a beam of protons is moving toward a target in a particle acceleratorthis beam constitutes a current whole value is 0.50a) each proton has a kinetic energy of 4.9 X 10^-12 Jsuppose the target is a 15 gram block of aluminum, and all the kinetic energy of the proton goes into heating it upwhat is the change in temperature of the block that results from the 15 s bombardment of protons?
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I don't know what a current whose value is 0.50 means: you might mean 0.5 A, or 0.5 mA, or you might mean 0.5 protons/second, or you might mean any of a number of different thingsI will tell you right at the beginning that 0.5 amperes is way way way too much, but I will use it as the baseline to solve this problemIf we assume that you meant 0.5 A, then we can start with the current: I 0.5 A 0.5 coulombs / secondWe know the energy per proton, and we know that energy is conserved, so that all of the kinetic energy of the protons is dissipated as heatWhat we don't know yet is how many protons collide with the targetWe first off use the proton's charge to figure out how many protons there are in one coulomb: 1 proton 1.602 · 10^{-19} coulombs, so 1 coulomb 6.2415 · 10^18 protonsNow we substitute that value in for the unit coulombs: I 0.5 C/s 0.5 · (6.2415 · 10^18 protons) / second 3.121 · 10^18 protons/secondIf we let this run for 15 seconds, we will therefore see a total charge accumulation of: Q 3.121 · 10^18 protons/s · 15 s 4.681 · 10^19 protonsNow we use their kinetic energy to find the total energy in these 4.681 · 10^19 protons: E (4.681 · 10^{19} protons) · (4.9 · 10^{-12} J / proton) 2.294 · 10^8 J Per reference [1], we have that for aluminium, the specific heat is: c 0.900 J / (g K) And thus 15 grams would have a heat capacity of: h 0.900 J / (g K) 15 g 13.5 J / KThe change in temperature would therefore be: ΔT E / h (2.294 · 10^8 J) / (13.5 J/K) 1.699 · 10^7 K We now round to two significant figures and express this as merely 1.7 · 10^7 KThis would probably vaporise the aluminium, and thus the assumption of a constant heat capacity is not valid if the current is 0.5 amperesHowever, if it were 0.5 μA or so, then this would only be an increase of 17 K 17 °C over the original heat, which is probably small enough to justify that assumption If instead of amperes A you had milliam
