Съдържание

USAPhO 2017, Полуфинал — Задача 2. Question A2

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2017

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

Внимание: Бележка към темата

The source gives different exam-security dates on its supervisor and student pages. Printed key inconsistencies at A1, A4(b), B1(c) and B2(f)/(g) are retained with their native wording and mathematics; no silent corrected formula is substituted. Exact original PDFs and detailed source-defect evidence accompany this transcription.

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

This examination consists of two parts: Part A has four questions and is allowed 90 minutes; Part B has two questions and is allowed 90 minutes.
The first page that follows is a cover sheet. Examinees may keep the cover sheet for both parts of the exam.
The parts are then identified by the center header on each page. Examinees are only allowed to do one part at a time, and may not work on other parts, even if they have time remaining.
Allow 90 minutes to complete Part A. Do not let students look at Part B. Collect the answers to Part A before allowing the examinee to begin Part B. Examinees are allowed a 10 to 15 minutes break between parts A and B.
Allow 90 minutes to complete Part B. Do not let students go back to Part A.
Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-6), Part B (pages 8-10). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!
The supervisor must collect all examination questions, including the cover sheet, at the end of the exam, as well as any scratch paper used by the examinees. Examinees may not take the exam questions. The examination questions may be returned to the students after April 15, 2017.
Examinees are allowed calculators, but they may not use symbolic math, programming, or graphic features of these calculators. Calculators may not be shared and their memory must be cleared of data and programs. Cell phones, PDA’s or cameras may not be used during the exam or while the exam papers are present. Examinees may not use any tables, books, or collections of formulas.

USA Physics Olympiad Exam — Instructions

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN - Work Part A first. You have 90 minutes to complete all four problems. Each question is worth 25 points. Do not look at Part B during this time. - After you have completed Part A you may take a break. - Then work Part B. You have 90 minutes to complete both problems. Each question is worth 50 points. Do not look at Part A during this time. - Show all your work. Partial credit will be given. Do not write on the back of any page. Do not write anything that you wish graded on the question sheets. - Start each question on a new sheet of paper. Put your proctor’s AAPT ID, your AAPT ID, your name, the question number and the page number/total pages for this problem, in the upper right hand corner of each page. For example, Doe, Jamie student AAPT ID # proctor AAPT ID # A1 - 1/3 - 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, PDA’s 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. - Questions with the same point value are not necessarily of the same difficulty. - In order to maintain exam security, do not communicate any information about the questions (or their answers/solutions) on this contest until after April 8, 2017.

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 $\vert \theta\vert \ll1$
$m_p$	$1.673\times10^{-27}\ \mathrm{kg}=938\ \mathrm{MeV}/c^2$
$\ln(1+x)$	$\approx x$ for $\vert x\vert \ll1$

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

Ideally the test supervisor will divide the question paper into 3 parts: the cover sheet (page 2), Part A (pages 3-13), Part B (pages 15-23). Examinees should be provided parts A and B individually, although they may keep the cover sheet. The answer sheets should be printed single sided!

Условие

Consider two objects with equal heat capacities $C$ and initial temperatures $T_1$ and $T_2$ . A Carnot engine is run using these objects as its hot and cold reservoirs until they are at equal temperatures. Assume that the temperature changes of both the hot and cold reservoirs is very small compared to the temperature during any one cycle of the Carnot engine.

a. Find the final temperature $T_f$ of the two objects, and the total work $W$ done by the engine.

Now consider three objects with equal and constant heat capacity at initial temperatures $T_1=100\ \mathrm K$ , $T_2=300\ \mathrm K$ , and $T_3=300\ \mathrm K$ . Suppose we wish to raise the temperature of the third object.

To do this, we could run a Carnot engine between the first and second objects, extracting work $W$ . This work can then be dissipated as heat to raise the temperature of the third object. Even better, it can be stored and used to run a Carnot engine between the first and third object in reverse, which pumps heat into the third object.

Assume that all work produced by running engines can be stored and used without dissipation.

b. Find the minimum temperature $T_L$ to which the first object can be lowered.

c. Find the maximum temperature $T_H$ to which the third object can be raised.

Решение

a.

Since a Carnot engine is reversible, it produces no entropy,

dS_1+dS_2=\frac{dQ_1}{T_1}+\frac{dQ_2}{T_2}=0.

By the definition of heat capacity, $dQ_i=CdT_i$ , so

\frac{dT_1}{T_1}=-\frac{dT_2}{T_2}.

Integrating this equation shows that $T_1T_2$ is constant, so the final temperature is

T_f=\sqrt{T_1T_2}.

The change in thermal energy of the objects is

C(T_f-T_1)+C(T_f-T_2)=C\left[2\sqrt{T_1T_2}-T_1-T_2\right].

By the First Law of Thermodynamics, the missing energy has been used to do work, so

W=C\left[T_1+T_2-2\sqrt{T_1T_2}\right].

b.

By the Second Law of Thermodynamics, we must have $T_L=100\ \mathrm K$ . Otherwise, we would have a process whose sole effect was a net transfer of heat from a cold body to a warm one.

c.

The entropy of an object with constant heat capacity is

S=\int\frac{dQ}{T}=C\int\frac{dT}{T}=C\ln T.

Since the total entropy remains constant, $T_1T_2T_3$ is constant; this is a direct generalization of the result for $T_f$ found in part (a). Energy is also conserved, as it makes no sense to leave stored energy unused, so $T_1+T_2+T_3$ is constant.

When one object is at temperature $T_H$ , the other two must be at the same lower temperature $T_0$ , or else further work could be extracted from their temperature difference, so

T_1+T_2+T_3=T_H+2T_0,\qquad T_1T_2T_3=T_HT_0^2.

Plugging in temperatures with values divided by 100 for convenience, and eliminating $T_0$ gives

T_H(7-T_H)^2=36.

We know that $T_H=1$ is one (spurious) solution, since this is the minimum possible final temperature as found in part (b). The other roots are $T_H=4$ and $T_H=9$ by the quadratic formula. The solution $T_H=9$ is impossible by energy conservation, so

T_H=400\ \mathrm K.

It is also possible to solve the problem more explicitly. For example, one can run a Carnot cycle between the first two objects until they are at the same temperature, then run a Carnot cycle in reverse between the last two objects using the stored work. At this point, the first two objects will no longer be at the same temperature, so we can repeat the procedure; this yields an infinite series for $T_H$ . Some students did this, and took only the first term of the series. This yields a fairly good approximation of $T_H\approx395\ \mathrm K$ .

Another explicit method is to continuously switch between running one Carnot engine forward and another Carnot engine in reverse; this yields three differential equations for $T_1$ , $T_2$ , and $T_3$ . Solving the equations and setting $T_1=T_2$ yields $T_3=T_H$ .

Exam Statistics

Question	A1	A2	A3	A4	B1	B2	Total
Mean	11	3	5	6	12	15	53
Standard Deviation	7	4	5	7	10	14	29
Maximum	25	21	25	25	42	50	163
Upper Quartile	18	4	9	8	18	26	73
Median	10	2	5	5	10	12	49
Lower Quartile	4	0	0	0	3	1	30
Minimum	0	0	0	0	0	0	5

Some Trivia:

California had 121 test takers
New Jersey had 42
Texas had 38
Florida, Illinois, Massachusetts, Maryland, New York, and Virginia each had between one dozen and two dozen test takers
Alabama, Alaska, Colorado, Idaho, Iowa, Louisiana, Montana, North Dakota, South Dakota, and Wyoming did not have a test taker this year.

Оригинал в Архива: 2017-USAPhO-exam-2.pdf · официални решения: USAPhO-2017-Solutions.pdf