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# Thermodynamics An Engireeng Approach 8Th Edition By SI Units – Test Bank

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Thermodynamics An Engireeng Approach 8Th Edition By SI Units – Test Bank

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##### Multiple-Choice Test Problems

Chapter 2:  Energy, Energy Transfer, and General Energy Analysis

Çengel/Boles – Thermodynamics: An Engineering Approach, 8th Edition

(Numerical values for solutions can be obtained by copying the EES solutions given and pasting them on a blank EES screen, and pressing the Solve command. Similar problems and their solutions can be obtained easily by modifying numerical values.)

Chap2-1 Heating by Resistance Heater

A 1.5-kW electric resistance heater in a room is turned on and kept on for 20 min. The amount of energy transferred to the room by the heater is

(a) 1.5 kJ                (b) 60 kJ                 (c) 750 kJ               (d) 1800 kJ             (e) 3600 kJ

Answer  (d) 1800 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

We= 1.5 “kJ/s”

time=20*60 “s”

E_total=We*time “kJ”

“Some Wrong Solutions with Common Mistakes:”

W1_Etotal=We*time/60 “using minutes instead of s”

W2_Etotal=We “ignoring time”

Chap2-2 Heat Supplied by Vacuum Cleaner

A 200 W vacuum cleaner is powered by an electric motor whose efficiency is 70%. (Note that the electric motor delivers 200 W of net mechanical power to the fan of the cleaner). The rate at which this vacuum cleaner supplies energy to the room when running is

(a) 140 W                (b) 200 W               (c)  286 W               (d) 360 W               (e) 86 W

Answer  (c)  286 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

Eff=0.70

W_vac=0.2 “kW”

E=W_vac/Eff  “kJ/s”

“Some Wrong Solutions with Common Mistakes:”

W1_E=W_vac*Eff “Multiplying by efficiency”

W2_E=W_vac “Ignoring efficiency”

W3_E=E-W_vac “Heat generated by the motor”

Chap2-3 Heat Convection

A 40-cm-long, 0.6-cm-diameter electric resistance wire is used to determine the convection heat transfer coefficient in air at 25°C experimentally. The surface temperature of the wire is measured to be 150°C when the electric power consumption is 90 W. If the radiation heat loss from the wire is calculated to be 30 W, the convection heat transfer coefficient is

(a)  0.48 W/m2.°C        (b) 127 W/m2.°C      (c) 63.7 W/m2.°C       (d) 95 W/m2.°C           (e) 200 W/m2.°C ”

Answer  (c) 63.7 W/m2.°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

L=0.4 “m”

D=0.006 “m”

A=pi*D*L “m^2”

We=90 “W”

Ts=150 “C”

Tf=25 “C”

We-30= h*A*(Ts-Tf) “W”

“Some Wrong Solutions with Common Mistakes:”

We-30= W1_h*(Ts-Tf) “Not using area”

We-30= W2_h*(L*D)*(Ts-Tf) “Using D*L for area”

We+30= W3_h*A*(Ts-Tf) “Adding Q_rad instead of subtracting”

We= W4_h*A*(Ts-Tf) “Disregarding Q_rad”

Chap2-4 Heat Convection and Radiation

A 1.5-m2 black surface at 120°C is losing heat to the surrounding air at 30°C by convection with a convection heat transfer coefficient of 18 W/m2.°C, and by radiation to the surrounding surfaces at 10°C. The total rate of heat loss from the surface is

(a) 1483 W              (b) 2430 W              (c) 2448 W              (d) 3913 W              (e) 2609 W

Answer  (d) 3913 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

sigma=5.67E-8 “W/m^2.K^4”

eps=1

A=1.5 “m^2”

h_conv=18 “W/m^2.C”

Ts=120 “C”

Tf=30 “C”

Tsurr=10 “C”

Q_conv=h_conv*A*(Ts-Tf)  “W”

Q_rad=eps*sigma*A*((Ts+273)^4-(Tsurr+273)^4) “W”

Q_total=Q_conv+Q_rad “W”

“Some Wrong Solutions with Common Mistakes:”

W1_Ql=Q_conv “Ignoring radiation”

W2_Q=Q_rad “ignoring convection”

W3_Q=Q_conv+eps*sigma*A*(Ts^4-Tsurr^4) “Using C in radiation calculations”

W4_Q=Q_total/A “not using area”

Chap2-5 Heat Conduction

Heat is transferred steadily through a 0.15-m thick 3 m by 5 m wall whose thermal conductivity is 1.2 W/m.°C. The inner and outer surface temperatures of the wall are measured to be 18°C to 4°C. The rate of heat conduction through the wall is

(a) 112 W        (b) 3360 W      (c)  2640 W              (d) 38 W          (e) 1680 W

Answer  (e) 1680 W

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

A=3*5 “m^2”

L=0.15 “m”

T1=18 “C”

T2=4 “C”

k=1.2  “W/m.C”

Q=k*A*(T1-T2)/L “W”

“Some Wrong Solutions with Common Mistakes:”

W1_Q=k*(T1-T2)/L “Not using area”

W2_Q=k*2*A*(T1-T2)/L  “Using areas of both surfaces”

W3_Q=k*A*(T1+T2)/L  “Adding temperatures instead of subtracting”

W4_Q=k*A*L*(T1-T2) “Multiplying by thickness instead of dividing by it”

##### Multiple-Choice Test Problems

Chapter 4:  Energy Analysis of Closed Systems

Çengel/Boles – Thermodynamics: An Engineering Approach, 8th Edition

(Numerical values for solutions can be obtained by copying the EES solutions given and pasting them on a blank EES screen, and pressing the Solve command. Similar problems and their solutions can be obtained easily by modifying numerical values.)

Chap4-1 Ideal Gas on Cp-Cv

A frictionless piston-cylinder device and a rigid tank contain 5 kg of helium gas at the same temperature, pressure and volume. Now heat is transferred, and the temperature of both systems is dropped by 12°C. The amount of extra heat that must be supplied to helium gas in the cylinder that is maintained at constant pressure is

(a) 125 kJ               (b) 0 kJ                  (c) 312 kJ               (d) 187 kJ               (e) 499 kJ ”

Answer  (a) 125 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

“Note that Cp-Cv=R, and thus Q_diff=m*R*dT=N*Ru*dT”

m=5 “kg”

R=2.0769 “kJ/kg.K”

Cp=5.1926 “kJ/kg.K”

Cv=3.1156 “kJ/kg.K”

Ru=8.314 “kJ/kg.K”

T_change=12

Q=m*R*T_change

“Some Wrong Solutions with Common Mistakes:”

W1_Q=0 “Assuming they are the same”

W2_Q=m*Cp*T_change  “using Cp of He”

W3_Q=m*Cv*T_change “using Cv of He”

W4_Q=m*Ru*T_change “using Ru instead of R”

Chap4-2 Boundary Work for V=constant

A 1.2-m3 rigid tank contains nitrogen gas at 200 kPa and 350 K. Now heat is transferred to the nitrogen in the tank and the pressure of nitrogen rises to 900 kPa. The work done during this process is

(a) 840 kJ               (b)  0 kJ                 (c) 156 kJ               (d) 240 kJ                (e) 1080 kJ

Answer  (b)  0 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

V=1.2 “m^3”

P1=200 “kPa”

T1=350 “K”

P2=900 “kPa”

W=0 “since constant volume”

“Some Wrong Solutions with Common Mistakes:”

R=0.297

W1_W=V*(P2-P1) “Using W=V*DELTAP”

W2_W=V*P1

W3_W=V*P2

W4_W=R*T1*ln(P2/P1)

Chap4-3 Isothermal Compression of Air

An 8-L cylinder contains air at 300 kPa and 300 K. Now air is compressed isothermally to a volume of 2 L. The work done on air during this compression process is

(a) 119 kJ               (b) 1.8 kJ                (c) 3.3 kJ                (d) 4.5 kJ                 (e) 0 kJ

Answer  (c) 3.3 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

R=0.287 “kJ/kg.K”

V1=0.008 “m^3”

V2=0.002 “m^3”

P1=300 “kPa”

T1=300 “K”

T2=T1

P1*V1=m*R*T1

W=m*R*T1* ln(V2/V1) “constant temperature”

“Some Wrong Solutions with Common Mistakes:”

W1_W=R*T1* ln(V2/V1)   “Forgetting m”

W2_W=P1*(V1-V2) “Using V*DeltaP”

P1*V1/T1=P2*V2/T2

W3_W=(V1-V2)*(P1+P2)/2 “Using P_ave*Delta V”

W4_W=P1*V1-P2*V2 “Using W=P1V1-P2V2”

Chap4-4 P=constant Compression of Steam

A cylinder contains 2 kg of steam at 200 kPa and 200°C. Now steam is compressed at constant pressure until its temperature rises to 250°C. The work done during this compression process is

(a) 24 kJ                 (b)  0 kJ                 (c) 92 kJ                 (d) 138 kJ                (e) 47 kJ

Answer   (e) 47 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

m=2 “kg”

P1=200 “kPa”

P2=P1

T1=200 “C”

T2=250 “C”

v1=VOLUME(Steam_IAPWS,T=T1,P=P1)

v2=VOLUME(Steam_IAPWS,T=T2,P=P2)

W=m*P1*(v2-v1) “kJ, constant pressure”

“Some Wrong Solutions with Common Mistakes:”

W1_W=P1*(v2-v1)   “Forgetting m”

R=0.4615 “kJ/kg.K”

P1*Vol1=m*R*(T1+273)

P2*Vol2=m*R*(T2+273)

W2_W=m*P1*(Vol2-Vol1) “Assuming ideal gas”

P1*Vol11=m*R*T1

P2*Vol22=m*R*T2

W3_W=m*P1*(Vol22-Vol11) “Assuming ideal gas with deg. C”

W4_W=0 “Assuming work to be zero”

Chap4-5 P=constant Expansion of R134

A piston-cylinder device contains 3.2 kg of R-134a at 400 kPa and 50°C. Now R-134a is allowed to expand at constant pressure until its temperature drops to 10°C. The work done during this expansion process is

(a) 13 kJ                 (b)  0 kJ                 (c) 4.2 kJ                (d) 33 kJ                  (e) 58 kJ

Answer  (a) 13 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

m=3.2 “kg”

P1=400 “kPa”

P2=P1

T1=50 “C”

T2=10 “C”

v1=VOLUME(R134a,T=T1,P=P1)

v2=VOLUME(R134a,T=T2,P=P2)

W=m*P1*(v2-v1) “kJ, constant pressure”

“Some Wrong Solutions with Common Mistakes:”

W1_W=P1*(v2-v1)   “Forgetting m”

R=0.08149 “kJ/kg.K”

P1*Vol1=m*R*(T1+273)

P2*Vol2=m*R*(T2+273)

W2_W=m*P1*(Vol2-Vol1) “Assuming ideal gas”

P1*Vol11=m*R*T1

P2*Vol22=m*R*T2

W3_W=m*P1*(Vol22-Vol11) “Assuming ideal gas with deg. C”

W4_W=0 “Assuming work to be zero”

Chap4-6 Polytropic Compression of Air

Air in a piston-cylinder device is compressed from 25°C and 100 kPa to 500 kPa during a polytropic process for which Pv1.3 = constant. The air temperature after compression is

(a) 33°C                  (b)  156°C               (c) 1207°C              (d) 115°C                 (e) 182°C

Answer  (b)  156°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

R=0.287 “kJ/kg.K”

P1=100 “kPa”

T1=25+271 “K”

P2=500 “kPa”

“PV^1.3 = constant gives V1/V2=(P2/P1)^(1/1.3).

Substituting into P1V1/T1=P2V2/T2 gives (P1/T1)*(P2/P1)^(1/1.3)=P2/T2 ”

(P1/T1)*(P2/P1)^(1/1.3)=P2/T2

T2_C=T2-273 “C”

“Some Wrong Solutions with Common Mistakes:”

(P1/(T1-273))*(P2/P1)^(1/1.3)=P2/W1_T2   “Usinf C instead of K”

P1/T1=P2/(W2_T2+273) “Assuming V=constant and using K”

P1/(T1-273)=P2/(W3_T2) “Assuming V=constant, and using C”

(P1/T1)*(P2/P1)^(1.3)=P2/W4_T2 “Using the exponent 1.3 instead of 1/1.3”

Chap4-7 Electric Heating of a Room with a Fan

A well-sealed room contains 90 kg of air at 1 atm and 20°C. Now a 1-kW electric heater and a 200-W fan are turned on. If heat transfer through the walls is negligible, the air temperature in the room in 15 min will be

(a) 36.7°C               (b)  31.9°C              (c) 33.9°C               (d) 20.3°C                (e) 49.5°C

Answer  (a) 36.7°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

R=0.287 “kJ/kg.K”

Cv=0.718 “kJ/kg.K”

m=90 “kg”

P1=1 “atm”

T1=20 “C”

We=1 “kJ/s”

time=15*60 “s”

Wfan=0.2 “kJ/s”

“Applying energy balance E_in-E_out=dE_system gives”

time*(Wfan+We)=m*Cv*(T2-T1)

“Some Wrong Solutions with Common Mistakes:”

Cp=1.005 “kJ/kg.K”

time*(Wfan+We)=m*Cp*(W1_T2-T1)  “Using Cp instead of Cv ”

time*(-Wfan+We)=m*Cv*(W2_T2-T1) “Subtracting Wfan instead of adding”

time*We=m*Cv*(W3_T2-T1) “Ignoring Wfan”

time*(Wfan+We)/60=m*Cv*(W4_T2-T1) “Using min for time instead of s”

Chap4-8 Electric Heating of a Room

An air-tight room contains 80 kg of air, and a 2-kW baseboard electric resistance heater in the room is turned on and kept on for 15 min. The  temperature rise of air at the end of 15 min is

(a) 22.4°C               (b) 31.3°C               (c) 0.52°C               (d) 13.4°C               (e) 42.8°C

Answer  (b) 31.3°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

R=0.287 “kJ/kg.K”

Cv=0.718 “kJ/kg.K”

m=80 “kg”

time=15*60 “s”

W_e=2 “kJ/s”

“Applying energy balance E_in-E_out=dE_system gives”

time*W_e=m*Cv*DELTAT “kJ”

“Some Wrong Solutions with Common Mistakes:”

Cp=1.005 “kJ/kg.K”

time*W_e=m*Cp*W1_DELTAT “Using Cp instead of Cv”

time*W_e/60=m*Cv*W2_DELTAT “Using min for time instead of s”

Chap4-9 A Room with Electric Heater and a Refrigerator

A room contains 25 kg of air at 100 kPa and 10°C. The room has a 250-W refrigerator (the refrigerator consumes 250 W of electricity when running) and 1-kW electric resistance heater. During a cold winter day, it is observed that both the refrigerator and the resistance heater are running continuously but the air temperature in the room remains constant. The rate of heat loss from the room during that day is

(a) 3600 kJ/h           (b)  2700 kJ/h         (c) 4500 kJ/h           (d) 1000 kJ/h           (e) 1250 kJ/h

Answer  (c) 4500 kJ/h

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

R=0.287 “kJ/kg.K”

Cv=0.718 “kJ/kg.K”

m=25 “kg”

P1=100 “kPa”

T1=10 “C”

time=30*60 “s”

W_ref=0.250 “kJ/s”

W_heater=1 “kJ/s”

“Applying energy balance E_in-E_out=dE_system gives E_out=E_in since T=constant and dE=0”

E_gain=W_ref+W_heater

Q_loss=E_gain*3600 “kJ/h”

“Some Wrong Solutions with Common Mistakes:”

E_gain1=-W_ref+W_heater “Subtracting Wrefrig instead of adding”

W1_Qloss=E_gain1*3600 “kJ/h”

E_gain2=W_heater “Ignoring the Refrigerator”

W2_Qloss=E_gain2*3600 “kJ/h”

Chap4-10 Air in a Cylinder w/Paddle-Wheel

A piston-cylinder device contains 3 kg of air at 300 kPa and 25°C. During a quasi-equilibrium isothermal expansion process, 20 kJ of boundary work is done by the system, and 8 kJ of paddle-wheel work is done on the system. The heat transfer during this process is

(a) 28 kJ                 (b)  20 kJ                (c) 36 kJ                 (d) 12 kJ                  (e) 4 kJ

Answer  (d) 12 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

R=0.287 “kJ/kg.K”

Cv=0.718 “kJ/kg.K”

m=3 “kg”

P1=300 “kPa”

T=25 “C”

Wout_b=20 “kJ”

Win_pw=8 “kJ”

“Noting that T=constant and thus dE_system=0, applying energy balance E_in-E_out=dE_system gives”

Q_in+Win_pw-Wout_b=0

“Some Wrong Solutions with Common Mistakes:”

W1_Qin=m*Q_in “Multiplying by mass”

W2_Qin=Win_pw+Wout_b “Adding both quantities”

W3_Qin=Q_in/m “Dividing by mass”

W4_Qin=Wout_b “Setting it equal to boundary work”

Chap4-11 Heating of Steam in a Cylinder

A piston-cylibder device equipped with a resistance heater is initially filled with 0.4 kg of saturated water vapor at 200°C. Now the heater is turned on, the steam is compressed, and there is heat loss to the surrounding air. At the end of the process, the temperature and pressure of steam in the container are measured to be 300°C and 0.5 MPa. The net energy transfer to the steam during this process is

(a) 208 kJ               (b) 56 kJ                 (c) 75 kJ                 (d) 109 kJ               (e) 83 kJ

Answer  (e) 83 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

m=0.4 “kg”

T1=200 “C”

x1=1 “saturated vapor”

P2=500 “kPa”

T2=300 “C”

u1=INTENERGY(Steam_IAPWS,T=T1,x=x1)

u2=INTENERGY(Steam_IAPWS,T=T2,P=P2)

“Noting that Eout=0 and dU_system=m*(u2-u1), applying energy balance E_in-E_out=dE_system gives”

E_out=0

E_in=m*(u2-u1)

“Some Wrong Solutions with Common Mistakes:”

Cp_steam=1.8723 “kJ/kg.K”

Cv_steam=1.4108 “kJ/kg.K”

W1_Ein=m*Cp_Steam*(T2-T1) “Assuming ideal gas and using Cp”

W2_Ein=m*Cv_steam*(T2-T1) “Assuming ideal gas and using Cv”

W3_Ein=u2-u1 “Not using mass”

h1=ENTHALPY(Steam_IAPWS,T=T1,x=x1)

h2=ENTHALPY(Steam_IAPWS,T=T2,P=P2)

W4_Ein=m*(h2-h1)  “Using enthalpy”

Chap4-12 Heating of Steam in a Cylinder

A bucket of water with a mass of 20 kg at 18°C is to be cooled to 0°C by dropping ice cubes at 0°C into it. The latent heat of fusion of ice is 334 kJ/kg, and the specific heat of water is 4.18 kJ/kg.°C. The amount of ice that needs to be added is

(a) 4.5 kg                (b) 0.25 kg              (c) 10 kg                 (d) 20 kg                (e) 8.6 kg

Answer  (a) 4.5 kg

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

C=4.18 “kJ/kg.K”

h_melting=334 “kJ/kg.K”

m_w=20 “kg”

T1=18 “C”

T2=0 “C”

DELTAT=T2-T1 “C”

“Noting that there is no energy transfer with the surroundings and the latent heat of melting of ice is

transferred form the water, and applying energy balance E_in-E_out=dE_system to ice+water gives”

dE_ice+dE_w=0

dE_ice=m_ice*h_melting

dE_w=m_w*C*DELTAT “kJ”

“Some Wrong Solutions with Common Mistakes:”

W1_mice*h_melting*(T1-T2)+m_w*C*DELTAT=0  “Multiplying h_latent by temperature difference”

W2_mice=m_w “taking mass of water to be equal to the mass of ice”

Chap4-13 Electric Heating of Water w/Heat Loss

A 1-kW electric resistance heater submerged in 10-kg water is turned on and kept on for 15 min. During the process, 400 kJ of heat is lost from the water. The temperature rise of water is

(a) 21.5°C               (b) 12.0°C               (c) 31.1°C               (d) 50.0°C               (e) 8.8°C ”

Answer  (b) 12.0°C

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

C=4.18 “kJ/kg.K”

m=10 “kg”

Q_loss=400 “kJ”

time=15*60 “s”

W_e=1 “kJ/s”

“Applying energy balance E_in-E_out=dE_system gives”

time*W_e-Q_loss = dU_system

dU_system=m*C*DELTAT “kJ”

“Some Wrong Solutions with Common Mistakes:”

time*W_e = m*C*W1_T “Ignoring heat loss”

time*W_e+Q_loss = m*C*W2_T  “Adding heat loss instead of subtracting”

time*W_e-Q_loss = m*1.0*W3_T “Using specific heat of air or not using specific heat”

Chap4-14 Electric Heating  of Water in a Teapot

1.5 kg of liquid water initially at 10°C is heated to 60°C in a teapot equipped with a 1000 W electric heating element inside.  The minimum time it takes to heat the water to the desired temperature is

(a) 3.5 min              (b) 7.3 min              (c) 5.2 min              (d) 16.0 min            (e) 11.4 min

Answer  (c) 5.2 min

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

C=4.18 “kJ/kg.K”

m=1.5 “kg”

T1=10 “C”

T2=60 “C”

Q_loss=0 “kJ”

W_e=1 “kJ/s”

“Applying energy balance E_in-E_out=dE_system gives”

(time*60)*W_e-Q_loss = dU_system  “time in minutes”

dU_system=m*C*(T2-T1) “kJ”

“Some Wrong Solutions with Common Mistakes:”

W1_time*60*W_e-Q_loss = m*C*(T2+T1) “Adding temperatures instead of subtracting”

W2_time*60*W_e-Q_loss = C*(T2-T1) “Not using mass”

Chap4-15 Cooling Rate of Eggs

Eggs with a mass of 0.12 kg per egg and a specific heat of 3.32 kJ/kg.°C are cooled from 30°C to 10°C at a rate of 800 eggs per minute. The rate of heat removal from the eggs is

(a) 8.0 kW               (b) 53 kW               (c) 478 kW              (d) 106 kW              (e) 159 kW

Answer  (d) 106 kW

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

C=3.32 “kJ/kg.K”

m_egg=0.12 “kg”

T1=30 “C”

T2=10 “C”

n=800 “eggs/min”

m=n*m_egg/60 “kg/s”

“Applying energy balance E_in-E_out=dE_system gives”

“-E_out = dU_system”

Qout=m*C*(T1-T2) “kJ/s”

“Some Wrong Solutions with Common Mistakes:”

W1_Qout = m*C*T1  “Using T1 only”

W2_Qout = m_egg*C*(T1-T2)  “Using one egg only”

W3_Qout = m*C*T2  “Using T2 only”

W4_Qout=m_egg*C*(T1-T2)*60   “Finding kJ/min”

Chap4-16 Cooling of an Orange

An orange with an average mass of 0.20 kg and average specific heat of 3.70 kJ/kg.C is cooled from 20°C to 5°C. The amount of heat transferred from the orange is

(a) 7.4 kJ                (b) 24.8 kJ              (c) 18.5 kJ              (d) 55.5 kJ              (e) 11.1 kJ

Answer  (e) 11.1 kJ

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

C=3.70 “kJ/kg.K”

m=0.2 “kg”

T1=20 “C”

T2=5 “C”

“Applying energy balance E_in-E_out=dE_system gives”

-Q_out = dU_system

dU_system=m*C*(T2-T1) “kJ”

“Some Wrong Solutions with Common Mistakes:”

-W1_Qout =C*(T2-T1) “Not using mass”

-W2_Qout =-m*C*(T2+T1) “adding temperatures”

Chap4-17 Quenching Steel Balls

Steel balls at 150°C with a specific heat of 0.50 kJ/kg.°C are quenched in an oil bath to an average temperature of 70°C at a rate of 15 balls per minute. If the average mass of steel balls is 1.2 kg, the rate of heat transfer from the balls to oil is

(a) 22.5 kJ/s           (b) 12 kJ/s              (c) 48 kJ/s              (d) 10.5 kJ/s           (e) 2880 kJ/s

Answer  (b) 12 kJ/s

Solution Solved by EES Software. Solutions can be verified by copying-and-pasting the following lines on a blank EES screen.

C=0.50 “kJ/kg.K”

m1=1.2 “kg”

T1=150 “C”

T2=70 “C”

n=15 “balls/min”

m=n*m1/60 “kg/s”

“Applying energy balance E_in-E_out=dE_system gives”

“-E_out = dU_system”

Qout=m*C*(T1-T2) “kJ/s”

“Some Wrong Solutions with Common Mistakes:”

W1_Qout = m*C*T1  “Using T1 only”

W2_Qout = m1*C*(T1-T2)  “Using one ball only”

W3_Qout = m*C*T2  “Using T2 only”

W4_Qout=m1*C*(T1-T2)*60   “Finding kJ/min”

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