A bowl of cereal heated to 90° in a 25° room. After 1 minute the temperature of the cereal drops to 60° after removing from heat. How long does it take the cereal to cool to 40°?

Let the difference of the temperature between the cereal and the room be θ. Let the time and the cooling constant be t and k, respectively.

Since the rate of cooling is proportional to the difference temperature between the cereal and the room, the following differential equation can be used.

Rearranging this equation gives

     dθ/θ = -kdt

Next, Integrate both side of the equation gives

     lnθ = -kt + c₁

Rearrange again, and using a new constant c₂

     θ = e^-kt + c₂

This expression can also be expressed as

     θ = be^-kt

where b is also a constant. This result can be checked by substituting it into the original differential equation. Substituting t = 0 into the above equation, gives θ = b. It has stated that θ is the difference of the temperature between the cereal and the room. When the time is 0, θ = 90 - 25 = 65

Therefore the expression of θ is

     θ = 65e^-kt

Substituting t = 1, θ = 60 - 25 = 35 into the above equation gives the value of cooling constant.

     35 = 65e^-k(1)

     e^-k = 35/65

     e^k = 65/35

     k = ln(65/35) = 0.619

So the equation can be simplified to

     θ = 65e^-0.619t

Substituting θ = 40 - 25 = 15 into the above equation gives

     15 = 65e^-0.619t

Rearrange it gives

     t = 2.369 min

Therefore, it takes 2.37 minutes before Ms. Jamison can start eating.