Question

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?

Answer 1

It will take at lease `150.5` minutes to cool down to close to `72^oF`

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. `(dT)/(dt)=-kT`, 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(t)=ce^(-kt)`.

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

Now as `T(0)=ce^(-kxx0)=c=375-72=303` and

`T(15)=303xxe^(-15k)=375^oF-215^oF=160^oF` or `e^(-15k)=160/303`

and `-15k=ln(160/303)=-0.638559` or `k=0.638559/15=0.0425706`

If pie cools to `72.5^oF` in `t` minutes, then

`303e^(-0.0425706xxt)=0.5` or `e^(-0.0425706xxt)=0.5/303=0.00165017`

or `-0.0425706t=ln0.00165017=-6.40688`

or `t=6.40688/0.0425706=150.5`

Hence, it will take at lease `150.5` minutes to cool down to close to `72^oF`