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USAPhO 2019, Полуфинал — Задача 2. Question A2

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Всички материали от USAPhO 2019

USA Physics Olympiad Exam — Part A / Part B · 25 т.

Внимание: Бележка към темата

Original question scan and matching revised key have been read in full. Printed sign/variable/wording defects in A1, A2, A3 and B1 are retained and documented; key alternatives and all answer sheets are preserved. The scanned B2(c) typo remains source faithful, with revised-key wording difference explicitly recorded.

Important Instructions for the Exam Supervisor — DO NOT DISTRIBUTE THIS PAGE

This examination consists of two parts. Part A has three questions and is allowed 90 minutes. Part B also has three questions and is allowed 90 minutes.
Divide the exam paper into 4 parts: the instructions (pages 2–3), Part A (pages 4–6), Part B (pages 8–10), and answer sheets for one of the questions in Part A (pages 12–13). The exam should be printed single-sided to facilitate dividing the test and scanning the answer sheets.
Provide students with the instructions for the competition (pages 2–3). Students can keep the pages for both parts of the exam, as they contain a reference list of physical constants.
Provide students with blank sheets of paper as scratch paper. Students are not allowed to bring their own papers.
Then provide students with Part A and the associated answer sheets, and allow 90 minutes to complete Part A. Do not give students Part B during this time, even if they finish with time remaining. At the end of the 90 minutes, collect the solutions to Part A along with the answer sheets and questions.
Students are allowed a 10 to 15 minute break between Parts A and B. Then allow 90 minutes to complete Part B. Do not let students go back to Part A.
At the end of the exam, the supervisor must collect all papers, including the questions, the instructions, and any scratch paper used by the students. Students may not take the exam questions. The examination questions may be returned to the students after April 8th, 2019.
Students are allowed calculators, but they may not use symbolic math, programming, or graphical features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDAs, or cameras may not be used during the exam or while the exam papers are present. Students may not use any tables, books, or collections of formulas.

We acknowledge the following people for their contributions to this year’s exam (in alphabetical order):

JiaJia Dong, Abijith Krishnan, Brian Skinner, and Kevin Zhou.

Instructions for the Student

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

At the beginning of the exam, you shall be provided with the instruction sheets, blank papers (both for your answers and scratch work), and the exam packet.
Work on Part A first. You have 90 minutes to complete three problems. Each question is worth 25 points, but they are not necessarily of the same difficulty. Do not look at Part B during this time.
After you have completed Part A you may take a break. You may consider checking your answers to Part A with the remaining time as you will not be allowed to return to Part A once you start Part B.
Then work on Part B. You have 90 minutes to complete three problems. Each question is worth 25 points. Do not look at Part A during this time.
Show your work and reasoning. Partial credit will be given if you make your reasoning clear. Do not write on the back of any page. Further guidance is given on the next page.
A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones, PDAs or cameras may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
In order to maintain exam security, do not communicate any information about the questions of this exam, or their solutions until after April 8th, 2019.

Possibly useful information. You may use this sheet for both parts of the exam.

$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ for $\vert \theta\vert \ll1$
$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2/kg^2}$
$k_m=\mu_0/(4\pi)$	$10^{-7}\ \mathrm{T\cdot m/A}$
$k_B$	$1.38\times10^{-23}\ \mathrm{J/K}$
$R=N_Ak_B$	$8.31\ \mathrm{J/(mol\cdot K)}$
$e$	$1.602\times10^{-19}\ \mathrm C$
$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}=4.14\times10^{-15}\ \mathrm{eV\cdot s}$
$(1+x)^n$	$\approx1+nx$ for $\vert x\vert \ll1$
$\cos\theta$	$\approx1-\frac12\theta^2$ for $\vert \theta\vert \ll1$

Instructions for the Student — Further Guidance

Following is some further guidance on formatting your solutions. Start each question on a new sheet of paper. Put your AAPT ID number, your proctor’s AAPT ID number, the problem number and the page number and total number of pages for this problem, in the upper right hand corner of each page. As an example, the second page of your solution to B3 might look as follows.

student AAPT #29485 proctor AAPT #1038 B3: 2/2

Remember to also write the AAPT ID numbers on the provided answer sheets. Write single- sided to facilitate scanning. You may use either pencil or pen, but in either case, make sure to write sufficiently clearly so your work will be legible after scanning. To preserve anonymity of grading, do not write your name on any sheet.

End of Instructions for the Student

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Important Instructions for the Exam Supervisor — DO NOT DISTRIBUTE THIS PAGE

Divide the exam paper into 4 parts: the instructions (pages 2–3), Part A (pages 4–12), Part B (pages 14–23), and answer sheets for one of the questions in Part A (pages 25–26). The exam should be printed single-sided to facilitate dividing the test and scanning the answer sheets.

Условие

Green Revolution[^1]

In this problem, we will investigate a simple thermodynamic model for the conversion of solar energy into wind. Consider a planet of radius $R$ , and assume that it rotates so that the same side always faces the Sun. The bright side facing the Sun has a constant uniform temperature $T_1$ , while the dark side has a constant uniform temperature $T_2$ . The orbit radius of the planet is $R_0$ , the Sun has temperature $T_s$ , and the radius of the Sun is $R_s$ . Assume that outer space has zero temperature, and treat all objects as ideal blackbodies.

[^1]: This question inspired by De Vos, Alexis, and Guust Flater, American Journal of Physics 59.8 (1991): 751-754.

a. Find the solar power $P$ received by the bright side of the planet. (Hint: the Stefan-Boltzmann law states that the power emitted by a blackbody with area $A$ is $\sigma AT^4$ .)

In order to keep both $T_1$ and $T_2$ constant, heat must be continually transferred from the bright side to the dark side. By viewing the two hemispheres as the two reservoirs of a reversible heat engine, work can be performed from this temperature difference, which appears in the form of wind power. For simplicity, we assume all of this power is immediately captured and stored by windmills.

b. The equilibrium temperature ratio $T_2/T_1$ depends on the heat transfer rate between the hemispheres. Find the minimum and maximum possible values of $T_2/T_1$ . In each case, what is the wind power $P_w$ produced?

c. Find the wind power $P_w$ in terms of $P$ and the temperature ratio $T_2/T_1$ .

d. Estimate the maximum possible value of $P_w$ as a fraction of $P$ , to one significant figure. Briefly explain how you obtained this estimate.

Решение

a.

The intensity of solar radiation at the surface of the sun is $\sigma T_s^4$ , so the intensity at the planet's orbit radius is

I=\sigma T_s^4\frac{R_s^2}{R_0^2}.

The area subtended by the planet is $\pi R^2$ , so

P=\pi\sigma T_s^4\frac{R^2R_s^2}{R_0^2}.

b.

The minimum value is simply zero; in this case zero heat is transferred to the dark side of the planet. Since no heat is transferred, the heat engine can't run, so $P_w=0$ . (To show this a bit more carefully, note that the entropy exhausted by the heat engine is $Q_2/T_2\propto T_2^3$ by the Stefan-Boltzmann law. In the limit $T_2\to0$ , the entropy out goes to zero, so the entropy in and hence the heat intake also goes to zero.)

The maximum value is $T_2/T_1=1$ . It cannot be any higher by the second law of thermodynamics. In this case, there is no temperature difference, so the heat engine has zero efficiency and $P_w=0$ . Power $P/2$ is simply transferred from the bright side to the dark side as heat.

c.

Let heat be transferred from the bright side at a rate $Q_1$ and transferred to the dark side at a rate $Q_2$ . Then by conservation of energy,

Q_1=P_w+Q_2.

Since the two hemispheres have constant temperatures, energy balance for each gives

P=Q_1+(2\pi R^2\sigma)T_1^4,\qquad Q_2=AT_2^4.

Finally, since the engine is reversible,

\frac{Q_1}{T_1}=\frac{Q_2}{T_2}.

By combining the first three equations, and defining $x=T_2/T_1$ , we have

P_w=Q_1-Q_2=P-(2\pi R^2\sigma)(T_1^4+T_2^4)=P-(2\pi R^2\sigma)T_1^4(1+x^4).

This is not yet in terms of $P$ and $x$ , so now we use the reversibility condition,

\frac{P-AT_1^4}{T_1}=\frac{AT_2^4}{T_2}

which simplifies to

P=(2\pi R^2\sigma)(T_1^4+T_1T_2^3)=(2\pi R^2\sigma)T_1^4(1+x^3).

Plugging this in above, we find

P_w=P-\frac P{1+x^3}(1+x^4)=\frac{x^3(1-x)}{1+x^3}P.

d.

There are many ways to get the required answer. For example, by sketching the function, one can see that there is a unique maximum at an intermediate value of $x$ , and furthermore that this maximum is at $x>0.5$ , because of the rapid rise of the $x^3$ factor. One could then compute $P_w/P$ with a calculator at a few trial values such as $x=0.5,0.7,0.9$ , which are already enough to get the desired accuracy.

The optimum value is $x=0.69$ , at which point

P_w^{\max}=0.077P.

Hence in this model, at most 7.7% of solar energy can be converted into wind energy. Any answer within 15% of this value was accepted.

Оригинал в Архива: 2019_USAPhO.pdf · официални решения: 2019_USAPHO_revised-solutions.pdf