluids, [ ] N/m 1000 kPa 1 m) )(5m/s )(9.81kg/m (1000m) )(5m/s )(9.81kg/m (850 )()( 2 2323 wateroilwateroiltotal kPa 90.7=         += +=∆+∆=∆ ghghPPP ρρ 1-44 1-107 The pressure of a gas contained in a vertical piston-cylinder device is measured to be 250 kPa. The mass of the piston is to be determined. Assumptions There is no friction between the piston and the cylinder. P Patm W = mg Analysis Drawing the free body diagram of the piston and balancing the vertical forces yield         ⋅ ×−= −= −= − kPa1 skg/m 1000 )m10kPa)(30 100(250)m/s (9.81)( )( 2 242 atm atm m APPmg APPAW It yields m = 45.9 kg 1-108 The gage pressure in a pressure cooker is maintained constant at 100 kPa by a petcock. The mass of the petcock is to be determined. Assumptions There is no blockage of the pressure release valve. P Patm W = mg Analysis Atmospheric pressure is acting on all surfaces of the petcock, which balances itself out. Therefore, it can be disregarded in calculations if we use the gage pressure as the cooker pressure. A force balance on the petcock (ΣFy = 0) yields kg 0.0408=         ⋅× == = − kPa 1 skg/m 1000 m/s 9.81 )m10kPa)(4 (100 2 2 26 gage gage g AP m APW 1-109 A glass tube open to the atmosphere is attached to a water pipe, and the pressure at the bottom of the tube is measured. It is to be determined how high the water will rise in the tube. Properties The density of water is given to be ρ = 1000 kg/m3. Water Patm= 92 kPa h Analysis The pressure at the bottom of the tube can be expressed as tubeatm )( hgPP ρ+= Solving for h, m 2.34=                 ⋅− = − = kPa1 N/m 1000 N 1 m/s kg1 )m/s )(9.81 kg/m(1000 kPa92)(115 22 23 atm g PP h ρ 1-45 1-110 The average atmospheric pressure is given as where z is the altitude in km. The atmospheric pressures at various locations are to be determined. 256.5 atm )02256.01(325.101 zP −= Analysis The atmospheric pressures at various locations are obtained by substituting the altitude z values in km into the relation P zatm = −101325 1 0 02256 5 256. ( . ) . Atlanta: (z = 0.306 km): Patm = 101.325(1 - 0.02256×0.306)5.256 = 97.7 kPa Denver: (z = 1.610 km): Patm = 101.325(1 - 0.02256×1.610)5.256 = 83.4 kPa M. City: (z = 2.309 km): Patm = 101.325(1 - 0.02256×2.309)5.256 = 76.5 kPa Mt. Ev.: (z = 8.848 km): Patm = 101.325(1 - 0.02256×8.848)5.256 = 31.4 kPa 1-111 The air pressure in a duct is measured by an inclined manometer. For a given vertical level difference, the gage pressure in the duct and the length of the differential fluid column are to be determined. Assumptions The manometer fluid is an incompressible substance. Properties The density of the liquid is given to be ρ = 0.81 kg/L = 810 kg/m3. Analysis The gage pressure in the duct is determined from Pa 636=                 ⋅ = =−= 22 23 atmabsgage N/m 1 Pa 1 m/skg 1 N 1 m) )(0.08m/s )(9.81kg/m (810 ghPPP ρ Fresh Water L 8 cm 35° The length of the differential fluid column is cm 13.9=°== 35sin/)cm 8(sin/ θhL Discussion Note that the length of the differential fluid column is extended considerably by inclining the manometer arm for better readability. 1-112E Equal volumes of water and oil are poured into a U-tube from different arms, and the oil side is pressurized until the contact surface of the two fluids moves to the bottom and the liquid levels in both arms become the same. The excess pressure applied on the oil side is to be determined. Assumptions 1 Both water and oil are incompressible substances. 2 Oil does not mix with water. 3 The cross-sectional area of the U-tube is constant. Oil Water Blown air 30 in Properties The density of oil is given to be ρoil = 49.3 lbm/ft3. We take the density of water to be ρw = 62.4 lbm/ft3. Analysis Noting that the pressure of both the water and the oil is the same at the contact surface, the pressure at this surface can be expressed as wwatmaablowcontact ghPghPP ρρ +=+= Noting that ha = hw and rearranging, psi 0.227=                 ⋅ = −=−= 2 2 2 23 atmblowblowgage, in 144 ft 1 ft/slbm 32.2 lbf 1 ft) )(30/12ft/s 2.32()lbm/ft 49.3-(62.4 )( ghPPP oilw ρρ Discussion When the person stops blowing, the oil will rise and some water will flow into the right arm. It can be shown that when the curvature effects of the tube are disregarded, the differential height of water will be 23.7 in to balance 30-in of oil. 1-48 1-115 A gasoline line is connected to a pressure gage through a double-U manometer. For a given reading of the pressure gage, the gage pressure of the gasoline line is to be determined. Pgage = 180 kPa Air 50 cm Oil 10 cm Water Mercury 22 cm Gasoline 45 cm Assumptions 1 All the liquids are incompressible. 2 The effect of air column on pressure is negligible. Properties The specific gravities of oil, mercury, and gasoline are given to be 0.79, 13.6, and 0.70, respectively. We take the density of water to be ρw = 1000 kg/m3. Analysis Starting with the pressure indicated by the pressure gage and moving along the tube by adding (as we go down) or subtracting (as we go up) the ghρ terms until we reach the gasoline pipe, and setting the result equal to Pgasoline gives gasolinegasolinegasolineHgHgoiloilw PghghghghP wgage =−−+− ρρρρ Rearranging, )SGSGSG( gasolinegasolineHgHgoiloilwgagegasoline hhhhgPP w ++−−= ρ Substituting, kPa 164.6=               ⋅ × ++−= 22 23 gasoline kN/m 1 kPa 1 m/skg 1000 kN 1 m)] 22.0(70.0m) 1.0(6.13m) 5.0(79.0m) )[(0.45m/s (9.807)kg/m (1000- kPa 180P Therefore, the pressure in the gasoline pipe is 15.4 kPa lower than the pressure reading of the pressure gage. Discussion Note that sometimes the use of specific gravity offers great convenience in the solution of problems that involve several fluids. 1-49 1-116E A water pipe is connected to a double-U manometer whose free arm is open to the atmosphere. The absolute pressure at the center of the pipe is to be determined. Oil Assumptions 1 All the liquids are incompressible. 2 The solubility of the liquids in each other is negligible. Properties The specific gravities of mercury and oil are given to be 13.6 and 0.80, respectively. We take the density of water to be ρw = 62.4 lbm/ft3. Analysis Starting with the pressure at the center of the water pipe, and moving along the tube by adding (as we go down) or subtracting (as we go up) the ghρ terms until we reach the free surface of oil where the oil tube is exposed to the atmosphere, and setting the result equal to Patm gives atmPghghghghP =−−+− oiloilHgHgoiloilwaterwaterpipewater ρρρρ Water 60 in 15 in 40 in 35 in Oil Mercury Solving for Pwater pipe, )( oiloilHgHgoiloilwaterwaterpipewater hSGhSGhSGhgPP atm ++−+= ρ Substituting, psia 22.3=                 ⋅ ×+ +−+= 2 2 2 23 pipewater in 144 ft 1 ft/slbm 32.2 lbf 1 ft)] (40/128.0 ft) (15/126.13ft) (60/128.0ft) )[(35/12ft/s 2.32()lbm/ft(62.4psia14.2P Therefore, the absolute pressure in the water pipe is 22.3 psia. Discussion Note that jumping horizontally from one tube to the next and realizing that pressure remains the same in the same fluid simplifies the analysis greatly. 1-50 1-117 The temperature of the atmosphere varies with altitude z as T zT β−= 0 , while the gravitational acceleration varies by . Relations for the variation of pressure in atmosphere are to be obtained (a) by ignoring and (b) by considering the variation of g with altitude. 2 0 )320,370,6/1/()( zgzg += Assumptions The air in the troposphere behaves as an ideal gas. Analysis (a) Pressure change across a differential fluid layer of thickness dz in the vertical z direction is gdzdP ρ−= From the ideal gas relation, the air density can be expressed as )( 0 zTR P RT P β ρ − == . Then, gdz zTR PdP )( 0 β− −= Separating variables and integrating from z = 0 where 0PP = to z = z where P = P, )( 000 zTR gdz P dP zP P β− −= ∫∫ Performing the integrations. 0 0 0 lnln T zT R g P P β β − = Rearranging, the desired relation for atmospheric pressure for the case of constant g becomes R g T zPP ββ       −= 0 0 1 (b) When the variation of g with altitude is considered, the procedure remains the same but the expressions become more complicated, dz z g zTR PdP 2 0 0 )320,370,6/1()( +− −= β Separating variables and integrating from z = 0 where 0PP = to z = z where P = P, 2 0 0 0 )320,370,6/1)((0 zzTR dzg P dP zP P +− −= ∫∫ β Performing the integrations, z P P zT kz kTkzkTR g P 00 2 00 0 1ln )/1( 1 )1)(/1( 1ln 0 ββββ − + + − ++ = where R = 287 J/kg⋅K = 287 m2/s2⋅K is the gas constant of air. After some manipulations, we obtain               − + + + ++ −= 000 0 0 /1 1ln /1 1 /11 1 )( exp Tz kz kTkzkTR g PP βββ where T0 = 288.15 K, β = 0.0065 K/m, g0 = 9.807 m/s2, k = 1/6,370,320 m-1, and z is the elevation in m.. Discussion When performing the integration in part (b), the following expression from integral tables is used, together with a transformation of variable zTx β−= 0 , x bxa abxaabxax dx + − + = +∫ ln 1 )( 1 )( 22 Also, for z = 11,000 m, for example, the relations in (a) and (b) give 22.62 and 22.69 kPa, respectively. 1-53 Fundamentals of Engineering (FE) Exam Problems 1-120 Consider a fish swimming 5 m below the free surface of water. The increase in the pressure exerted on the fish when it dives to a depth of 45 m below the free surface is (a) 392 Pa (b) 9800 Pa (c) 50,000 Pa (d) 392,000 Pa (e) 441,000 Pa Answer (d) 392,000 Pa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1000 "kg/m3" g=9.81 "m/s2" z1=5 "m" z2=45 "m" DELTAP=rho*g*(z2-z1) "Pa" "Some Wrong Solutions with Common Mistakes:" W1_P=rho*g*(z2-z1)/1000 "dividing by 1000" W2_P=rho*g*(z1+z2) "adding depts instead of subtracting" W3_P=rho*(z1+z2) "not using g" W4_P=rho*g*(0+z2) "ignoring z1" 1-121 The atmospheric pressures at the top and the bottom of a building are read by a barometer to be 96.0 and 98.0 kPa. If the density of air is 1.0 kg/m3, the height of the building is (a) 17 m (b) 20 m (c) 170 m (d) 204 m (e) 252 m Answer (d) 204 m Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1.0 "kg/m3" g=9.81 "m/s2" P1=96 "kPa" P2=98 "kPa" DELTAP=P2-P1 "kPa" DELTAP=rho*g*h/1000 "kPa" "Some Wrong Solutions with Common Mistakes:" DELTAP=rho*W1_h/1000 "not using g" DELTAP=g*W2_h/1000 "not using rho" P2=rho*g*W3_h/1000 "ignoring P1" P1=rho*g*W4_h/1000 "ignoring P2" 1-54 1-122 An apple loses 4.5 kJ of heat as it cools per °C drop in its temperature. The amount of heat loss from the apple per °F drop in its temperature is (a) 1.25 kJ (b) 2.50 kJ (c) 5.0 kJ (d) 8.1 kJ (e) 4.1 kJ Answer (b) 2.50 kJ Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). Q_perC=4.5 "kJ" Q_perF=Q_perC/1.8 "kJ" "Some Wrong Solutions with Common Mistakes:" W1_Q=Q_perC*1.8 "multiplying instead of dividing" W2_Q=Q_perC "setting them equal to each other" 1-123 Consider a 2-m deep swimming pool. The pressure difference between the top and bottom of the pool is (a) 12.0 kPa (b) 19.6 kPa (c) 38.1 kPa (d) 50.8 kPa (e) 200 kPa Answer (b) 19.6 kPa Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). rho=1000 "kg/m^3" g=9.81 "m/s2" z1=0 "m" z2=2 "m" DELTAP=rho*g*(z2-z1)/1000 "kPa" "Some Wrong Solutions with Common Mistakes:" W1_P=rho*(z1+z2)/1000 "not using g" W2_P=rho*g*(z2-z1)/2000 "taking half of z" W3_P=rho*g*(z2-z1) "not dividing by 1000" 1-55 1-124 At sea level, the weight of 1 kg mass in SI units is 9.81 N. The weight of 1 lbm mass in English units is (a) 1 lbf (b) 9.81 lbf (c) 32.2 lbf (d) 0.1 lbf (e) 0.031 lbf Answer (a) 1 lbf Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). m=1 "lbm" g=32.2 "ft/s2" W=m*g/32.2 "lbf" "Some Wrong Solutions with Common Mistakes:" gSI=9.81 "m/s2" W1_W= m*gSI "Using wrong conversion" W2_W= m*g "Using wrong conversion" W3_W= m/gSI "Using wrong conversion" W4_W= m/g "Using wrong conversion" 1-125 During a heating process, the temperature of an object rises by 20°C. This temperature rise is equivalent to a temperature rise of (a) 20°F (b) 52°F (c) 36 K (d) 36 R (e) 293 K Answer (d) 36 R Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen. (Similar problems and their solutions can be obtained easily by modifying numerical values). T_inC=20 "C" T_inR=T_inC*1.8 "R" "Some Wrong Solutions with Common Mistakes:" W1_TinF=T_inC "F, setting C and F equal to each other" W2_TinF=T_inC*1.8+32 "F, converting to F " W3_TinK=1.8*T_inC "K, wrong conversion from C to K" W4_TinK=T_inC+273 "K, converting to K" 1-126 … 1-129 Design, Essay, and Experiment Problems