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Unformatted text preview: Exercises 3.3 (b) We repeat the arguments from part (a), but use the halflife 5600 years given in Problem 21 instead of 5550 years, to find that 1 2 M = M e (5600) 1 2 = e (5600) = ln(0 . 5) 5600 . 00012378 , and so the decay is governed by M ( t ) = M e . 00012378 t . Therefore, 3% of the original amount of carbon14 remains in the campfire when t satisfies . 03 M = M e . 00012378 t . 03 = e . 00012378 t t = ln 0 . 03 . 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 Newtons 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 3min 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 Chapter 3 Finally, after 20 min, the temperature of the beer will be T (20) = 70 35 6 7 20 / 3 57 . 5 (F ) ....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.
 Spring '08
 MAZMANI
 Equations

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