PHY 126-07 Final Exam

1
PHY 126-07 Final Exam

• Problem 1 (20 points)

A small ball of mass 2.00 kg is suspended by two
wires, wire 1 and wire 2. If two wires make 90o
and wire 2 makes q30o with respect to the
vertical, find the tension T1 and T2.
P
Q
wire 2
q 30o
wire 1
90o
T2
T1
mg
Solution
Lets denote the length of wire 2 by L. Then the
condition that the ball does not rotate about the
pivot point of wire 2, Q, is -L mg sinq LT10
(5 points), namely, T1mg sinq (2.00 kg)(9.80
m/s2)sin30o9.80 N (5 points). The condition that
the net force is zero in the horizontal direction
gives T1cosq - T2sinq 0 (5 points),
namely, T2T1/tan30o 17.0 N (5 points). (The
condition that the net force is zero in the
vertical direction gives T1cosq T2sinq mg
can be used too.)
2
Problem 2 (1010 points)
An innovative musician is making a new music
instrument using cylindrical glasses of
height h1 m with some water in them where the
top (bottom) of each glass is open (closed).
Right now he is adjusting the height of the water
in one of the glasses so that sound of middle
C note at frequency 256 Hz creates resonance.
Assume that the speed of sound is 344 m/s. (a)
What is the maximum height of water for the
instrument to resonate with middle C note? (b)
What is the next lowest frequency of sound that
this glass instrument can resonate with if the
water level is the same as in Part (a).
open

• Solution
• This instrument works as a stopped pipe.
  Therefore the
• frequencies of possible n-th harmonics are
  expressed by
• fnnv/(4Ln) where n1,3,5,, v is the speed of
  sound, and
• Ln is the height of air column. Since Ln
  nv/(4f), for f256 Hz,
• L1 becomes shortest at (344 m/s)/(4 x 256
  1/s)0.336 m
• when n1. So the maximum water height for middle
  C note is
• 1.000 m - 0.336 m0.664 m. For a higher
  harmonics, the height
• of the air column is larger.
• (b) When the water level is adjusted as in Part
  (a), the fundamental
• frequency f1 is 256 Hz. The next lowest
  frequency is when n3,
• namely the third harmonics f33f13 x 256 Hz
  768 Hz.

L
water
h
closed
3
Problem 3 (20 points)
Jane and John will host a big party. They are
preparing iced tea for the party. Their plan
is to have iced tea at 3.0oC in an insulated
container without ice left in it. They first made
15 liters of tea at 20oC. Then they plan to put
some ice at -5.0oC into the container to cool
the tea. How much of ice is needed to make this
iced tea according to their plan? Treat the
tea as water. If needed, use the following
constants the density of water rwater1000
kg/m3, the latent heat of fusion of water Lf3.33
x 105 J/kg, specific heat of ice cice2090 J/(kg
K) and the specific heat of water cwater4190
J/(kg K).
Solution
The mass of tea is mwater(15 L) x (1.0 kg/L)15
kg (3 points). The amount of heat needed to melt
the ice is Qmeltmice x (3.33 x 105 J/kg) (3
points). To heat the ice from -5.0oC to 0.0oC the
amount of heat needed is Qice mice x cice x
0.0oC (-5.0oC)micex(10.5x103 J/kg) (3
points). The heat needed to raise the temperature
of tea from the ice from 0.0oC to 3.0oC
is Qice-watermice x cwater x 3.0-(0.0oC)micex(
12.6x103 J/kg) (3 points). The heat the tea
acquires is Qwatermwater x cwater x 3.0oC
-(20.0oC) (15 kg)x cwaterx(-17.0oC)1068 kJ (3
points). From QmeltQiceQice-waterQwater0 (2
points), micex(333x 103 J/kg) micex(10.5x103
J/kg) micex (12.6x103 J/kg) (1068x103 J) 0.
So micex(356x103 J)1068x103 J and therefore the
amount of ice needed is mice3.0 kg (5 points).
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Problem 4 (5555 points)
You are experimenting with a converging lens with
a focal length of 10 cm. Find the image position
(whether on the left or right of the lens and how
far from the lens), magnification, and whether
the image is upright or inverted, and real or
virtual when an object is placed at (a) 25 cm,
(b) 10 cm, (c) 7.5 cm and (d) -20 cm. If the
image is formed at infinity, just stating that
gets you 5 points. Assume that the incoming light
rays come from the left of the lens toward the
lens.

• Solution
• 1/s 1/s 1/f where s is the object distance,
  s is the image distance and f is the focal
  length.
• f10 cm and s 25 cm, so s17 cm. The
  magnification m-s/s16.7 cm/25 cm-2/3.
• So the image is formed at 17 cm to the right of
  the lens, real and inverted.
• f10 cm and s10 cm, so sinfinity. So the
  image is formed at infinity.
• f10 cm and s7.5 cm, so s-30 cm. The
  magnification m-s/s4. So the image is
• formed at 30 cm to the left of the lens virtual
  and upright.
• f10 cm and s-20 cm, so s6.7 cm. The
  magnification m-s/s1/3. So the image is
• formed at 6.7 cm to the right of the lens real
  and upright. (Note the object is a virtual
• image.)

5

• Problem 5 (55555 points)

An experimental power plant at the Natural Energy
Lab generates electricity from the temperature
gradient of the ocean. The surface and deep-water
temperatures are 27oC and 6oC, respectively. (a)
What is the maximum theoretical efficiency of
this power plant? (b) If the power plant is to
produce 210 kW of power, at what rate must heat
be extracted from the warm water? (c) If
the power plant is to produce 210 kW of power, at
what rate must heat be absorbed by the cold
water? (d) To raise the temperature of water
whose mass is dm by DT, how much of heat dQ is
needed? (For this question you may use the
symbol c for specific heat of water). (e) The
cold water that enters the plant leaves it at a
temperature of 10oC. What must be the flow
rate of cold water through the system in kg/h?
Use, if needed, the fact that specific heat of
water is c4190 J/(kg K).
Solution
(a) The maximum efficiency is realized by a
Carnot engine. Therefore
(b)
(c)
(d)
(e)
6
Problem 6 (3333332 points)

• Three moles of an ideal gas are taken around the
  cycle acb shown in Figure. For this gas
• Cp 29.1 J/(mol K). Process ac is at constant
  pressure, process ba is at constant volume,
• and process cb is adiabatic. The temperature of
  the gas in states a,c, and b is Ta300 K,
• Tc492 K, and Tb600 K. The universal gas
  constant R is 8.31 J/(mol K).
• For a process at constant pressure, when the
  temperature changes by DT, how much
• does the volume change DV?
• For process ac, how much work is done (plug-in
  numbers)?
• For an ideal gas, find the molar heat capacity at
  constant volume CV in terms of Cp and
• the universal gas constant R (no numbers).
• (d) For process cb, how much is the change of
  internal energy, DU in terms of the number of
• moles n, CV, and the change of temperature DT
  (no numbers)?
• (e) For process cb, how much work is done
  (plug-in numbers)?
• For process ba, how much work is done (plug-in
  numbers)?
• Find the total work done for the cycle acba?

p
b
Solution

• From the ideal gas equation, pVnRT. So pDVnRDT.
• WacpDVnRDT(3.00 mol)8.31 J/(mol K)(492 K
  300 K)
• 4.79x103 J.
• (c) CVCp-R.
• (d) DUnCVDT.
• (e) As Qcb0 for an adiabatic process,
  WcbQcb-DU-nCVDT-n(Cp-R)DT
• -(3.00 mol)29.1 J/(mol K)-8.31 J/(mol
  K)(600 K 492 K)-6.74x103 J.
• (f) For a process at constant volume, the work
  done is zero. So Wba0.00 J.
• (g) The total work WtotWacWcbWba4.78x103 J
  6.74x103 J-1.95x103 J.

a
c
O
V