# newtons law of cooling

• March 29th 2008, 03:09 PM
batman121
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?
• March 29th 2008, 05:39 PM
topsquark
Quote:

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.)

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

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

-Dan
• March 30th 2008, 02:59 AM
batman121
$-0.02t= ln(15)$
$t= ln(15)/-0.02$

$t=58.804sec$

I dont think thats right, im not sure hoe to rearange when $ln$ is in the equation.
• March 30th 2008, 06:49 AM
topsquark
Quote:

Originally Posted by batman121
$-0.02t= ln(15)$
$t= ln(15)/-0.02$

$t=58.804sec$

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

Quote:

Originally Posted by topsquark
What's that X in there? (And please start using parenthesis.)

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

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

-Dan

A minor sign mistake here
$0.02t = ln(15)$

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

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
• March 30th 2008, 08:08 AM
batman121
Thanks for the help greatly appreciated (Rofl)