# 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_{i}e^{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.