A cup of coffee with temperature 104F is placed in a freezer with temperature 0F. After 7 minutes, the temperature of the coffee is 60.2

Question

A cup of coffee with temperature 104F is placed in a freezer with temperature 0F. After 7 minutes, the temperature of the coffee is 60.2F. What will the temperature be after 19 minutes?

Answer 1

Hi Kela:

 

  Newton's law of cooling, stating that the rate of cooling is proportional to the difference in temperature between an object and its surroundings (in the absence of a phase change), creates a differential equation that can be solved for temperature of the object as a function of time (assuming the ambient temperature of the surroundings stays constant).  That function is of the form:

 

T(t) = T_a+(T_o-T_a)e^-kt

 

where:

 

T_a is the (constant) ambient temperature (of the surroundings)

T_o is the initial temperature of the object

k is a proportionality constant

 

This problem gives us the constant ambient temperature (0 F), the initial temperature of the object (104 F), and the temperature at one given point in time (60.2 F at 19 min).  This information will allow us to solve for k, and then apply that knowledge to solve for the temperature at 19 minutes later.  We can keep degrees in F and time in minutes.

 

T(7 min) = 60.2 F = 0 F + (104 F - 0 F)e^{-k(7 min)}

 

I will solve this for k (in the units I have chosen):

 

60.2 = 0 + 104e^-7k

 

0.579 = e^-7k

 

-0.547 = -7k

 

k = 0.078

 

Then we have:

 

T(t) = 104e^-(0.078)t

 

Since this was solved for time in minutes, put in t = 19 min at solve for T (which will come out in degrees Fahrenheit).

 

I hope this helps!  Let me know if you would like to check an answer or have any questions at all.