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MATHEMATICS - CASE STUDY SOLUTION


The Hot Cereal
 

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.

     
          dθ/dt = -kθ

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

     
Symbol Value Meaning
θ
60 - 25 = 35
Temperature difference
b
90 - 25 = 65
k
?
Cooling constant
t
1
Time
 

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

     ek = 65/35

     k = ln(65/35) = 0.619

     
Symbol
Value
Meaning
θ
40 - 25 = 15
Temperature difference
b
90 - 25 = 65
k
0.619
Cooling constant
t
?
Time
 

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.