# newton's law of cooling

• Dec 20th 2008, 01:37 AM
aero-me
newton's law of cooling
hey everyone.
i hve a doubt in dis question: A body at temperature T higher than its surroundings cools down according to the formula T=(T-T(with base 0))e(raise to the power of -kt), (newton's law of cooling) where T(with base 0) is the temperature of the surroundings, t is the time in hours, and k is a constant. It is found that the temperature of the body falls from 80degrees celsius to 60 degrees celsius in 2 hours. If T(with base 0)=20 degress celsius, how long does it take for the temperature of the body to fall to 30 degrees celsius?....well is the answer coming out to be 8 hrs 50 mins...i just wana confirm...
thankx for seeing dis qs....hope 2 c a reply soon...
tc pple
• Dec 20th 2008, 02:04 AM
running-gag
Hi

Newton's law is
$T = T_0 + (T_i - T_0) e^{-kt}$
where $T_0$ is the temperature of the surroundings and $T_i$ the initial temperature of the body

Let $T_1$ be the temperature at $t_1$
and $T_2$ the temperature at $t_2$

Then
$T_1 - T_0 = (T_i - T_0) e^{-kt_1}$
$T_2 - T_0 = (T_i - T_0) e^{-kt_2}$

Dividing the 2 expressions

$k = \frac{1}{t_2-t_1}\,ln \frac{T_1 - T_0}{T_2 - T_0}$