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

It will take 43.37 mins for the cake to cook off to 90

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)`

`(58)/(285) = e^(25k)`

Take the natural log (ln) of both sides

`ln^{(58)/(285) } =ln^{ e^(25k)}`

`-1.5920 = 25k`

∴
`k =(-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)`

`(18)/(285) = e^(-0.06368t)`

`(6)/(95) =e^(-0.06368t)`

Take the natural log (ln) of both sides

`ln^{(6)/(95) } = ln^{e^(-0.06368t)}`

`-2.7621 = -0.06368t\\`

∴
`t = (-2.7621)/(-0.06368)`

`t = 43.37 mins`

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