NAME______________________________
PHYSICS 39
Practice Midterm Exam
Note: These are not the only types of problems that may occur on the midterm.
1. An aluminum rod and a pyrex rod are each 1.30 meter long at a temperature of 23.0
C. At what temperature will the pyrex rod be 0.100 mm longer than the aluminium
rod?
Coefficients of linear expansion: AL = 24 x 10-6 /C, PYREX = 9.0 x 10-6 /C
Ans: 17.9C
2. 3.0 moles of an ideal monatomic gas expands isothermally from 0.500 m3 to 1.25 m3 at
a constant temperature of 675 K. Find (a) the initial pressure of the gas, and (b) the
work done on the gas.
Ans: (a) 3.37 X 104 Pa (b) -1.54 X 104 J
3. A drop of molten lead at 327.3C is dropped into a tub containing 1.00 kg of water at
23.0C. If the final temperature of the mixture is 23.5C, find the mass of the lead.
(Ignore the specific heat of the tub.) (Note: mLPb is negative because the lead is losing
heat.)
Melting point of lead: 327.3C, specific heat of lead, clead = 128 J/kg-C
Specific heat of water, cH20 = 4186 J/kg-C,
Latent heat of fusion of lead, LPb = 2.45 X104 J/kg,
Ans: 33 grams
4. Due to heat entering the system, a piston rises in a cylinder containing 0.23 moles of
an ideal gas. The process is isobaric, during which the temperature increases from
214 K to 383 K. During the process, 1130 Joules of heat enters the gas. Is the gas
monatomic or diatomic? Prove your answer. (R = 8.31 J/mole-K)
Ans; diatomic
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5. How many times faster does a hydrogen molecule move, on average, than an oxygen
molecule, at the same temperature? Mass of O2 molecule = 32.2 a.m.u, mass of H2
molecule = 2.01 a.m.u.
Ans: 4X
6. An engine trnasfers 300 Joules of energy from a hot reservoir during its 0.300-second
cycle, and transfers 75 Joules as exhaust to a cold reservoir. What is the power output
of the engine?
Ans: 750 W
7. A solid that has a latent heat of fusion 323 J/kg melts at a temperature of 960C.
Calculate the change in entropy of this substance when 8.21 kg of the mass melts.
2.15 J/K
8. Light travels 10% slower in medium 1 than in light, and 25% slower in medium 2 than
in light. If a ray passes from medium 1 to medium 2 with and angle of incidence 1 =
38.0 (a) will the ray in medium 2 bend towards the normal or away from it? (b) What
will be the angle of refraction in medium 2?
Ans: 30.9
9. A very distant object in front of a converging lens makes an image 15.0 cm from the
lens on the other side. (a) Where will the image be formed if the object is 6.00 cm in
front of the lens? In this case, will the image be (b) upright or inverted? (c) real or
virtual? (d) Calculate the magnification of the image.
Ans: (a) -10 cm, (b) upright, (c) virtual, (d) 1.67X
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10. A small light bulb is sitting at the bottom of a swimming pool, 1.75 meters below the
sufarce, as shown. The light moving up at a small angle of incidence is refracted into
the air. The light at greater angles is totally internally reflected everywhere on the
bottom except near the light bulb, so the bulb is sitting at the center of a circle to which
no light is reflected. Calculate the diameter of that circle. (nwater = 1.33)
Ans: 8.00 m
11. Instead of having a normal near point of 25 cm, a certain near-sighted person can read
fine print when held as close as 15.0 cm from her eyes. When she uses a magnifying
lens whose focal length is 7.00 cm, what is the greatest magnification she can
experience?
Ans: 3.14X
12. Below is shown a candle situated on the axis of a diverging lens. Using the straight
edge provided, construct the three principle rays from the top of the candle flame and
locate the position of the image. Sketch the image at that position.
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13. In a Young’s double-slit experiment, light of a certain wavelength falls on two slits
separated by 0.120 mm. The interference pattern falls onto a screen located 1.10 meter
from the slits. If the third-order bright fringe (m = 3) is 12.0 mm above the central
maximum on the screen, what must be the wavelength of the light, in nanometers?
Ans: 436 nm
14. A nonreflective coating of a certain chemical (n = 1.68) covers the glass (n = 1.42) of
a pair of eyeglasses. If the thickness of the film is 175 nanometers, what wavelength
(in a vacuum) of light will be prevented from reflecting from the glass? (nair = 1.00)
588 nm
TC = TK - 273 C
number of moles, n =
9
TF  TC  32
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mass
molar mass
5
TC  (TF  32)
9
Avogadro’s number, NA = 6.02 x 10-23
Thermal Expansion
L   L0 T
Specific Heat:
A = 2T
Q  mc(T f  Ti )
V  V0 T
  3
c
Q
m T
(J/kg  C )
Mixtures: Qcold  Qhot
Equation of state for an ideal gas:
PV = nRT, R = 8.31 J/molK
Latent Heat:
phase change: Q   mL
Thermal conductivity:
P1V1 = P2V2
T1 = T2
Q/t = Power = A(Th  Tc)
(Li/ki)
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T(K) = T(C) +273
Average kinetic energy per molecule:
PV = nRT, where R = 8.31 J/mole-K
Kav = 3/2 KBT
First Law of Thermodynamics: U = Q + W
Total translational kinetic energy
of n moles: Ktot trans = 3/2 nRt = 3/2
NKBT
ISOBARIC:
W =  P V,
P = Constant
Q = n CP T
KB = 1.38 X 10-23 J/K
ISOVOLUMETRIC: V = 0, W = 0, U
= Q,
Eint = nCvT for all ideal gases
Q = n CV T , CV = 12.5 J/mol-K, Eint =
Q
vrms = 1.73
root-mean-square
vavg = 1.60
average
vmp = 1.41
most probable
ISOTHERMAL: T = 0, U = 0, W = Q,
Wgas = n R T ln(Vi/Vf),
T(K) = T(C) +273
ADIABATIC: Q = 0,
PV = constant,
TV-1 = constant
where  = Cp/Cv
monatomic
diatomic
(rotation
only)
diatomic
(rotation plus
vibration)
Cv
R
R
R
Cp
R
R
R
Cp = C v + R
Heat Engine: |Qh| = W + |Qc|
eengine = Wengine/|Qh|
COP (cooling mode) = |Qc|/W =
|Qc|/(|Qh|-|Qc|)
Power = Work/time
Carnot efficiency, eC = 1 – Tc/Th
Heat Pump:
PV = nRT, where R = 8.31 J/mole-K
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n =c/v
First Law of Thermodynamics:
n1/n2 = 2/1
Law of reflection: i = r
U = Q + W
Entropy: s =
Snell’s Law: n1 sin1 = n2 sin2
Speed of light in a vacuum, c = 2.99 X
108 m/s
Critical angle = sin-1(n2/n1)
Sign Conventions for Mirrors________________________________________________
Quantity
Object location
Image location
Radius
Focal length
Image height
Magnification
Symbol
p
q
R
f
h’
In Front
In Back
Upright




+
+
+
+
+
+
Inverted___


Sign Conventions for Lenses________________________________________________
Quantity
Object location
Image location
Radius
Image height
Magnification
Symbol
p
q
R
h’
In Front
In Back
Upright
Inverted___

+
+
+


+
+


f positive for converging lenses and negative for diverging lenses.
Simple magnifying lens: mmax = 1 + 25cm/f where 25 cm is the near point for persons
with normal vision.
Double Slits:
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d sin bright = m
m = 0, 1, 2, 3 ….
d sin dark = (m+1/2)
m = 0, 1, 2, 3 ….
Ybright = Lm/d m = 0, 1, 2, 3 ….
(A similar expression can be derived for dark fringes)
Interference in thin films:
n = /n
Equation
(m = 0, 1, …)
1 phase
reversal
0 or 2 phase
reversal
2t = (m + ½ )n
Constructive
Destructive
2t = m n
Destructive
Constructive
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