Physics 2C Summer Session I
Quiz #3
Multiple Choice. Choose the answer that best completes the
statement or answers the question.
1. An ideal Carnot engine operates between 52 C and 269 C with a heat
input of 250J per cycle. Assuming that the acceleration due to gravity is g =
10m=s2 ; how many cycles are required to lift a 400g rock 50m?
A) 1
B) 2
C) 4
D) 5
E) 8
The e¢ ciency of this engine is given by
e=1
Tc
=1
Th
52 + 273
= :4:
269 + 273
Hence the work output per cycle is 100J: The work required to lift a 400g rock
50m is 200J. Hence 2 cycles are required to lift the rock.! B
2. Given that the gas in problem 1 expands by a factor of 5 : 1 during the
isothermal expansion process at T = 269 C; how many moles of an inert gas
are required in the Carnot cycle of problem 1 to produce the 250J per cycle?
A) :086moles B) :098moles C) :112moles D) :136moles
The net work by the gas in a Carnot cycle is given by
W = nRTh ln VB =VA
E) :156moles
nRTc ln VC =VD ;
where VB =VA is the expansion ratio during the isothermal expansion process at
T = Th . The expansion ratio VC =VD for the expansion ratio at T = Tc is found
from
T h VB
1
T h VA 1
VB =VA
= Tc VC
1
= Tc VD 1
= VC =VD :
Hence the expansion ratios are equal and
W
n
= nR (Th Tc ) ln VB =VA ! 250 = n (8:314) (217) ln 5
= 250= ((8:314) (217) ln 5) = :086moles ! A
3. How much heat does a refrigerator with a COP of 4:0 exhaust as it freezes
800g of water already at the freezing point?
A) 117kJ
B) 214kJ
C) 334kJ
1
D) 1069kJ
E) 1366kJ
The heat exhausted to the high temperature reservoir is found from
COP
=
Qh
=
Qc
Qc
=
! COP Qh = (COP + 1) Qc
W
Qh Qc
COP + 1
5
Qc = :8 334kJ = 334kJ ! C
COP
4
4. What is the entropy increase during an adiabatic free expansion from an
initial volume of Vi = 2L to a …nal volume of Vf = 10L for :36 moles of an ideal
diatomic gas at an initial temperature T = 100 C?
A) 2:44J=K
B) 4:82J=K
C) 7:23J=K
The entropy increase is found from
s=
P V
nRT ln (Vf =Vi )
Q
=
=
= :36
T
T
T
D) 12:0J=K
E) 0J=K
8:314 ln (5) = 4:82J=K ! B
:
5. At the surface of Titan (Saturn’s largest moon) the atmosphere is essentially all nitrogen gas, (N2 ) ; which has a pressure of P = 1:5 105 P a and a molar
density of n=V = :014mol=L: What is the speed of sound in this atmosphere
(note that atomic nitrogen has an atomic number of A = 14)?
A) 23:1km=s
B) 1035m=s
C) 732km=s
D) 799m=s
E) 732m=s
The temperature on the surface of Titan is found from the ideal gas law,
P V = nRT ! RT = P V =n
The speed of sound is
p
p
p
vs =
P= =
kT =m =
RT =mmol :
From the expression for RT we …nd
p
p
P V =nmmol = 1:4
vs =
vs = 732m=s ! E
1:5
105 = (:014
103
28
10
3)
6. At a distance of 4m from a sound source the intensity level is 76dB: What
is the intensity level (in dB) at a distance of 10m?
A) 68dB
B) 58dB
C) 8dB
D) 18dB
E) 60dB
The intensity falls o¤ as 1=r2 : Hence I=Io = :42 = :16: The change in dB is
= 10 log I=Io = 10 log (:16) =
Hence the intensity at 10m is 76dB
8dB
8dB = 68:0dB ! A
2
7. One end of a 700g nylon rope is suspended at the top of a vertical mine
shaft (analogous to a well) that is 60m deep. The rope is attached to a box of
mineral samples with mass 110kg. Assuming an acceleration due to gravity of
g = 10m=s2 ; how long does it take for a transverse wave to propagate from the
top to the bottom of the rope?
A) 5:12s
B) 265ms
C) 307ms
D) 6:17s
E) 195ms
The speed of a wave on the rope is
p
p
v = T = = 110 10= (:7=60) = 307m=s
The propragation time is
t = `=v = 60=307 = 195ms ! E
8. What is the frequency of the third harmonic on the nylon rope described
in problem 6?
3
A) 40Hz
B) 4:0Hz
C) 5:1Hz
D) 7:7Hz
E) 15:4Hz
The wavelength of the mth harmonic is determined from m = 2`=m. Hence
= 2(60)=3 = 40m: This means that the frequency of this harmonic is
f3 = v=
3
= 307=40 = 7:675Hz ! D
9. Consider the …gure that plots an harmonic wave as a function of position
(a) at t = 0 and as a function of time (b) at x = 3 =2 where is the wavelength
of the wave. The mathematical expression for this wave is
Problem 9
3
A) y = 2 sin (x 2t) =3 B) y = 2 sin 2 (x + 2t) =3 C) y = 2 sin 2 (x + 2t) =3
D) y = 2 sin 2 (x 2t) =3 E) y = 2 cos 2 (x + 2t + 3=2) =3
The most general expression for a simple harmonic wave is
y = A sin (kx
!t + ) = A sin (2 (x=
t=T ) + ) :
: From the …gures we know that A = 2; = 3; and T = 1:5: From …gure a
(t = 0)
y = 2 sin (2 x=3 + ) ! = 0
When x = 3 =2 the waveform becomes
y = A sin (3
2 t=T ) =
A sin ( 2 t=T ) ! +sign
Hence the waveform is
y = 2 sin
2
(x + 2t) ! C
3
10. Given the wave train along a string shown in Figure 3 obtain an expression for the total energy in this wave train, in terms of the tension, F , the wave
amplitude, A, and the wavelength :
Problem 10
A) E = 2 A2 F= B) E = 2 2 A2 F= C) E = 4 2 A2 F= D) E = 2
E) E = 2 AF
The total energy is given by
Z
Z 4
2
E =
P dt = F kA
sin2 (kx !t) (!dt) = 2 F kA2
0
E
=
2
(2 ) F A2 = = 4
2
A2 F= ! C
Some useful constants and expressions
Universal gas constant
speci…c heat (water)
heat of fusion (water)
power transmitted along a wave
4
R = 8:314J=Kmol
c = 4184J=kgK
Lf = 334kJ=kg
P = F !kA2 sin2 (kx
!t)
2
AF
Some useful equations
Z
cp
= N kT = nRT; U = Q W; W = P dV; =
= 1:4(diatomic)
cv
Z
I
dQ
; = 10 log ; Q = mc T; Q = mLf ; W = Qh Qc = eQh ;
s =
T
I0
p
p
3
v =
F= ; P V = const; hKEi = kT; vs =
P= ; COP = Qc =W
2
e = 1 Tc =Th ; m = 2`=m
PV
5