Question

An a my baker is making a cake for a Terran visitor. The cake is removed from the over at 357 and cools to 130 after 25 min in a room at 72. How long will it take for the cake to cook off to 90?

Answer 1

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

$T_{(t)} = T_{s} + (T_{o} - T_{s})e^{kt}$

Where

$t =$ time

$T_{(t)}$ = Temperature of the given body at time $(t)$

$T_{s}$ = Surrounding temperature

$T_{o}$ = Initial temperature of the body

and $k$ = constant

From the question,

$T_{o}$ = 357

$T_{s}$ = 72

$T_{o} - T_{s}$ = 357 - 72 = 285

∴ $T_{(t)} = 72 + 285e^{kt}$

From the question, the cake cools to 130 after 25 min. Then, we can write that

$T_{(25)} = 72 + 285e^{k(25)}$

∴ $130 = 72 + 285e^{25k}$

Then,

$130 - 72 = 285e^{25k}$

$58 = 285e^{25k}$

$\frac{58}{285} = e^{25k}$

Take the natural log (ln) of both sides

$ln^{\frac{58}{285} } =ln^{ e^{25k}}$

$-1.5920 = 25k$

∴ $k =\frac{-1.5920}{25}$

$k = -0.06368$

Now, to determine how long it take for the cake to cook off to 90

That is,

$T_{(t)} = 72 + 285e^{kt}$

$90 = 72 + 285e^{-0.06368t}$

$90 - 72= 285e^{-0.06368t}$

$18 = 285e^{-0.06368t}$

$\frac{18}{285} = e^{-0.06368t}$

$\frac{6}{95} =e^{-0.06368t}$

Take the natural log (ln) of both sides

$ln^{\frac{6}{95} } = ln^{e^{-0.06368t}}$

$-2.7621 = -0.06368t$

∴ $t = \frac{-2.7621}{-0.06368}$

$t = 43.37 mins$

Hence, it will take 43.37 mins for the cake to cook off to 90.