# Thread: newtons law of cooling

1. ## newtons law of cooling

Question given: newtons law of cooling is given by
(Θ=theta)

Θ=Θoe^-kt

where the excess of tempreature at 0 deg = Θo deg and the time T is Θ deg.
given that Θo=15 deg and K=-0.02 find the time when the rate of change of tempture is 1 deg

my work

1=15e^-0.02XT
i assume u must use log to rearence the forumla in oder to get a value then differentiate it to find the rate of change of T.
is what iv done so far correct?

2. Originally Posted by batman121
Question given: newtons law of cooling is given by
(Θ=theta)

Θ=Θoe^-kt

where the excess of tempreature at 0 deg = Θo deg and the time T is Θ deg.
given that Θo=15 deg and K=-0.02 find the time when the rate of change of tempture is 1 deg

my work

1=15e^-0.02XT
i assume u must use log to rearence the forumla in oder to get a value then differentiate it to find the rate of change of T.
is what iv done so far correct?
What's that X in there? (And please start using parenthesis.)

$\displaystyle 1 = 15e^{-0.02t}$

Yes, take the ln of both sides:
$\displaystyle 0 = ln(15) - 0.02t$
and solve for t.

-Dan

3. $\displaystyle -0.02t= ln(15)$
$\displaystyle t= ln(15)/-0.02$

$\displaystyle t=58.804sec$

I dont think thats right, im not sure hoe to rearange when $\displaystyle ln$ is in the equation.

4. Originally Posted by batman121
$\displaystyle -0.02t= ln(15)$
$\displaystyle t= ln(15)/-0.02$

$\displaystyle t=58.804sec$

I dont think thats right, im not sure hoe to rearange when $\displaystyle ln$ is in the equation.
Originally Posted by topsquark
What's that X in there? (And please start using parenthesis.)

$\displaystyle 1 = 15e^{-0.02t}$

Yes, take the ln of both sides:
$\displaystyle 0 = ln(15) - 0.02t$
and solve for t.

-Dan
A minor sign mistake here
$\displaystyle 0.02t = ln(15)$

$\displaystyle t = \frac{ln(15)}{0.02} \approx 135.403$

I'm not sure why you made the numerical mistake.

ln( ) is like any other function: treat ln(15) as a constant and you'll do fine with it.

-Dan

5. Thanks for the help greatly appreciated