Multiple-Choice Test Problems
Chapter
2: Properties
of Pure Substances
Cengel/Boles
- Thermodynamics: An Engineering Approach, 4th 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 IDEAL GAS Discharged (V=const)
A rigid tank
contains 7 kg of an ideal gas at 5 atm and 30°C. Now a valve is opened,
and half of mass of the gas is allowed to escape. If the final pressure in the
tank is 1.5 atm, the final temperature in the tank is
(a) -91°C (b) °18C (c) -182°C (d) 91°C (e) 15°C
Answer (a) -91°C
Solution Solved by EES Software.
Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.
"When R=const
and V= constat, P1/P2=m1*T1/m2*T2"
m1=7 "kg"
P1=5 "atm"
P2=1.5 "atm"
T1=30+273 "K"
m2=0.5*m1 "kg"
P1/P2=m1*T1/(m2*T2)
T2_C=T2-273 "C"
“Some Wrong
Solutions with Common Mistakes:”
P1/P2=m1*(T1-273)/(m2*W1_T2)
"Using C instead of K"
P1/P2=m1*T1/(m1*(W2_T2+273))
"Disregarding the decrease in mass"
P1/P2=m1*T1/(m1*W3_T2)
"Disregarding the decrease in mass, and not converting to deg.
C"
W4_T2=(T1-273)/2 "Taking
T2 to be half of T1 since half of the mass is discharged"
Chap2-2 Temp Rise of AIR in Tank (V=const)
The pressure in an air storage tank is
measured to be 250 kPa (gage) in the morning, and 275 kPa (gage) in the
afternoon as a result of solar heating. The local atmospheric pressure is 100
kPa. If the temperature of air in the tank is 15°C in the morning, the air
temperature in the afternoon is
(a) 15.9°C (b) 31.5°C (c) 38.0°C (d) 16.2°C (e) 15°C
Answer (b) 31.5°C
Solution Solved by EES Software.
Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.
"When R, V,
and m are constant, P1/P2=T1/T2"
Patm=100
P1=250+Patm "kPa"
P2=270+Patm "kPa"
T1=15+273 "K"
P1/P2=T1/T2
T2_C=T2-273 "C"
“Some Wrong
Solutions with Common Mistakes:”
P1/P2=(T1-273)/W1_T2
"Using C instead of K"
(P1-Patm)/(P2-Patm)=T1/(W2_T2+273)
"Using gage pressure instead of absolute pressure"
(P1-Patm)/(P2-Patm)=(T1-273)/W3_T2
"Making both of the mistakes above"
W4_T2=T1-273 "Assuming
the temperature to remain constant"
Chap2-3 Saturated H2O mixture in Tank
A 230-m3 rigid tank is
filled with saturated liquid-vapor mixture of water at 120°C. If 25% of the mass is
liquid and the 75% of the mass is vapor, the total mass in the tank is
(a) 252 kg (b)
824 kg (c) 344 kg (d) 230 kg (e)
683 kg
Answer (c) 344 kg
Solution Solved by EES Software.
Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.
V_tank=230 "m3"
T1=120 "C"
x=0.75
v_f=VOLUME(Steam_NBS,
x=0,T=T1)
v_g=VOLUME(Steam_NBS,
x=1,T=T1)
v=v_f+x*(v_g-v_f)
m=V_tank/v "kg"
“Some Wrong
Solutions with Common Mistakes:”
R=0.4615 "kJ/kg.K"
P1=PRESSURE(Steam_NBS,T=T1,x=0)
P1*V_tank=W1_m*R*(T1+273)
"Treating steam as ideal gas"
P1*V_tank=W2_m*R*T1
"Treating steam as ideal gas and using deg.C"
W3_m=V_tank "Taking
the density to be 1 kg/m^3"
Chap2-4 Pressure Cooker Heat Transfer Rate
Water is boiled at 300 kPa pressure in
a pressure cooker. The cooker initially contains 3 kg of water. Once boiling
started, it is observed that half of the water in the cooker evaporated in 30
minutes. If the heat loss from the cooker is negligible, the average rate of
energy transfer to the cooker is
(a) 2.3 kW (b) 136 kW (c) 3.6 kW (d) 1.80 kW
(e) 0.9 kW
Answer (d) 1.80 kW
Solution Solved by EES Software.
Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.
m_1=3 "kg"
P=300 "kPa"
time=30*60 "s"
m_evap=0.5*m_1
Power*time=m_evap*h_fg
"kJ"
h_f=ENTHALPY(Steam_NBS,
x=0,P=P)
h_g=ENTHALPY(Steam_NBS,
x=1,P=P)
h_fg=h_g-h_f
“Some Wrong
Solutions with Common Mistakes:”
W1_Power*time=m_evap*h_g
"Using h_g"
W2_Power*time/60=m_evap*h_g
"Using minutes instead of seconds for time"
W3_Power=2*Power "Assuming all the water
evaporates"
Chap2-5
Temperature in H2O Tank
A 0.2-m3 rigid tank
contains 8 kg of water (in any phase or phases) at 500 kPa. The temperature in
the tank is
(a) 27°C (b) 102°C (c) 381°C (d) 246°C (e) 152°C
Answer (e) 152°C
Solution Solved by EES Software.
Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.
V_tank=0.2 "m^3"
m=8 "kg"
v=V_tank/m
P=500 "kPa"
T=TEMPERATURE(Steam_NBS,v=v,P=P)
“Some Wrong
Solutions with Common Mistakes:”
R=0.4615 "kJ/kg.K"
P*V_tank=m*R*W1_T "Treating
steam as ideal gas and using deg.C"
Chap2-6
Boiling Water in a Pan
Water is boiling at a temperature of
95°C
in a pan. If heat is transferred to the pan at a rate of 2 kW, the amount of
water that evaporates in 30 minutes is
(a) 1. 6 kg (b) 1.4 kg (c)
9.0 kg (d) 0.022 kg (e) 0.87 kg
Answer (a) 1. 6 kg
Solution Solved by EES Software.
Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.
T=95 "C"
time=30*60 "s"
Q=2 "kJ/s"
Q*time=m*h_fg "kJ"
h_f=ENTHALPY(Steam_NBS,
x=0,T=T)
h_g=ENTHALPY(Steam_NBS,
x=1,T=T)
h_fg=h_g-h_f
“Some Wrong
Solutions with Common Mistakes:”
Q*time=W1_m*h_g "Using
h_g"
Q*time/60=W2_m*h_g
"Using minutes instead of seconds for time"
Q*time=W3_m*h_f "Using
h_f"
Chap2-7
Boiling Water in a Pan at Sea Level
Water is boiled in a pan on a stove at
sea level. During 15 min of boiling, it is observed that 150 g of water has evaporated.
Then the rate of heat transfer to the water is
(a) 26.8 kJ/min (b)
22.6 kJ/min (c) 0.45 kJ/min (d) 4.2
kJ/min (e) 16.9
kJ/min
Answer (b) 22.6 kJ/min
Solution Solved by EES Software.
Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.
m_evap=0.15 "kg"
P=101.325 "kPa"
time=15 "min"
Q*time=m_evap*h_fg
"kJ"
h_f=ENTHALPY(Steam_NBS,
x=0,P=P)
h_g=ENTHALPY(Steam_NBS,
x=1,P=P)
h_fg=h_g-h_f
“Some Wrong
Solutions with Common Mistakes:”
W1_Q*time=m_evap*h_g
"Using h_g"
W2_Q*time*60=m_evap*h_g
"Using s instead of min for time"
W3_Q*time=m_evap*h_f
"Using h_f"
Chap2-8 Steam
Condensing on Brass Ball
A 6-kg brass ball whose specific heat
is 0.40 kJ/kg.°C is brought in to a room filled with
saturated steam at 90°C. The temperature of the
ball rises from 20°C to 90°C while some steam
condenses on the ball as it loses heat (but it still remains at 90°C). The mass of steam that
condensed during this process is
(a) 63 g (b) 574 g (c)
74 g (d)
446 g (e) 217 g
Answer (c) 74 g
Solution Solved by EES Software.
Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.
m_ball=6 "kg"
T=90 "C"
T1=20 "C"
T2=90 "C"
Cp=0.4 "kJ/kg.C"
Q=m_ball*Cp*(T2-T1)
Q=m_steam*h_fg "kJ"
h_f=ENTHALPY(Steam_NBS,
x=0,T=T)
h_g=ENTHALPY(Steam_NBS,
x=1,T=T)
h_fg=h_g-h_f
“Some Wrong
Solutions with Common Mistakes:”
Q=W1m_steam*h_g "Using
h_g"
Q=W2m_steam*4.18*(T2-T1)
"Using m*C*DeltaT = Q for water"
Q=W3m_steam*h_f "Using
h_f"
Chap2-9 SH Steam
in a Tank
A rigid tank contains 0.8 kg of steam
at 10 MPa and 300°C. The volume of the tank is
(a) 11 m3 (b) 21 m3 (c) 4.3 m3 (d) 1.1 m3 (e) 0.23 m3
Answer (d) 1.1 m3
Solution Solved by EES Software.
Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.
m=0.8 "kg"
V=m*v1
"m^3"
P1=10000
"kPa"
T1=300
"C"
v1=VOLUME(Steam_NBS,T=T1,P=P1)
“Some Wrong
Solutions with Common Mistakes:”
R=0.4615
"kJ/kg.K"
P1*W1_V=m*R*(T1+273)
"Treating steam as ideal gas"
P1*W2_V=m*R*T1
"Treating steam as ideal gas and using deg.C"
Chap2-10
R134a Saturation Temperature
Consider a sealed can that is filled
with refrigerant-134a at 20°C. Now a leak develops, and
the pressure in the can drops to the local atmospheric pressure of 80 kPa. The
temperature of the refrigerant in the can is expected to drop to (rounded to
the nearest integer)
(a) 0°C (b) -31°C (c) -18°C (d)
3°C (e) 20°C
Answer (b)
-31°C
Solution Solved by EES Software.
Solutions can be verified by copying-and-pasting the following lines on a blank
EES screen.
P2=80 "kPa"
T2=TEMPERATURE(R134a,x=0,P=P2)
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