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nagle_differential_equations_ISM_Part18

nagle_differential_equations_ISM_Part18 - Exercises 3.3(b...

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Exercises 3.3 (b) We repeat the arguments from part (a), but use the half-life 5600 years given in Problem 21 instead of 5550 years, to find that 1 2 M 0 = M 0 e - α (5600) 1 2 = e - α (5600) α = ln(0 . 5) - 5600 0 . 00012378 , and so the decay is governed by M ( t ) = M 0 e - 0 . 00012378 t . Therefore, 3% of the original amount of carbon-14 remains in the campfire when t satisfies 0 . 03 M 0 = M 0 e - 0 . 00012378 t 0 . 03 = e - 0 . 00012378 t t = ln 0 . 03 - 0 . 00012378 28328 . 95 (years) . (c) Comparing the results obtained in parts (a) and (b) with the answer to Problem 21, that is, 31606 years, we conclude that the model is more sensitive to the percent of the mass remaining. EXERCISES 3.3: Heating and Cooling of Buildings 2. Let T ( t ) denote the temperature of the beer at time t (in minutes). According to the Newton’s law of cooling (see (1)), dT dt = K [70 - T ( t )] , where we have taken H ( t ) U ( t ) 0 and M ( t ) 70 F, with the initial condition T (0) = 35 C. Solving this initial value problem yields dT T - 70 = - K dt ln | T - 70 | = - Kt + C 1 T ( t ) = 70 - Ce - Kt ; 35 = T (0) = 70 - Ce - K (0) C = 35 T ( t ) = 70 - 35 e - Kt . To find K , we use the fact that after 3 min the temperature of the beer was 40 F. Thus, 40 = T (3) = 70 - 35 e - K (3) K = ln(7 / 6) 3 , and so T ( t ) = 70 - 35 e - ln(7 / 6) t/ 3 = 70 - 35 6 7 t/ 3 . 81
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Chapter 3 Finally, after 20 min, the temperature of the beer will be T (20) = 70 - 35 6 7 20 / 3 57 . 5 (F ) . 4. Let T ( t ) denote the temperature of the wine at time t (in minutes). According to the Newton’s law of cooling, dT dt = K [23 - T ( t )] , where we have taken the outside (room’s) temperature M ( t ) 23 C, with the initial
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