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Answer :
Answer:
It will take 43.37 mins for the cake to cook off to 90
Step-by-step explanation:
From Newton's law of cooling
[tex]T_{(t)} = T_{s} + (T_{o} - T_{s})e^{kt}[/tex]
Where
[tex]t =[/tex] time
[tex]T_{(t)}[/tex] = Temperature of the given body at time [tex](t)[/tex]
[tex]T_{s}[/tex] = Surrounding temperature
[tex]T_{o}[/tex] = Initial temperature of the body
and [tex]k[/tex] = constant
From the question,
[tex]T_{o}[/tex] = 357
[tex]T_{s}[/tex] = 72
[tex]T_{o} - T_{s}[/tex] = 357 - 72 = 285
∴ [tex]T_{(t)} = 72 + 285e^{kt}[/tex]
From the question, the cake cools to 130 after 25 min. Then, we can write that
[tex]T_{(25)} = 72 + 285e^{k(25)}[/tex]
∴ [tex]130 = 72 + 285e^{25k}[/tex]
Then,
[tex]130 - 72 = 285e^{25k}[/tex]
[tex]58 = 285e^{25k}[/tex]
[tex]\frac{58}{285} = e^{25k}[/tex]
Take the natural log (ln) of both sides
[tex]ln^{\frac{58}{285} } =ln^{ e^{25k}}[/tex]
[tex]-1.5920 = 25k[/tex]
∴ [tex]k =\frac{-1.5920}{25}[/tex]
[tex]k = -0.06368[/tex]
Now, to determine how long it take for the cake to cook off to 90
That is,
[tex]T_{(t)} = 72 + 285e^{kt}[/tex]
[tex]90 = 72 + 285e^{-0.06368t}[/tex]
[tex]90 - 72= 285e^{-0.06368t}[/tex]
[tex]18 = 285e^{-0.06368t}[/tex]
[tex]\frac{18}{285} = e^{-0.06368t}[/tex]
[tex]\frac{6}{95} =e^{-0.06368t}[/tex]
Take the natural log (ln) of both sides
[tex]ln^{\frac{6}{95} } = ln^{e^{-0.06368t}}[/tex]
[tex]-2.7621 = -0.06368t\\[/tex]
∴ [tex]t = \frac{-2.7621}{-0.06368}[/tex]
[tex]t = 43.37 mins[/tex]
Hence, it will take 43.37 mins for the cake to cook off to 90.
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