2016 USA Physics Olympiad Exam AAPT UNITED STATES PHYSICS TEAM AIP 2016 USA Physics Olympiad Exam DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor 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 3-12),Part B(pages 14-19),and several answer sheets for one of the questions in Part A(pages 21-22).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. Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam 1 AAPT AIP 2016 UNITED STATES PHYSICS TEAM USA Physics Olympiad Exam DO NOT DISTRIBUTE THIS PAGE Important Instructions for the Exam Supervisor • 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 3-12), Part B (pages 14-19), and several answer sheets for one of the questions in Part A (pages 21-22). 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. Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Cover Sheet 2 AAPT UNITED STATES PHYSICS TEAM AIP 2016 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. g=9.8 N/kg G=6.67×10-11N·m2/kg2 k=1/4πe0=8.99×109N.m2/C2 km=40/4r=10-7T·m/A c=3.00×103m/s kB=1.38×10-23J/K NA=6.02×1023(mol)-1 R=NAkB =8.31 J/(mol.K) σ=5.67×10-8J/(s·m2.K4) e=1.602×10-19C 1eV=1.602×10-19J h=6.63×10-34J.s=4.14×10-15eV.s me=9.109×10-31kg=0.511MeV/c2(1+x)n≈1+na for≤1 sin0≈0-03forl0l<1 cos0≈1-号02forl0l<1 Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Cover Sheet 2 AAPT AIP 2016 UNITED STATES PHYSICS TEAM 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. g = 9.8 N/kg G = 6.67 × 10−11 N · m2/kg2 k = 1/4π0 = 8.99 × 109 N · m2/C 2 km = µ0/4π = 10−7 T · m/A c = 3.00 × 108 m/s kB = 1.38 × 10−23 J/K NA = 6.02 × 1023 (mol)−1 R = NAkB = 8.31 J/(mol · K) σ = 5.67 × 10−8 J/(s · m2 · K4 ) e = 1.602 × 10−19 C 1eV = 1.602 × 10−19 J h = 6.63 × 10−34 J · s = 4.14 × 10−15 eV · s me = 9.109 × 10−31 kg = 0.511 MeV/c 2 (1 + x) n ≈ 1 + nx for |x| 1 sin θ ≈ θ − 1 6 θ 3 for |θ| 1 cos θ ≈ 1 − 1 2 θ 2 for |θ| 1 Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A Part A Question Al The Doppler effect for a source moving relative to a stationary observer is described by fo f=1-(v/c)cos0 where f is the frequency measured by the observer,fo is the frequency emitted by the source,v is the speed of the source,c is the wave speed,and 6 is the angle between the source velocity and the line between the source and observer.(Thus 0=0 when the source is moving directly towards the observer and 6=when moving directly away.) A sound source of constant frequency travels at a constant velocity past an observer,and the observed frequency is plotted as a function of time: 450 000 448 00 00 oObserved Frequency 0 446 444 442 440 到 438 436 434 0 432 430 0 428 0 426 424 0 0 0 422 42 0 2 345 67 891011121314 Time (s) The experiment happens in room temperature air,so the speed of sound is 340 m/s. a.What is the speed of the source? Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 3 Part A Question A1 The Doppler effect for a source moving relative to a stationary observer is described by f = f0 1 − (v/c) cos θ where f is the frequency measured by the observer, f0 is the frequency emitted by the source, v is the speed of the source, c is the wave speed, and θ is the angle between the source velocity and the line between the source and observer. (Thus θ = 0 when the source is moving directly towards the observer and θ = π when moving directly away.) A sound source of constant frequency travels at a constant velocity past an observer, and the observed frequency is plotted as a function of time: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 420 422 424 426 428 430 432 434 436 438 440 442 444 446 448 450 Time (s) Frequency (Hz) Observed Frequency The experiment happens in room temperature air, so the speed of sound is 340 m/s. a. What is the speed of the source? Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 4 Solution For 0=0 we have a≈fo/1-v/c) and for0=π， f6=fo/(1+v/c): Read fa and fo off the early and late time portions of the graph and use fa/f6=(1+v/c)/(1-v/c) giving an answer of v=10.7 m/s. Alternatively,we can see that v<c and approximate a/f6≈1+2v/c which makes the calculation of v slightly faster.This is acceptable because the error terms are of order (v/c)2~0.1%. b.What is the smallest distance between the source and the observer? Solution Let d be the (fixed)distance between the observer and the path of the source;let x be the displacement along the path,with x=0 at closest approach.Then for d, cos0≈cot0=x/d so we have f fo/(1-(v/c)(x/d))fo(1+(v/c)(z/d)). Taking the time derivative,and noting that z'is simply v, f'=fo(v2/c)d Therefore we can read f'off the center region of the graph.We still need to find fo,which we can do using our result from part (a)or simply by averaging fa and fo,since v<c,giving fo =435 Hz and an answer of d =17.8 m. There's also a nice trick to speed up this computation.Draw lines at the asymptotic values and through the central data points.The two horizontal lines are 2fo(v/c)apart in frequency, so the time between their intersections with the third line is simply 2d/v. Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 4 Solution For θ = 0 we have fa ≈ f0/(1 − v/c) and for θ = π, fb = f0/(1 + v/c). Read fa and fb off the early and late time portions of the graph and use fa/fb = (1 + v/c)/(1 − v/c) giving an answer of v = 10.7 m/s. Alternatively, we can see that v c and approximate fa/fb ≈ 1 + 2v/c which makes the calculation of v slightly faster. This is acceptable because the error terms are of order (v/c) 2 ∼ 0.1%. b. What is the smallest distance between the source and the observer? Solution Let d be the (fixed) distance between the observer and the path of the source; let x be the displacement along the path, with x = 0 at closest approach. Then for |x| d, cos θ ≈ cot θ = x/d so we have f = f0/(1 − (v/c)(x/d)) ≈ f0(1 + (v/c)(x/d)). Taking the time derivative, and noting that x 0 is simply v, f 0 = f0(v 2 /c)d Therefore we can read f 0 off the center region of the graph. We still need to find f0, which we can do using our result from part (a) or simply by averaging fa and fb, since v c, giving f0 = 435 Hz and an answer of d = 17.8 m. There’s also a nice trick to speed up this computation. Draw lines at the asymptotic values and through the central data points. The two horizontal lines are 2f0(v/c) apart in frequency, so the time between their intersections with the third line is simply 2d/v. Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 5 Question A2 A student designs a simple integrated circuit device that has two inputs,Va and V,and two outputs, Vo and V.The inputs are effectively connected internally to a single resistor with effectively infinite resistance.The outputs are effectively connected internally to a perfect source of emf &The integrated circuit is configured so that &=G(Va-V),where G is a very large number somewhere between 107 and 109.The circuits below are chosen so that the precise value of G is unimportant. On the left is an internal schematic for the device;on the right is the symbol that is used in circuit diagrams. 0 Solution The key idea is that if is finite,then VaVo,since G is so large.If we work exactly,then the answers will contain terms like (V-Va)/G which are negligible.Thus we can find the same answers by just setting Va =Vo. a.Consider the following circuit.R1 =8.2 kn and R2 =560 are two resistors.Terminal g and the negative side of Vin are connected to ground,so both are at a potential of 0 volts. Determine the ratio Vout/Vin Vout Solution For this first part,we will not assume Va =V.Since terminal g is grounded,V=0 and Va =Vin,so Vout =G(Vin-V).No current runs between a and b,so any current through Ri also flows through R2.Then Ohm's law gives Vout Vout=G1 R2 → -Vout R1+R2 and solving for Vout gives Vou V But since G>RI/R2,we can neglect the 1/G term,giving Vout≈ 1+B2 a R2 This circuit is an amplifier with feedback. Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 5 Question A2 A student designs a simple integrated circuit device that has two inputs, Va and Vb, and two outputs, Vo and Vg. The inputs are effectively connected internally to a single resistor with effectively infinite resistance. The outputs are effectively connected internally to a perfect source of emf E. The integrated circuit is configured so that E = G(Va − Vb), where G is a very large number somewhere between 107 and 109 . The circuits below are chosen so that the precise value of G is unimportant. On the left is an internal schematic for the device; on the right is the symbol that is used in circuit diagrams. Va Vb Vo Vg a b o g Solution The key idea is that if E is finite, then Va ≈ Vb, since G is so large. If we work exactly, then the answers will contain terms like (Vb − Va)/G which are negligible. Thus we can find the same answers by just setting Va = Vb. a. Consider the following circuit. R1 = 8.2 kΩ and R2 = 560 Ω are two resistors. Terminal g and the negative side of Vin are connected to ground, so both are at a potential of 0 volts. Determine the ratio Vout/Vin. a b o g R1 R2 Vout Vin Solution For this first part, we will not assume Va = Vb. Since terminal g is grounded, Vg = 0 and Va = Vin, so Vout = G(Vin − Vb). No current runs between a and b, so any current through R1 also flows through R2. Then Ohm’s law gives Vb R2 = Vout R1 + R2 ⇒ Vout = G Vin − Vout R2 R1 + R2 and solving for Vout gives Vout = Vin 1 1 G + R2 R1+R2 . But since G R1/R2, we can neglect the 1/G term, giving Vout Vin ≈ R1 + R2 R2 . This circuit is an amplifier with feedback. Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 6 b.Consider the following circuit.All four resistors have identical resistance R.Determine Vout in terms of any or all of Vi,V2,and R. R R Vout R Solution For this part,we will assume Va =V.Again Va=0,and if current I flows through the bottom resistor(below the b and g terminals)then V2=2V,since the voltage drop across the bottom two resistors must be equal.Similarly,the voltage drop across the top two resistors is equal, so Vi+Vout =2Va.Then Vout 2Va-Vi=V2-Vi. This circuit is a subtractor. c.Consider the following circuit.The circuit has a capacitor C and a resistor R with time constant RC =r.The source on the left provides variable,but bounded voltage.Assume Vin is a function of time.Determine Vout as a function of Vin,and any or all of time t and T. R Vout Solution We again set Va=V=0.Then the capacitor charge and current satisfy Q=CVin, 9--6u R where the second result follows from Ohm's law.Then → Vin R dt Vout=-T- dt This circuit is a differentiator. Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 6 b. Consider the following circuit. All four resistors have identical resistance R. Determine Vout in terms of any or all of V1, V2, and R. a b o g R R Vout V1 V2 R R Solution For this part, we will assume Va = Vb. Again Vg = 0, and if current I flows through the bottom resistor (below the b and g terminals) then V2 = 2Vb, since the voltage drop across the bottom two resistors must be equal. Similarly, the voltage drop across the top two resistors is equal, so V1 + Vout = 2Va. Then Vout = 2Va − V1 = V2 − V1. This circuit is a subtractor. c. Consider the following circuit. The circuit has a capacitor C and a resistor R with time constant RC = τ . The source on the left provides variable, but bounded voltage. Assume Vin is a function of time. Determine Vout as a function of Vin, and any or all of time t and τ . a b o g R Vout Vin C Solution We again set Va = Vb = 0. Then the capacitor charge and current satisfy Q = CVin, Q˙ = − Vout R where the second result follows from Ohm’s law. Then Vout R = −C dVin dt ⇒ Vout = −τ dVin dt . This circuit is a differentiator. Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A Question A3 Throughout this problem the inertial rest frame of the rod will be referred to as the rod's frame, while the inertial frame of the cylinder will be referred to as the cylinder's frame. A rod is traveling at a constant speed of v=c to the right relative to a hollow cylinder.The rod passes through the cylinder,and then out the other side.The left end of the rod aligns with the left end of the cylinder at time t=0 and x =0 in the cylinder's frame and time t'=0 and x'=0 in the rod's frame. The left end of the rod aligns with the left end of the cylinder at the same time as the right end of the rod aligns with the right end of the cylinder in the cylinder's frame;in this reference frame the length of the cylinder is 15 m. For the following,sketch accurate,scale diagrams of the motions of the ends of the rod and the cylinder on the graphs provided.The horizontal axis corresponds to t,the vertical axis corresponds to ct,where c is the speed of light.Both the vertical and horizontal gridlines have 5.0 meter spacing. a.Sketch the world lines of the left end of the rod(RL),left end of the cylinder (CL),right end of the rod(RR),and right end of the cylinder(CR)in the cylinder's frame. b.Do the same in the rod's frame. c.On both diagrams clearly indicate the following four events by the letters A,B,C,and D. A:The left end of the rod is at the same point as the left end of the cylinder B:The right end of the rod is at the same point as the right end of the cylinder C:The left end of the rod is at the same point as the right end of the cylinder D:The right end of the rod is at the same point as the left end of the cylinder d.At event B a small particle P is emitted that travels to the left at a constant speed vp=e in the cylinder's frame. i.Sketch the world line of P in the cylinder's frame. ii.Sketch the world line of P in the rod's frame. e.Now consider the following in the cylinder's frame.The right end of the rod stops instanta- neously at event B and emits a flash of light,and the left end of the rod stops instantaneously when the light reaches it.Determine the final length of the rod after it has stopped.You can assume the rod compresses uniformly with no other deformation. Any computation that you do must be shown on a separate sheet of paper,and not on the graphs.Graphical work that does not have supporting computation might not receive full credit. Solution The graphs are shown below,where the yellow line is the particle P and the green line is the flash of light.Solutions that used Galilean relativity received partial credit,as long as they were self-consistent.The final length of the rod is simply the distance between the line CR and the intersection of RL and the green line,i.e.25/3 m.There's no need to apply length contraction,as we're already in the rest frame of the rod at this point.Nonetheless,the way we have chosen to stop the rod has squeezed it shorter. Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 7 Question A3 Throughout this problem the inertial rest frame of the rod will be referred to as the rod’s frame, while the inertial frame of the cylinder will be referred to as the cylinder’s frame. A rod is traveling at a constant speed of v = 4 5 c to the right relative to a hollow cylinder. The rod passes through the cylinder, and then out the other side. The left end of the rod aligns with the left end of the cylinder at time t = 0 and x = 0 in the cylinder’s frame and time t 0 = 0 and x 0 = 0 in the rod’s frame. The left end of the rod aligns with the left end of the cylinder at the same time as the right end of the rod aligns with the right end of the cylinder in the cylinder’s frame; in this reference frame the length of the cylinder is 15 m. For the following, sketch accurate, scale diagrams of the motions of the ends of the rod and the cylinder on the graphs provided. The horizontal axis corresponds to x, the vertical axis corresponds to ct, where c is the speed of light. Both the vertical and horizontal gridlines have 5.0 meter spacing. a. Sketch the world lines of the left end of the rod (RL), left end of the cylinder (CL), right end of the rod (RR), and right end of the cylinder (CR) in the cylinder’s frame. b. Do the same in the rod’s frame. c. On both diagrams clearly indicate the following four events by the letters A, B, C, and D. A: The left end of the rod is at the same point as the left end of the cylinder B: The right end of the rod is at the same point as the right end of the cylinder C: The left end of the rod is at the same point as the right end of the cylinder D: The right end of the rod is at the same point as the left end of the cylinder d. At event B a small particle P is emitted that travels to the left at a constant speed vP = 4 5 c in the cylinder’s frame. i. Sketch the world line of P in the cylinder’s frame. ii. Sketch the world line of P in the rod’s frame. e. Now consider the following in the cylinder’s frame. The right end of the rod stops instantaneously at event B and emits a flash of light, and the left end of the rod stops instantaneously when the light reaches it. Determine the final length of the rod after it has stopped. You can assume the rod compresses uniformly with no other deformation. Any computation that you do must be shown on a separate sheet of paper, and not on the graphs. Graphical work that does not have supporting computation might not receive full credit. Solution The graphs are shown below, where the yellow line is the particle P and the green line is the flash of light. Solutions that used Galilean relativity received partial credit, as long as they were self-consistent. The final length of the rod is simply the distance between the line CR and the intersection of RL and the green line, i.e. 25/3 m. There’s no need to apply length contraction, as we’re already in the rest frame of the rod at this point. Nonetheless, the way we have chosen to stop the rod has squeezed it shorter. Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 8 The Cylinder's Frame ct. C RL RR Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 8 The Cylinder’s Frame x ctCL CR RL RR Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 9 The Rod's Frame RR 2 CR CL Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 9 The Rod’s Frame x 0 ct0 RL RR CL CR x Copyright c 2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 10 Question A4 The flow of heat through a material can be described via the thermal conductivity k.If the two faces of a slab of material with thermal conductivity K,area A,and thickness d are held at temperatures differing by AT,the thermal power P transferred through the slab is P=AAT 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 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 C by the cold air (an infinitely large constant temperature heat sink).The bottom of the lake remains at a fixed 4.0C 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. Solution 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 contri- butions.The water right at the bottom of the ice is at 0 Co.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 Co/5 m.Multiplying by the conductivity,we get a power of P。=4C0.57W=0.456w/m2 5m mCo delivered through the water.The same calculation for the ice gives power B=10C22W =2200W/m2 .01 mm.Co delivered through the ice.Thus P is negligible and can be ignored. Now,each square meter of water directly underneath the ice loses 2200J of energy per second. That is enough energy to freeze 2200W/(330,000J/kg)=6.7×10-3kg/s Copyright C2016 American Association of Physics Teachers
2016 USA Physics Olympiad Exam Part A 10 Question A4 The flow of heat through a material can be described via the thermal conductivity κ. If the two faces of a slab of material with thermal conductivity κ, area A, and thickness d are held at temperatures differing by ∆T, the thermal power P transferred through the slab is P = κA∆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 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 ◦C by the cold air (an infinitely large constant temperature heat sink). The bottom of the lake remains at a fixed 4.0 ◦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. Solution 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 C◦ . 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 C◦/5 m. Multiplying by the conductivity, we get a power of Pw = 4 C◦ 5 m 0.57 W mC◦ = 0.456 W/m2 delivered through the water. The same calculation for the ice gives power Pi = 10 C◦ .01 m 2.2 W m · C◦ = 2200 W/m2 delivered through the ice. Thus Pw is negligible and can be ignored. Now, each square meter of water directly underneath the ice loses 2200 J of energy per second. That is enough energy to freeze 2200 W/(330, 000 J/kg) = 6.7 × 10−3 kg/s Copyright c 2016 American Association of Physics Teachers