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A large object with an initial temperature of 150 degrees Fahrenheit is dropped into the ocean. The current temperature of the ocean water in the area is 77 degrees. The function f(t)=Ce(−kt)+77 represents the situation, where t is time in minutes, C is a constant, and k is a constant. After 10 minutes the object has a temperature of 120 degrees. After how many minutes will the temperature of the object be equal to 80 degrees?

Respuesta :

Answer:

t = 60.3 minutes.

Step-by-step explanation:

The function [tex]f(t) = Ce^{-kt} +77[/tex] ........ (1)  

Now, at t = 0, f(0) = 150 = C + 77

⇒ C = 73

So, the function (1) becomes [tex]f(t) = 73e^{-kt} +77[/tex] ......... (2)

Now, it is given that at t = 10 minutes, f(10) = 120 degree.

Therefore, from equation (2),  [tex]120 = 73e^{-10k} +77[/tex]

⇒ [tex]73e^{-10k} = 43[/tex]

⇒ [tex]e^{-10k} = 0.589[/tex]

Now, taking ln both sides we get -10k (ln e) =ln (0.589)

⇒ k = 0.0529

Therefore, the equation (2) becomes  [tex]f(t) = 73e^{-0.0529t} +77[/tex] ......(3)

Now, putting f(t) = 80 degree, we have fro equation (3),

[tex]80 = 73e^{-0.0529t} +77[/tex]

⇒ [tex]3 = 73e^{-0.0529t}[/tex]

⇒ [tex]e^{-0.0529t} = 0.041[/tex]

Taking ln both sides we get, -0.0529t = - 3.19, ⇒ t = 60.3 minutes. (Answer)

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