
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=(TT(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

Hi
Newton's law is
$\displaystyle T = T_0 + (T_i  T_0) e^{kt}$
where $\displaystyle T_0$ is the temperature of the surroundings and $\displaystyle T_i$ the initial temperature of the body
Let $\displaystyle T_1$ be the temperature at $\displaystyle t_1$
and $\displaystyle T_2$ the temperature at $\displaystyle t_2$
Then
$\displaystyle T_1  T_0 = (T_i  T_0) e^{kt_1}$
$\displaystyle T_2  T_0 = (T_i  T_0) e^{kt_2}$
Dividing the 2 expressions
$\displaystyle k = \frac{1}{t_2t_1}\,ln \frac{T_1  T_0}{T_2  T_0}$