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g When a cake is removed from an oven and placed on a kitchen counter, its temperature is measured at 300◦F. Three minutes later its temperature is 250◦F. Write the initial value problem describing the change in the temperature of the cake, if the temperature of the kitchen is 70◦F.

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Answer:

[tex]T = 70 +230 (\frac{18}{23})^{\frac{t}{3}}[/tex]

Explanation:

We use T to represent the temperature of the cake at time "t"

Also [tex]T_k[/tex] to be the kitchen temperature :

By differentiation:

[tex]\frac{dT}{dt}=-k (T-T_k)\\ \\\frac{dT}{dt}=-k (T-70)\\\\[/tex]

where;

T(0) = 300° F

T(3) = 250° F

[tex]T = 70 + Ce^{-kt}[/tex]

[tex]300 = 70 + Ce^0\\\\300-70 = C*1\\\\C = 230\\\\T = 70 + 230 e^{-kt}[/tex]

T(3) = 250

[tex]250 = 70 + 230 e^{-3k}[/tex]

[tex]250 - 70 = 230 e^{-3k}\\\\180 = 230 e^{-3k}\\\\\frac{180}{230}= e^{-3k}\\\\\frac{18}{23}= e^{-3k}\\\\-k = \frac{1}{3}In(\frac{18}{23})\\\\T = 70 + 230 e^{\frac{t}{3}In \frac{18}{23}}\\\\T = 70 +230 (\frac{18}{23})^{\frac{t}{3}}[/tex]

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