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USAPhO 2016, Полуфинал — Задача 4. Question A4

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Всички материали от USAPhO 2016

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

Внимание: Бележка към темата

The native worked key contains printed mathematical inconsistencies at A1(b), B1 and B2(a)(iii)/(c)(ii); the source formulas and adjacent reasoning are retained as printed. The A3 key graphs omit the requested event-letter labels and do not draw the stopped-rod world-line changes; no new answer labels or lines have been invented. Exact originals 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 4 parts: the cover sheet (page 2), Part A (pages 3-6), Part B (pages 8-9), and several answer sheets for one of the questions in Part A (pages 11-12). 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, 2016.
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 AAPT ID number, 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,

AAPT ID # Doe, Jamie 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 15, 2016.

Possibly Useful Information. You may use this sheet for both parts of the exam.

Information	Value
$g$	$9.8\ \mathrm{N/kg}$
$k=1/(4\pi\varepsilon_0)$	$8.99\times10^9\ \mathrm{N\cdot m^2/C^2}$
$c$	$3.00\times10^8\ \mathrm{m/s}$
$N_A$	$6.02\times10^{23}\ (\mathrm{mol})^{-1}$
$\sigma$	$5.67\times10^{-8}\ \mathrm{J/(s\cdot m^2\cdot K^4)}$
$1\ \mathrm{eV}$	$1.602\times10^{-19}\ \mathrm J$
$m_e$	$9.109\times10^{-31}\ \mathrm{kg}=0.511\ \mathrm{MeV}/c^2$
$\sin\theta$	$\approx\theta-\frac16\theta^3$ 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 $\vert \theta\vert \ll1$

Условие

The flow of heat through a material can be described via the thermal conductivity $\kappa$ . If the two faces of a slab of material with thermal conductivity $\kappa$ , area $A$ , and thickness $d$ are held at temperatures differing by $\Delta T$ , the thermal power $P$ transferred through the slab is

P=\frac{\kappa A\Delta T}{d}

A large, flat lake in the upper Midwest has a uniform depth of $5.0$ meters of water that is covered by a uniform layer of $1.0\ \mathrm{cm}$ of ice. Cold air has moved into the region so that the upper surface of the ice is now maintained at a constant temperature of $-10\ ^\circ\mathrm C$ by the cold air (an infinitely large constant temperature heat sink). The bottom of the lake remains at a fixed $4.0\ ^\circ\mathrm C$ because of contact with the earth (an infinitely large constant temperature heat source). It is reasonable to assume that heat flow is only in the vertical direction and that there is no convective motion in the water.

a. Determine the initial rate of change in ice thickness.

b. Assuming the air stays at the same temperature for a long time, find the equilibrium thickness of the ice.

c. Explain why convective motion can be ignored in the water.

Some important quantities for this problem:

Specific heat capacity of water	$C_{\mathrm{water}}$	$4200\ \mathrm{J/(kg\cdot C^\circ)}$
Specific heat capacity of ice	$C_{\mathrm{ice}}$	$2100\ \mathrm{J/(kg\cdot C^\circ)}$
Thermal conductivity of water	$\kappa_{\mathrm{water}}$	$0.57\ \mathrm{W/(m\cdot C^\circ)}$
Thermal conductivity of ice	$\kappa_{\mathrm{ice}}$	$2.2\ \mathrm{W/(m\cdot C^\circ)}$
Latent heat of fusion for water	$L_f$	$330,000\ \mathrm{J/kg}$
Density of water	$\rho_{\mathrm{water}}$	$999\ \mathrm{kg/m^3}$
Density of ice	$\rho_{\mathrm{ice}}$	$920\ \mathrm{kg/m^3}$

STOP: Do Not Continue to Part B

If there is still time remaining for Part A, you should review your work for Part A, but do not continue to Part B until instructed by your exam supervisor.

Решение

a.

The main effect is that the ice radiates heat into the air due to the temperature gradient through it, and this freezes the water next to the ice. However, there are many other effects that slightly change the answer.

i. There is another contribution to the thermal power from the temperature gradient in the water.

ii. As the water freezes, it lifts the ice above it.

iii. When a layer of water freezes into ice, all of the other water and ice becomes slightly colder.

The first point should be addressed for full credit. To do this, we will calculate both contributions. The water right at the bottom of the ice is at $0\ \mathrm C^\circ$ . The temperature gradients in the water and ice are both uniform since the system is in quasi-equilibrium; physically, if the temperature gradient were not uniform, there would be a net flow of heat to or away from some regions, quickly making the gradient uniform again.

The temperature gradient in the water is $4\ \mathrm C^\circ/5\ \mathrm m$ . Multiplying by the conductivity, we get a power of

P_w=\frac{4\ \mathrm C^\circ}{5\ \mathrm m}\frac{0.57\ \mathrm W}{\mathrm{m C^\circ}}=0.456\ \mathrm{W/m^2}

delivered through the water. The same calculation for the ice gives power

P_i=\frac{10\ \mathrm C^\circ}{.01\ \mathrm m}\frac{2.2\ \mathrm W}{\mathrm{m\cdot C^\circ}}=2200\ \mathrm{W/m^2}

delivered through the ice. Thus $P_w$ is negligible and can be ignored.

Now, each square meter of water directly underneath the ice loses $2200\ \mathrm J$ of energy per second. That is enough energy to freeze

2200\ \mathrm W/(330,000\ \mathrm{J/kg})=6.7\times10^{-3}\ \mathrm{kg/s}

of water into ice. Converting to volume, we have

\left(6.7\times10^{-3}\ \mathrm{kg/s}\right)/\left(920\ \mathrm{kg/m^3}\right)=7.2\times10^{-6}\ \mathrm{m^3/s}

of ice formed for each square meter of ice, which means the ice is growing at a rate

r=7.2\times10^{-6}\ \mathrm{m/s}=2.6\ \mathrm{cm/hr}.

Next, we will account for the second and third points; these are not necessary for full credit. First consider the rising of the water. Each square meter of ice initially weighs $9.2\ \mathrm{kg}$ . A power of $2200\ \mathrm W$ is enough to lift this ice about $24\ \mathrm{m/s}$ against gravity. In reality, the ice is lifted at a much slower rate, so this accounts for a negligible portion of the energy.

The third point requires some more explanation. In an appropriate coordinate system, the temperature profile of the ice is

T(x,d)=\left(1-\frac xd\right)\delta T,\quad x\in[0,d]

where $d$ is the thickness and $\delta T=-10\ \mathrm C^\circ$ . As the thickness $d$ increases, all of the ice must decrease slightly in temperature to maintain a linear temperature gradient,

\frac{\partial T}{\partial d}=\frac x{d^2}\delta T.

By drawing a graph, one can see this contribution is equal to the heat that would be needed to cool the new ice formed by $5\ \mathrm C^\circ$ , which gives a $3\%$ correction to the answer. There is also a similar contribution from cooling the water, which is negligible. Finally, we neglected the sublimation of the ice.

b.

This part is independent of the previous part. For convenience, define $h_0$ to be the depth of the lake if all the water were in liquid form. Accounting for the centimeter of ice, $h_0=5.01\ \mathrm m$ to the number of significant digits we're using.

The ice will stop getting thicker when the energy flux through the water equals that through the ice,

\frac{\Delta T_w}{h_w}\kappa_w=\frac{\Delta T_i}{h_i}\kappa_i.

Since the thickness of the water is $h_w$ , the amount of water that has frozen into ice had a thickness of $h_0-h_w$ . Setting the mass of water frozen equal to the mass of the ice,

h_i\rho_i=(h_0-h_w)\rho_w\quad\Rightarrow\quad h_w=\frac{h_0\rho_w-h_i\rho_i}{\rho_w}.

Plugging this into the previous expression gives

\frac{\Delta T_w\kappa_w\rho_w}{h_0\rho_w-h_i\rho_i}=\frac{\Delta T_i\kappa_i}{h_i}.

Solving for $h_i$ and plugging in numbers,

h_i=h_0\frac{\Delta T_i\kappa_i\rho_w}{\Delta T_w\kappa_w\rho_w+\Delta T_i\kappa_i\rho_i}=4.89\ \mathrm m.

c.

Convection occurs when boiling a pot of water because the hot water at the bottom of the pot has lower density than the colder water higher up. This means gravitational energy can be released when that hot, low-density water rises and cold, high-density water falls. When the hot water rises, it releases heat, cools, gets denser, and falls back down again, in a convection cycle. This phenomenon relies on the hotter water having lower density.

However, water reaches its maximum density at $4\ \mathrm C^\circ$ , so the water at the bottom of the lake, though warmer, is more dense than the water above it. Convection does not occur because moving the water around vertically would not release any gravitational potential energy.

Оригинал в Архива: USAPhO-2016.pdf · официални решения: USAPhO-2016-Solutions.pdf