20160624

Physics quiz question: thermal contraction of James Webb Space Telescope mirror segment

Physics 205A Quiz 7, fall semester 2015
Cuesta College, San Luis Obispo, CA

"Reflections on JWST - May 4, 2016"
James Webb Space Telescope (JWST)
youtu.be/R-ldmlCkAoU

The James Webb Space Telescope will use beryllium mirror segments (linear expansion coefficient 7.0×10–6 K–1), each 1.32 m wide.[*][**] During testing it is cooled from room temperature (293 K) down to its operational temperature of 53 K. The width of a mirror segment will contract by:
(A) 2.2×10–3 m.
(B) 2.7×10–3 m.
(C) 3.2×10–3 m.
(D) 4.2×10–3 m.

[*] P. Hidnert, W. T. Sweeney, "Thermal Expansion of Beryllium and Aluminum Beryllium Alloys," Scientific Papers of the Bureau of Standards, vol. 22 (1927), p. 533, dx.doi.org/10.6028/nbsscipaper.251.
[**] jwst.nasa.gov/mirrors.html.

Correct answer (highlight to unhide): (A)

The relation between the change ∆L of a linear dimension L (here, width) due to a temperature change ∆T is given by:

α·∆T = ∆L/L,

such that:

L = L·α·T = (1.32 m)·(7.0×10–6 K–1)·(53 K – 293 K) = -2.2176×10–3 m,

or to two significant figures, the width of the mirror segment decreases by 2.2×10–3 m.

(Response (B) is L·α·(293 K); response (C) is L·α·(53 K + 293 K); response (D) is 1/∆T.)

Sections 70854, 70855, 73320
Exam code: quiz07zSOL
(A) : 62 students
(B) : 5 students
(C) : 1 student
(D) : 1 student

Success level: 90%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.22

Physics quiz question: rate of thermal loss through sleeping pad

Physics 205A Quiz 7, fall semester 2015
Cuesta College, San Luis Obispo, CA

"Therm-a-Rest® Closed-Cell Mattresses"
Therm-a-Rest
youtu.be/SwWevyXpibg

A Thermarest® Z Lite SOL™ polyethylene insulating sleeping pad[*] 
(area of 0.933 m2 and thickness of 0.020 m) has a thermal resistance of 0.43 K/watt. If there is a 20° C temperature difference between the top (in contact with a sleeping person) and bottom (in contact with the cold ground) of the sleeping pad, the rate of heat per time conducted through the pad will be:
(A) 1.1×10–3 watts.
(B) 2.2×10–2 watts.
(C) 8.6 watts.
(D) 47 watts.

[*] cascadedesigns.com/therm-a-rest/mattresses/fast-and-light/z-lite/product.

Correct answer (highlight to unhide): (D)

The rate of heat per time conducted is given by:

Power = ∆T/R,

where the temperature difference ∆T and the thermal resistance R of the sleeping pad are given, such that:

Power = (20° C)/(0.43 K/watt) = 46.511627907 watts,

or to two significant figures, the rate of heat conducted through the sleeping pad is 47 watts.

(Response (A) is d/(A·∆T); response (B) is R/∆T; response (C) is R·∆T.)

Sections 70854, 70855, 73320
Exam code: quiz07zSOL
(A) : 3 students
(B) : 4 students
(C) : 8 students
(D) : 54 students

Success level: 78%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.46

20160622

Physics quiz question: comparing surface temperatures of Bellatrix and Pollux

Physics 205A Quiz 7, fall semester 2015
Cuesta College, San Luis Obispo, CA

Stars can be modeled as spherical blackbodies that radiate heat out to an environment assumed to have a temperature of 0 K. Pollux is a star that radiates less heat per time, but is larger in size than Bellatrix. Pollux has a surface temperature __________ Bellatrix.
(A) cooler than.
(B) same as.
(C) hotter than.
(D) (Not enough information given.)

Correct answer: (A)

The net power radiated by a star with an environment assumed to be at absolute zero is given by:

Power = –e·σ·A·((T)4 – (Tenv)4) = –e·σ·A·((T)4 – (0 K)4).

where the emissitivity e = 1 (for an ideal blackbody), the Stefan-Boltzmann constant is σ = 5.670×10–8 watts/(m2·K4), A is the surface area, and the surface temperature T is unknown, and can be solved for:

(T)4 = –Power/(e·σ·A),

T = (–Power/(e·σ·A))(1/4).

This allows a ratio to be set up to compare the surface temperatures of Bellatrix and Pollux:

TBellatrix = (–PowerBellatrix/(e·σ·ABellatrix))(1/4),

TPollux = (–PowerPollux/(e·σ·APollux))(1/4),

such that:

(TBellatrix/TPollux) = (PowerBellatrix/PowerPollux))·(APollux/ABellatrix)(1/4).

Since Pollux radiates less heat per time than Bellatrix, then the ratio (PowerBellatrix/PowerPollux)) is greater than 1; and since Pollux is larger in size than Bellatrix, the ratio (APollux/ABellatrix)(1/4) is also greater than 1. Thus the ratio (TBellatrix/TPollux) must then be greater than 1, so Pollux has a cooler surface temperature than Bellatrix.

Sections 70854, 70855
Exam code: quiz07zSOL
(A) : 50 students
(B) : 12 students
(C) : 7 students
(D) : 0 students

Success level: 72%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.70

Physics quiz question: surface temperature of Bellatrix

Physics 205A Quiz 7, fall semester 2015
Cuesta College, San Luis Obispo, CA

Stars can be modeled as spherical blackbodies that radiate heat out to an environment assumed to have a temperature of 0 K. Bellatrix is a star that radiates heat per time at a rate of 2.5×1030 watts, and has a surface area of 2.2×1020 m2. The surface temperature of Bellatrix is:
(A) 4.2×103 K.
(B) 2.1×104 K.
(C) 1.1×105 K.
(D) 6.1×105 K.

Correct answer: (B)

The net power radiated by Bellatrix (–2.5×1030 watts, where the negative sign indicates that heat is continuously leaving the surface of the star) given by:

Power = –e·σ·A·((T)4 – (Tenv)4) = –e·σ·A·((T)4 – (0 K)4),

where the emissitivity e = 1 (for an ideal blackbody), the Stefan-Boltzmann constant σ = 5.670×10–8 watts/(m2·K4), A is the surface area, and the surface temperature T is unknown, and can be solved for:

(T)4 = –Power/(e·σ·A),

T = (–Power/(e·σ·A))(1/4),

T = (–(–2.5×1030 watts)/((1)·(5.670×10–8 watts/(m2·K4))·(2.2×1020 m2)))(1/4),

T = 2.11584362479×104 K,

or to two significant figures, the surface temperature of Bellatrix is 2.2×104 K.

(Response (A) is σ(–1/2); response (C) is (Power/A)(–1/2); response (D) is (Power·σ)(1/4).)

Sections 70854, 70855
Exam code: quiz07zSOL
(A) : 12 students
(B) : 31 students
(C) : 15 students
(D) : 10 students

Success level: 45%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.35

20160615

Physics quiz question: electron capture decay of iodine-125

Physics 205B Quiz 7, spring semester 2016
Cuesta College, San Luis Obispo, CA

I(125,53) is used as a radiation source placed surgically implanted next to tumors for treatment of eye cancer[*]. I(125,53) decays via electron capture (with a half life of 59.43 days) to become:
(A) Te(123, 52).
(B) Te(125, 52).
(C) Xe(125, 52).
(D) Xe(127, 52).

[*] cancer.ucsd.edu/care-centers/radiation-oncology/procedures/Pages/eye-plaques.aspx.

Correct answer (highlight to unhide): (B)

Electron capture takes in an electron to covert a proton into a neutron, which also leaves the nucleon number the same (a total of 125 protons and neutrons), but decreases the atomic number by one (from 53 protons to 52 protons).

(Response (C) corresponds to the daughter product of I(125,53) undergoing β decay.)

Sections 30882, 30883
Exam code: quiz07eYE5
(A) : 1 student
(B) : 32 students
(C) : 7 students
(D) : 0 students

Success level: 80%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.13

Physics quiz question: interpreting β+ decay Feynman diagram

Physics 205B Quiz 7, spring semester 2016
Cuesta College, San Luis Obispo, CA

This (valid) Feynman diagram depicts the __________ process.
(A) α.
(B) β.
(C) β+.
(D) electron capture.
(E) γ.

Correct answer (highlight to unhide): (C)

This diagram shows β+ decay, where a proton transforms into a neutron, and the intermediate W+ particle decays into a neutrino and a positron, as shown by the left-to-right neutrino arrow, and a right-to-left electron arrow (corresponding to a positron--the "β+" particle--progressing left-to-right as time progresses).

Sections 30882, 30883
Exam code: quiz07eYE5
(A) : 2 students
(B) : 3 students
(C) : 30 students
(D) : 5 students
(E) : 0 students

Success level: 75%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.60

20160614

Physics quiz question: ammeter reading, connected in series with voltmeter

Physics 205B Quiz 5, spring semester 2016
Cuesta College, San Luis Obispo, CA

An ideal 3.2 V emf source is connected to an ideal voltmeter and ammeter, a light bulb, and a resistor. The light bulb is lit. The ammeter reading is:
(A) zero.
(B) some finite, non-zero value below 0.67 A.
(C) exactly 0.67 A.
(D) some finite value above 0.67 A.
(E) ∞.

Correct answer (highlight to unhide): (A)

An ideal ammeter has zero resistance, but it is connected in series to an ideal voltmeter, which has an infinite resistance. Thus no current will flow through the upper loop of this circuit, such that the ammeter will read zero.

(Response (C) is the amount of current flowing through the emf, light bulb, and resistor in the lower loop of this circuit.)
Sections 30882, 30883
Exam code: quiz05Tt1p
(A) : 17 students
(B) : 5 students
(C) : 11 students
(D) : 4 students
(E) : 1 student

Success level: 45%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.26

Physics quiz question: power dissipated by light bulb in series circuit

Physics 205B Quiz 5, spring semester 2016
Cuesta College, San Luis Obispo, CA

An ideal 6.0 V emf source is connected to two light bulbs (each with a 1.2 Ω resistance), and a 0.50 Ω resistor. The electrical power used by the top light bulb is:
(A) 2.1 W.
(B) 5.1 W.
(C) 12 W.
(D) 30 W.

Correct answer (highlight to unhide): (B)

The equivalent resistance of this series circuit is the arithmetic sum of their resistances:

Req = 1.2 Ω + 1.2 Ω + 0.50 Ω = 2.9 Ω.

From Kirchhoff's junction rule, the amount of current flowing through the emf source and all three resistive circuit elements must be the same, as they are in series with each other, and the current flowing this circuit is given by applying Ohm's law to the entire equivalent circuit:

Ieq = ε/Req = (6.0 V)/(2.9 Ω).

Since we now know the resistance value of the top light bulb, and the amount of current flowing through it, the power dissipated by the top light bulb is then:

Ptop light bulb = Ieq·ΔVtop light bulb = Ieq·(Ieq·Rtop light bulb) = Ieq2·Rtop light bulb,

Ptop light bulb = ((6.0 V)/(2.9 Ω))2·(1.2 Ω) = 5.1367419737 W,

or to two significant figures, 5.1 W.

Response (A) is the numerical value for the current flowing through the circuit Ieq = ε/Req = (6.0 V)/(2.9 Ω); response (C) is ((6.0 V)/(2.9 Ω))2·(2.9 Ω), which is the amount of power supplied by the emf source to rest of the circuit; response (D) is (6.0 V)2/(1.2 &Ohms;), which would be the power dissipated by the top light bulb if it were to use up all of the voltage supplied by the emf (which would violate Kirchhoff's loop rule!).

Sections 30882, 30883
Exam code: quiz05Tt1p
(A) : 4 students
(B) : 16 students
(C) : 13 students
(D) : 5 students

Success level: 43%
Discrimination index (Aubrecht & Aubrecht, 1983): 0.79

20160613

Physics quiz question: step-down transformer output voltage

Physics 205B Quiz 6, spring semester 2016
Cuesta College, San Luis Obispo, CA

A step-up transformer with a turns ratio of 1:20 is used with an input voltage of 120 V, and an input current of 0.50 A in the primary coil. The current in the secondary coil is:
(A) 2.5×10–2 A.
(B) 6.0 A.
(C) 10 A.
(D) 2.4×103 A.

Correct answer (highlight to unhide): (A)

The transformer equations are:

ε21 = N2/N1 = I1/I2,

where the input (alternating root-mean-square) voltage and current in the primary coil are ε1 = 120 V and I1 = 0.50 A, respectively. Since the secondary coil has 20 times more turns than the primary, then the output voltage ε2 will be stepped up by a factor of 20, while the output current I2 will be stepped down by a factor of 20:

I2 = I1·(N1/N2),

I2 = (0.50 A)·(1/20) = 2.5×10–2 A.

Response (B) is ε1·(N1/N2); response (C) is I1·(N2/N1); response (D) is ε1·(N2/N1).

Sections 30882, 30883
Exam code: quiz06eL3k
(A) : 24 students
(B) : 3 students
(C) : 10 students
(D) : 2 students

Success level: 62%
Discrimination index (Aubrecht & Aubrecht, 1983): 1.00

20160609

Astronomy quiz question: Milky Way's oldest stars

Astronomy 210 Quiz 7, spring semester 2016
Cuesta College, San Luis Obispo, CA

According to the monolithic collapse model, the oldest stars in the Milky Way are located:
(A) in the halo.
(B) in the nuclear bulge.
(C) within the spiral arms.
(D) inside the central supermassive black hole.

Correct answer (highlight to unhide): (A)

According to the monolithic collapse model, the Milky Way evolved from a spherical shape to its current disk shape, leaving behind the globular clusters out in its halo. Thus globular clusters should all have the identical (old) ages (as determined from the lack of metal absorption lines, and their H-R diagram turn-off points), while the nuclear bulge and the spiral arms should be comprised of mixed-age (old and newer) stars.

Section 30674
Exam code: quiz07N4rK
(A) : 10 students
(B) : 1 student
(C) : 6 students
(D) : 0 students

"Success level": 61% (including partial credit for multiple-choice)
Discrimination index (Aubrecht & Aubrecht, 1983): 0.47