Tutoring calculus or differential equations, Newton’s Law of Cooling will surface. The tutor looks at a real-life example.

In yesterday’s post I mention that a casserole dish taken out of the oven cooled from 177C to about 40C during one hour.

Newton’s Law of Cooling can be used to calculate the temp of a cooling object:

T_f = T_ie^kt

where

T_f = final temp

T_i = initial temp

k = the constant of cooling (if cooling, k will turn out negative)

t = time (usually in seconds)

For this case, we have t=3600 (3600s in one hour):

40 = 177e^k3600

Dividing both sides by 177 gives

0.226=e^3600k

Now we ln both sides:

ln0.226 = 3600k

Finally we divide by 3600:

-4.13×10^-4 = k

Apparently the cooling constant of the casserole is -4.13×10^-4.

Source:

Larson, Roland E. and Robert P. Hostetler. Calculus. Toronto: D.C. Heath and Company, 1989.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.