newtons law of cooling

Question

You are making pudding for a dinner party, and you put it in the refrigerator at 5 p.m. Your refrigerator maintains a constant temperature of 40 degrees. The pudding will be ready to serve when it cools to 45 degrees. You measure the temperature of the pudding and it is 190 degrees when you put it in the refrigerator and at 6 p.m. the temperature of the pudding is 100 degrees. Based on Newton’s Law of Cooling, when is the earliest you can serve the pudding?
 
 6:17 pm
 7:59 pm
 8:43 pm
 11:48 pm

Answer 1

T(t) = A + (T0 - A)*e^(-kt)

T0 = initial temperature of pudding

 

A = refrigerator temperature = 40 deg

We need a value for k.

T goes from 190 deg to 100 deg in a time span of t = 1 hour.

 

Plugging into the T(t) equation,

 

100 = 40 + (190 - 40)*e^[-k(1)]

 

k = - ln(0.4) deg/hour = 0.91629 deg/hour

 

Now plug k into the T(t) equation with an initial temperature of 190 deg and a final temperature of 45 deg, and solve for t in hours, which will be the time elapsed in cooling form 190 deg to 45 deg.  Then add t to 5 PM to get clock time.

 

45 = 40 + (190 - 40)*e^(-0.91629 t)

 

t = - ln(0.033333) / 0.91629 hours = 3.71 hours = 3 hours + 42.6 minutes = about 3:43 hours

 

Clock time = 5 PM + 3:43 hours = 8:43 PM