Question 1119971
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The rate of change of temperature of the body at time *[tex \Large t] is *[tex \Large \frac{dT}{dt}].  By Newton's Law of Cooling, this rate of change is proportional to the instantaneous difference between the current temperature of the body and the constant ambient temperature.  Hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dT}{dt}\ =\ -k\(T(t)\ -\ T_a\)]


Let *[tex \Large y(t)\ =\ T(t)\ -\ T_a] where *[tex \Large T_a] is ambient.


Then *[tex \Large y_o\ =\ T(0)\ -\ T_a\ =\ T_o\ -\ T_a\ =\ 98.6\ -\ 55\ =\ 43.6^\circ]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dt}\ =\ \frac{d}{dt}\(T(t)\ -\ T_a\)\ =\ \frac{dT}{dt}\ -\ \frac{dT_a}{dt}\ =\ \frac{dT}{dt}]


because *[tex \Large T_a] is a constant.


Hence


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{dy}{dt}\ =\ -k(T(t)\ -\ T_a)\ =\ -ky]


But this function is well-known to be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y(t)\ =\ y_oe^{-kt}]


Since we are given *[tex \Large k\ =\ 0.1947] for time in hours, we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ T(t)\ =\ 43.6e^{\small{-0.1947t}}]


If *[tex \Large T(t)\ =\ 79^\circ], solve


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  43.6e^{\small{-0.1947t}}\ =\ 79]


for *[tex \Large t] hours.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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