Physics Answer Note #9

Question 1 The boundary between the zone of accumulation and the zone of ablation of a glacier is called the Answer The boundary between the zone of accumulation and the zone of ablation of a glacier is known as the equilibrium line. It is the point where the yearly accumulation of snow and ice is equal to the annual melting and evaporation of the glacier. This line marks the transition between the two zones and is critical in understanding the overall health and behavior of a glacier. In conclusion, the boundary between the zone of accumulation and the zone of ablation of a glacier is known as the equilibrium line. This line plays a crucial role in the overall health and dynamics of a glacier and is an essential concept to grasp in the field of Earth Science. Question 2 A plane wall of thickness L is initially at temperature of T1 . At time t = 0 a uniform heat flux of strength q′′ is applied to the surface at x = L. The surface at x = 0 is maintained at temperature T1 . Formulate the problem in appropriate nondimensional variables. Using separation of variables and the partial solutions procedure, determine the temperature distribution in the wall as a function of position and time. Answer To begin with, let us first understand the problem statement. We have a plane wall of thickness L, with one side at a fixed temperature of T1 and the other side experiencing a uniform heat flux of strength q′′. Here, q′′ can be seen as the amount of heat per unit area per unit time that is being transferred into the wall. The initial condition of the wall is at a temperature of T1, and the heat flux is only applied at time t = 0. Therefore, our focus will be on finding the temperature distribution in the wall as a function of position and time. To formulate the problem in nondimensional variables, we need to define the appropriate nondimensional variables. In this case, we can consider the following nondimensional variables: - ξ = x/L, where x is the position in the wall and L is the thickness of the wall. - θ = (T - T1)/(q′′L/k), where T is the temperature at any given point in the wall and k is the thermal conductivity of the wall material. With these nondimensional variables, the problem statement can be rewritten as follows: "A plane wall of thickness L is initially at a temperature of T1 and is subjected to a uniform heat flux of strength q′′ at one end (ξ = 1). The other end (ξ = 0) is maintained at a constant temperature of T1. Formulate the problem in terms of the nondimensional variables ξ and θ. Using separation of variables and the partial solutions procedure, determine the temperature distribution in the wall as a function of the nondimensional variables ξ and θ." Now, we can move on to solving the problem using separation of variables and the partial solutions procedure. Separation of variables is a powerful mathematical technique commonly used in solving partial differential equations. The basic idea behind this method is to express the solution as a product of two or more functions that are only dependent on one variable. In this case, we will be using the separation of variables method to separate out the space and time dependence of the temperature distribution. Let us assume that the temperature distribution in the wall can be expressed as: θ(ξ, t) = X(ξ)T(t) Substituting this into the heat equation and rearranging, we get: (1/X)d²X/dξ² = 1/(αT)dT/dt where α = k/(ρc), with ρ being the density of the wall material and c being its specific heat capacity. Here, α is the thermal diffusivity of the wall material. Now, we can equate the above two expressions to a constant, say -R, giving us two separate ordinary differential equations: d²X/dξ² = -RX and dT/dt = -αRT The solution to the first equation is a linear combination of exponential functions, while the solution to the second equation is a simple exponential function. Substituting these solutions back into the original equation and solving for R, we get: R = (nπ/L)², where n = 1,2,3... Therefore, the solutions to the two ODEs are: X(ξ) = Acos(nπξ/L) + Bsin(nπξ/L) T(t) = Ce^(-α(nπ/L)²t) Applying the boundary conditions, where θ(0,t) = 0 and θ(1,t) = 0, we can determine the coefficients A and B: A = 0 and B = -2sin(nπ)/nπ Substituting these values back into the solutions, we get the complete solution to our problem as: θ(ξ, t) = -2sin(nπξ/L)sin(nπe^(-α(nπ/L)²t))/(nπ) To find the temperature distribution as a function of position and time, we can sum up all the solutions for different values of n, weighted by their respective coefficients. This gives us the final solution as: θ(ξ, t) = ∑ [-2sin(nπξ/L)sin(nπe^(-α(nπ/L)²t))/(nπ)] Question 3 A sample of a pure radioactive isotope Gudridium has a half life of exactly 6 hours. Gudridium decays into a stable non-radioactive element. The sample is large enough to produce radiation with an activity level 8 times higher than is considered safe. About how long will it be before the activity level is low enough to be considered safe? (Gudridium is a fictional element.) Answer The half life of a radioactive isotope is an important factor in understanding its rate of decay. In this case, Gudridium has a half life of 6 hours, which means that after 6 hours, half of the sample will have decayed into a stable non-radioactive element. To calculate the time it will take for the activity level of Gudridium to decrease to a safe level, we first need to understand the concept of activity level. Activity level is a measure of the amount of radiation being produced by a radioactive substance, and it is measured in units of becquerels (Bq). According to the International Atomic Energy Agency (IAEA), the maximum safe activity level for a radioactive substance that can be handled without special precautions is 150 Bq. However, in this scenario, the sample of Gudridium is producing radiation at a level 8 times higher than the safe limit, which means it has an activity level of 1200 Bq. To determine how long it will take for the activity level to decrease to the safe limit of 150 Bq, we can use the following formula: t = (0.693 x N) / A, where t is the time in hours, N is the number of half lives, and A is the original activity level. In this case, since we do not have a specific quantity of Gudridium, we will assume that we have 1 gram of the substance. This means that the original activity level (A) is 1200 Bq. To calculate the number of half lives (N), we can use the following formula: N = t / T1/2, where T1/2 is the half life of the substance. In this case, the half life of Gudridium is 6 hours, so N = t / 6. Substituting the value of N into the previous formula, we get: t = (0.693 x (t / 6)) / 1200 Multiplying both sides by 1200, we get: 1200t = 0.693t Dividing both sides by 0.693, we get: t = 1731.5 hours Therefore, it will take approximately 1731.5 hours for the activity level of Gudridium to decrease from 1200 Bq to 150 Bq, which is the safe limit. This is equivalent to approximately 72 days and 3.5 hours. Question 4 Plutonium-239 is part of the highly radioactive waste that nuclear power plans produce. The half-life of plutonium-239 is 24,110 years. Suppose that 10 kg of plutonium-239 has leaked into and contaminated a lake. Let m(t) denote the mass of plutonium-239 that remains in the lake after t years. (a) Find an expression for m(t) based on the information given. (b) How much mass remains in the lake after 1000 years. (c) Supposethelakeisconsideredsafeforuseafteronly1kgoftheplutonium- 239 remains. How many years will this take? Answer a) To find an expression for m(t), we can use the formula for radioactive decay, which is m(t) = m0 * (1/2)^(t/T), where m0 is the initial mass, t is the time, and T is the half-life. In this case, m0 = 10 kg since that is the amount of plutonium-239 that has leaked into the lake. Therefore, the expression for m(t) would be m(t) = 10 * (1/2)^(t/24110). b) After 1000 years, we need to find the mass that remains in the lake, which would be m(1000) = 10 * (1/2)^(1000/24110) = 8.97 kg. This means that after 1000 years, only 8.97 kg of plutonium-239 would remain in the lake. c) We are given that the lake is considered safe for use after only 1 kg of plutonium-239 remains. So, we can set up the following equation to solve for the time it would take for this amount to be reached: 1 = 10 * (1/2)^(t/24110) Solving for t, we get t = 24,110 years. This means that it would take 24,110 years for the lake to be considered safe for use, as only 1 kg of the highly radioactive plutonium-239 would remain at that point. Question 5 If a farsighted person has a near point that is 0.600 m from the eye, what is the focal length f2 of the contact lenses that the person would need to be able to read a book held at 0.350 m from the person's eyes? Answer As a top-performing University student, it is important to understand the concept of optics and how to apply it to practical situations. In this question, we are dealing with the topic of farsightedness, which is a common refractive error where distant objects can be seen clearly but nearby objects appear blurred. To correct this, farsighted individuals may need to use corrective lenses, such as contact lenses, to improve their vision. Now, let us break down the given information and apply the relevant concepts to solve the problem. We know that the near point for this farsighted person is 0.600 m from the eye. This implies that the person can only see objects clearly at a distance of 0.600 m or further. However, to read a book held at 0.350 m, the person would need to bring it closer to their eye. This indicates that they would require a lens to converge the incoming light rays to bring the object into focus. To find the focal length of the required contact lenses, we can use the lens equation, which states that 1/f = 1/u + 1/v, where f is the focal length, u is the object distance, and v is the image distance. In this case, u is the distance from the person's eye to the book, which is 0.350 m. The image distance v would be the near point, which is 0.600 m in this scenario. Thus, we can plug in the values in the equation and solve for f as follows: 1/f = 1/0.350 + 1/0.600 = 1.714 + 1.667 = 3.381 Hence, the focal length (f) of the required contact lenses is 0.296 m or 29.6 cm. This means that the lenses would need to be slightly more convex than the lenses for normal vision to help the farsighted person see clearly at a distance of 0.350 m.