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Newton's Law of Cooling

A ta and dt k dt ta ln t a kt c with c the

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Unformatted text preview: ct’s temperature T as a function of time t. With specific time and temperature information we can derive a particular solution to be used in solving a problem related to that given information. dT dT dT = K (T ! A ) " = K # dt " \$ = K \$ dt dt T!A T!A and " dT = K " dt T!A # ln T ! A = Kt + C, with C the constant of integration #e ln T ! A = e Kt +C , and with T \$ A, we have T ! A = e Kt +C # T = e Kt +C + A In general, then, T = e Kt +C + A is the solution to the differential equation dT = K (T ! A) . dt Suppose, further, we know that the temperature of the object is 95° C, that the ambient temperature is 20° C, and that exactly 20 minutes after the object began to cool its temperature was 70° C. We have A = 20° C and (0,95) and (20,70) as known conditions. Wi...
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