2016 USA Physics Olympiad Exam Part A 11 of water into ice.Converting to volume,we have (6.7×10-3kg/s)/(920kg/m3)=7.2×10-6m3/s of ice formed for each square meter of ice,which means the ice is growing at a rate r=7.2×10-6m/s=2.6cm/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 kg.A power of 2200 W is enough to lift this ice about 24 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,=(-)证，x∈o,4 where d is the thickness and T=-10 Co.As the thickness d increases,all of the ice must decrease slightly in temperature to maintain a linear temperature gradient, 阳-系如 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 Co,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.Assuming the air stays at the same temperature for a long time,find the equilibrium thickness of the ice. Solution This part is independent of the previous part.For convenience,define ho to be the depth of the lake if all the water were in liquid form.Accounting for the centimeter of ice,ho=5.01 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, △T Ki hw hi Since the thickness of the water is hw,the amount of water that has frozen into ice had a thickness of ho-h.Setting the mass of water frozen equal to the mass of the ice, hipi=(ho-hw)pw→hr= hopw-hipi Pw Copyright C2016 American Association of Physics Teachers2016 USA Physics Olympiad Exam Part A 11 of water into ice. Converting to volume, we have (6.7 × 10−3 kg/s)/(920 kg/m3 ) = 7.2 × 10−6 m3 /s of ice formed for each square meter of ice, which means the ice is growing at a rate r = 7.2 × 10−6 m/s = 2.6 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 kg. A power of 2200 W is enough to lift this ice about 24 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) =  1 − x d  δT, x ∈ [0, d] where d is the thickness and δT = −10 C◦ . As the thickness d increases, all of the ice must decrease slightly in temperature to maintain a linear temperature gradient, ∂T ∂d = x d 2 δT. 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 C◦ , 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. Assuming the air stays at the same temperature for a long time, find the equilibrium thickness of the ice. Solution This part is independent of the previous part. For convenience, define h0 to be the depth of the lake if all the water were in liquid form. Accounting for the centimeter of ice, h0 = 5.01 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, ∆Tw hw κw = ∆Ti hi κi . Since the thickness of the water is hw, the amount of water that has frozen into ice had a thickness of h0 − hw. Setting the mass of water frozen equal to the mass of the ice, hiρi = (h0 − hw)ρw ⇒ hw = h0ρw − hiρi ρw . Copyright c 2016 American Association of Physics Teachers