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)=-kTdTdt=−kT, where tt is the time taken and TT is the difference of the temperatures between the object and its surroundings.
This gives us TT as a function of tt and is given by T(t)=ce^(-kt)T(t)=ce−kt.
With this it will take infinite time for object to cool down to room temperature, when T(t)=0T(t)=0. Still let us assume that it cools to 72.5^oF72.5oF or less, which is roundable to 72^oF72oF and work it out.
Now as T(0)=ce^(-kxx0)=c=375-72=303T(0)=ce−k×0=c=375−72=303 and
T(15)=303xxe^(-15k)=375^oF-215^oF=160^oFT(15)=303×e−15k=375oF−215oF=160oF or e^(-15k)=160/303e−15k=160303
and -15k=ln(160/303)=-0.638559−15k=ln(160303)=−0.638559 or k=0.638559/15=0.0425706k=0.63855915=0.0425706
If pie cools to 72.5^oF72.5oF in tt minutes, then
303e^(-0.0425706xxt)=0.5303e−0.0425706×t=0.5 or e^(-0.0425706xxt)=0.5/303=0.00165017e−0.0425706×t=0.5303=0.00165017
or -0.0425706t=ln0.00165017=-6.40688−0.0425706t=ln0.00165017=−6.40688
or t=6.40688/0.0425706=150.5t=6.406880.0425706=150.5
Hence, it will take at lease 150.5150.5 minutes to cool down to close to 72^oF72oF
