physics 111 spring 2006 exam # 1
1/8 PHYSICS 1308 SPRING 2008 EXAM # 1 Thursday, February 14, 2008
Solutions
CONCEPTUAL QUESTIONS 1. For which of the following adiabatic processes is the change in entropy bigger ? Explain why. P
a b
c
V Let us remember first the definition of entropy change : S=
Q T
. Since all processes are
adiabatic Q =0 . So we can conclude that the change in entropy for all of them is the same and equal to zero. 2. Sketch a PV diagram of the following process: 2.0 L of ideal gas at atmospheric pressure are cooled at constant pressure to a volume of 1.0 L, and then expanded isothermally back to 2.0 L, whereupon the pressure is increased at constant volume until the original pressure is reached. P 1 atm
2 L 1 L V 3. Imagine a test charge inside a conductor in electrostatic equilibrium. Now imagine that this this charge is moved very slowly. What is the work performed by the electric forces during this process? Explain.
2/8 The work done is zero. The electric field inside a conductor in electrostatic equilibrium is zero, thus the net electric force on the test charge is also zero. Hence no work is necessary to move the charge inside the conductor. 4. What can you say about the electric field in a region of space that has the same potential throughout? Explain. The electric field is given by E x =
−V x
, E y=
can conclude that the electric field is zero.
−V y
, E z=
− V z
. If V is constant throughout, we
5. When the voltage across a certain conductor is doubled, the current is observed to triple. What can you conclude about the conductor? Explain. Since the current is not proportional to the voltage the conductor is not ohmic.
NUMERICAL QUESTIONS An ideal gas expands at constant pressure of 5.0 atm from 400 mL to 660 mL. Heat then flows out the gas, at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. 6. The total work done in the process is ______atm·L A. 2.3
C. 1.3
B. 1.7
D. 1.0
E. 0.3
5 atm
0.4 L
0.660 L
V
The work done is the area below the curve. Thus W =5 atm0.660 L−0.400 L =1.3 atm⋅L
3/8
7. The net heat flow into the gas is ____J A. 230
C. 130
B. 170
D. 100
E. 30
Since the initial and final temperatures are the same we have U =0 which means by the first law of thermodynamics that Q=W . Therefore: Q=1.3 atm⋅L
1×105 Pa 1 atm
×
1 m3 10 3 L
=130 J
8. 5mol of an ideal monatomic gas are adiabatically expanded according to the diagram below. The net work done by the gas is _________J P A. 1800 B. 1248 C. 782 D. 672 E. 564
6 L, 3 atm
12 L, 0.94 atm
V By the first law of thermodynamics we have U =Q−W since the process is adiabatic the equation reduces to U =−W . So let us find the change in internal energy: P f V f Pi V i 3 3 3 U = n R T = n RT f −T i = n R − 2 2 2 nR nR
where we have used the equation of state for an ideal gas in the last equality. Thus the work performed by the gas is : W=
−3 −3 P f V f − P i T i = 12L⋅0.94 atm−6 L⋅3atm=6.72 L⋅atm=672 J 2 2
4/8 9. A Carnot engine extracts 890 J of work form a 550 K reservoir during each cycle and rejects 470 J to a cooler reservoir. The temperature of the cooler reservoir is _________K A. 190
B. 210
C.273
D. 330
E. 425
Since we are talking about a Carnot engine, the ideal efficiency is equal to the efficiency, so: 1−
TL TH
T H 550 K
W
=
QH
=
QL W QL
W QL W
=T L
470 J 890 J470 J
=190 K
10. A refrigerator has a coefficient of performance equal to 5.00. The refrigerator takes 120 J of energy from a cold reservoir in each cycle. The energy expelled to a hot reservoir is ______J B. 144
A. 24
COP=
C. 224
D. 600
E. 720
QL
W Q H =W QL QH =
1 COP
1Q L
1 Q H = 1120 J =144 J 5
11. A charge Q1 is fixed at the origin and another charge Q2 is located 2 m away. Charge Q2 is at rest, and the electrical potential energy of the system is 12 J. If charge Q2 moves away from the origin. Its kinetic energy when it is 4 m away from the origin is ____J A. 16
B. 12
C. 10
D. 8
E. 6
The total energy of the system is conserved and is equal to 12 J. This energy is also equal to the sum of the kinetic and potential energies. In particular the potential energy is equal to:
5/8 U =k
Q1 Q2 r
. At the initial configuration, all the energy is potential. When the second charge
moves away the energy is now distributed between kinetic and potential. Since the distance doubles we have U f = k
Q1 Q 2 rf
=k
Q1 Q 2 2r i
=
12J 2
thus the kinetic energy is 12J – 6 J= 6 J.
=6 J
The following diagram shows two charges. A positive charge +2 nC located at the origin and a – 5 nC located at y= 2 mm. 5 nC
y
E1
E2
x
2 nC
12. The electric field at point (0, 1) mm has magnitude ___MV/m and direction___°. A. 63, 90
B. 63, 90
C. 27,0
D. 27, 90
E. 27, 90
The total electric field is: E =k E=E 1 2
∣Q1∣
∣Q ∣ jk 2 j r 1 2 r 2 2
2 7×10−9 C 9 N⋅m E =9×10 j=63×106 N /C j 2 −3 2 C 1×10 m
13. The electric potential at the same point is ______kV. A. 63 B. 63 For the voltage we have: V =V 1 V 2 =k
Q1 r1
k
Q2 r2
C. 0
=9×10 9
D. 27
N⋅m 2 −3×10−9 C C
2
1×10 m −3
E. 27
=−27×10 3 V
6/8 14. A sphere of radius r = 5 mm has a charge of 5 µC distributed uniformly over its surface. A 3 nC point charge is being held at 10 mm over the surface of the sphere. The magnitude of the force that the charged sphere exerts over the point charge is _____N A. 5.4
B. 1.4
C. 0.6
D. 0.2
E. 0.6
We know by Gauss' Law that a field produced by a charged sphere is equal to the one produced by a point charge with the distance to other charges including the radius of the sphere, hence: ∣F∣=q∣ E∣=k
∣Q sphere Q point charge∣
= 9×10 2 r sphere r sphere− point charge
9
N⋅m 2 3×10 −9 C 5×10−6 C C
2
2
15×10 m −3
=0.6 N
15. A 2 nC charge is located between the parallel plates of a capacitor with a potential difference of 25V as shown in the figure below. The force experienced by the charge is___mN 0.1 mm
+ + + + +
2 nC
25 V
A. B. C. D. E.
100 i −100 i 0 i −50 i 50 i
x
V In a parallel plate capacitor the electric field is uniform and its magnitude is given by E= . d
The direction is from the positively charged plate to the negatively charged one. Therefore the −9 force is equal to: F=q E =−2×10 C
25 V 0.1×10 m −3
i =−5×10−4 N i
7/8 Refer to the following circuit for questions 16 and 17. All capacitances are in F.
15 V
2/3
1/2 1/6
1/4
16. The equivalent capacitance of the circuit above is _____µF B. 1
A. 13/6
C. 19/12
D. 12/19
E. 6/13
17. The voltage across the ¼ µF is _____V A. 0
B. 5
C. 10
D. 15 E. 20
Let us use the fact that capacitors in series have the same charge. Thus : Q1/ 4 =Q1/ 2 1
1 V 1= V 2 4 2
where we have used that C=
Q V
. We need an extra equation. We know that the sum of these
voltages has to be equal to the one of the source, so V 1V 2=15V . Solving for the required voltage we obtain: V 1=10 V
18. The power delivered in 0.2 ms by the total discharge of a 50 µF capacitor connected to 25 V battery is ______W A. 78
P=
B. 67 1 CV2 2
C. 52
D. 49
1 50×10−6 F 25V 2 E = = =78 W −3 t t 2 0.2×10 s
E. 32
8/8 19. A bird stands on an electric transmission line carrying 2500 A. The line has 2.5×10−5 resistance per meter and the bird's feet are 4.0 cm apart. The voltage that the bird feels is ____mV A. 4.0
B. 3.5
C. 3.0
D. 2.5
E. 2.0
R −5 −3 V =R⋅I = l⋅2500 A=2.5×10 ×0.04 m×2500 A=2.5×10 V l m
20. How many 100W light bulbs, connected in parallel to 120 V source can be used with a maximum current of 2.5 A?
A. 7
B. 6
C. 5
D. 4
E. 3 P
100 W
V
120 V
The current for one light bulb working between operating parameters is I = =
=
10 A 12
.
Since they are connected in parallel the total current is the sum of each individual current. Thus 10 the number of light bulbs is n=2.5/ =3 12
_____________________________________________________________________________
Write your answers here: 6.C
7.C
8.D
9.A
10.B
11.E
12.A
13.E
14.C
15.D
16.B
17.C
18.A
19.D
20. E
Solutions
CONCEPTUAL QUESTIONS 1. For which of the following adiabatic processes is the change in entropy bigger ? Explain why. P
a b
c
V Let us remember first the definition of entropy change : S=
Q T
. Since all processes are
adiabatic Q =0 . So we can conclude that the change in entropy for all of them is the same and equal to zero. 2. Sketch a PV diagram of the following process: 2.0 L of ideal gas at atmospheric pressure are cooled at constant pressure to a volume of 1.0 L, and then expanded isothermally back to 2.0 L, whereupon the pressure is increased at constant volume until the original pressure is reached. P 1 atm
2 L 1 L V 3. Imagine a test charge inside a conductor in electrostatic equilibrium. Now imagine that this this charge is moved very slowly. What is the work performed by the electric forces during this process? Explain.
2/8 The work done is zero. The electric field inside a conductor in electrostatic equilibrium is zero, thus the net electric force on the test charge is also zero. Hence no work is necessary to move the charge inside the conductor. 4. What can you say about the electric field in a region of space that has the same potential throughout? Explain. The electric field is given by E x =
−V x
, E y=
can conclude that the electric field is zero.
−V y
, E z=
− V z
. If V is constant throughout, we
5. When the voltage across a certain conductor is doubled, the current is observed to triple. What can you conclude about the conductor? Explain. Since the current is not proportional to the voltage the conductor is not ohmic.
NUMERICAL QUESTIONS An ideal gas expands at constant pressure of 5.0 atm from 400 mL to 660 mL. Heat then flows out the gas, at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. 6. The total work done in the process is ______atm·L A. 2.3
C. 1.3
B. 1.7
D. 1.0
E. 0.3
5 atm
0.4 L
0.660 L
V
The work done is the area below the curve. Thus W =5 atm0.660 L−0.400 L =1.3 atm⋅L
3/8
7. The net heat flow into the gas is ____J A. 230
C. 130
B. 170
D. 100
E. 30
Since the initial and final temperatures are the same we have U =0 which means by the first law of thermodynamics that Q=W . Therefore: Q=1.3 atm⋅L
1×105 Pa 1 atm
×
1 m3 10 3 L
=130 J
8. 5mol of an ideal monatomic gas are adiabatically expanded according to the diagram below. The net work done by the gas is _________J P A. 1800 B. 1248 C. 782 D. 672 E. 564
6 L, 3 atm
12 L, 0.94 atm
V By the first law of thermodynamics we have U =Q−W since the process is adiabatic the equation reduces to U =−W . So let us find the change in internal energy: P f V f Pi V i 3 3 3 U = n R T = n RT f −T i = n R − 2 2 2 nR nR
where we have used the equation of state for an ideal gas in the last equality. Thus the work performed by the gas is : W=
−3 −3 P f V f − P i T i = 12L⋅0.94 atm−6 L⋅3atm=6.72 L⋅atm=672 J 2 2
4/8 9. A Carnot engine extracts 890 J of work form a 550 K reservoir during each cycle and rejects 470 J to a cooler reservoir. The temperature of the cooler reservoir is _________K A. 190
B. 210
C.273
D. 330
E. 425
Since we are talking about a Carnot engine, the ideal efficiency is equal to the efficiency, so: 1−
TL TH
T H 550 K
W
=
QH
=
QL W QL
W QL W
=T L
470 J 890 J470 J
=190 K
10. A refrigerator has a coefficient of performance equal to 5.00. The refrigerator takes 120 J of energy from a cold reservoir in each cycle. The energy expelled to a hot reservoir is ______J B. 144
A. 24
COP=
C. 224
D. 600
E. 720
QL
W Q H =W QL QH =
1 COP
1Q L
1 Q H = 1120 J =144 J 5
11. A charge Q1 is fixed at the origin and another charge Q2 is located 2 m away. Charge Q2 is at rest, and the electrical potential energy of the system is 12 J. If charge Q2 moves away from the origin. Its kinetic energy when it is 4 m away from the origin is ____J A. 16
B. 12
C. 10
D. 8
E. 6
The total energy of the system is conserved and is equal to 12 J. This energy is also equal to the sum of the kinetic and potential energies. In particular the potential energy is equal to:
5/8 U =k
Q1 Q2 r
. At the initial configuration, all the energy is potential. When the second charge
moves away the energy is now distributed between kinetic and potential. Since the distance doubles we have U f = k
Q1 Q 2 rf
=k
Q1 Q 2 2r i
=
12J 2
thus the kinetic energy is 12J – 6 J= 6 J.
=6 J
The following diagram shows two charges. A positive charge +2 nC located at the origin and a – 5 nC located at y= 2 mm. 5 nC
y
E1
E2
x
2 nC
12. The electric field at point (0, 1) mm has magnitude ___MV/m and direction___°. A. 63, 90
B. 63, 90
C. 27,0
D. 27, 90
E. 27, 90
The total electric field is: E =k E=E 1 2
∣Q1∣
∣Q ∣ jk 2 j r 1 2 r 2 2
2 7×10−9 C 9 N⋅m E =9×10 j=63×106 N /C j 2 −3 2 C 1×10 m
13. The electric potential at the same point is ______kV. A. 63 B. 63 For the voltage we have: V =V 1 V 2 =k
Q1 r1
k
Q2 r2
C. 0
=9×10 9
D. 27
N⋅m 2 −3×10−9 C C
2
1×10 m −3
E. 27
=−27×10 3 V
6/8 14. A sphere of radius r = 5 mm has a charge of 5 µC distributed uniformly over its surface. A 3 nC point charge is being held at 10 mm over the surface of the sphere. The magnitude of the force that the charged sphere exerts over the point charge is _____N A. 5.4
B. 1.4
C. 0.6
D. 0.2
E. 0.6
We know by Gauss' Law that a field produced by a charged sphere is equal to the one produced by a point charge with the distance to other charges including the radius of the sphere, hence: ∣F∣=q∣ E∣=k
∣Q sphere Q point charge∣
= 9×10 2 r sphere r sphere− point charge
9
N⋅m 2 3×10 −9 C 5×10−6 C C
2
2
15×10 m −3
=0.6 N
15. A 2 nC charge is located between the parallel plates of a capacitor with a potential difference of 25V as shown in the figure below. The force experienced by the charge is___mN 0.1 mm
+ + + + +
2 nC
25 V
A. B. C. D. E.
100 i −100 i 0 i −50 i 50 i
x
V In a parallel plate capacitor the electric field is uniform and its magnitude is given by E= . d
The direction is from the positively charged plate to the negatively charged one. Therefore the −9 force is equal to: F=q E =−2×10 C
25 V 0.1×10 m −3
i =−5×10−4 N i
7/8 Refer to the following circuit for questions 16 and 17. All capacitances are in F.
15 V
2/3
1/2 1/6
1/4
16. The equivalent capacitance of the circuit above is _____µF B. 1
A. 13/6
C. 19/12
D. 12/19
E. 6/13
17. The voltage across the ¼ µF is _____V A. 0
B. 5
C. 10
D. 15 E. 20
Let us use the fact that capacitors in series have the same charge. Thus : Q1/ 4 =Q1/ 2 1
1 V 1= V 2 4 2
where we have used that C=
Q V
. We need an extra equation. We know that the sum of these
voltages has to be equal to the one of the source, so V 1V 2=15V . Solving for the required voltage we obtain: V 1=10 V
18. The power delivered in 0.2 ms by the total discharge of a 50 µF capacitor connected to 25 V battery is ______W A. 78
P=
B. 67 1 CV2 2
C. 52
D. 49
1 50×10−6 F 25V 2 E = = =78 W −3 t t 2 0.2×10 s
E. 32
8/8 19. A bird stands on an electric transmission line carrying 2500 A. The line has 2.5×10−5 resistance per meter and the bird's feet are 4.0 cm apart. The voltage that the bird feels is ____mV A. 4.0
B. 3.5
C. 3.0
D. 2.5
E. 2.0
R −5 −3 V =R⋅I = l⋅2500 A=2.5×10 ×0.04 m×2500 A=2.5×10 V l m
20. How many 100W light bulbs, connected in parallel to 120 V source can be used with a maximum current of 2.5 A?
A. 7
B. 6
C. 5
D. 4
E. 3 P
100 W
V
120 V
The current for one light bulb working between operating parameters is I = =
=
10 A 12
.
Since they are connected in parallel the total current is the sum of each individual current. Thus 10 the number of light bulbs is n=2.5/ =3 12
_____________________________________________________________________________
Write your answers here: 6.C
7.C
8.D
9.A
10.B
11.E
12.A
13.E
14.C
15.D
16.B
17.C
18.A
19.D
20. E
