# A pie is removed from a 375°F oven and cools to 215°F after 15 minutes in a room at 72°F. How long (from the time it is removed from the oven) will it take the pie to cool to 72°F?

Nov 8, 2016

It will take at lease $150.5$ minutes to cool down to close to ${72}^{o} F$

#### Explanation:

Newton's Law of Cooling states that the rate of cooling of an object is inversely proportional to the difference of temperatures between the object and its surroundings i.e. $\frac{\mathrm{dT}}{\mathrm{dt}} = - k T$, where $t$ is the time taken and $T$ is the difference of the temperatures between the object and its surroundings.

This gives us $T$ as a function of $t$ and is given by $T \left(t\right) = c {e}^{- k t}$.

With this it will take infinite time for object to cool down to room temperature, when $T \left(t\right) = 0$. Still let us assume that it cools to ${72.5}^{o} F$ or less, which is roundable to ${72}^{o} F$ and work it out.

Now as $T \left(0\right) = c {e}^{- k \times 0} = c = 375 - 72 = 303$ and

$T \left(15\right) = 303 \times {e}^{- 15 k} = {375}^{o} F - {215}^{o} F = {160}^{o} F$ or ${e}^{- 15 k} = \frac{160}{303}$

and $- 15 k = \ln \left(\frac{160}{303}\right) = - 0.638559$ or $k = \frac{0.638559}{15} = 0.0425706$

If pie cools to ${72.5}^{o} F$ in $t$ minutes, then

$303 {e}^{- 0.0425706 \times t} = 0.5$ or ${e}^{- 0.0425706 \times t} = \frac{0.5}{303} = 0.00165017$

or $- 0.0425706 t = \ln 0.00165017 = - 6.40688$

or $t = \frac{6.40688}{0.0425706} = 150.5$

Hence, it will take at lease $150.5$ minutes to cool down to close to ${72}^{o} F$