# Newton's Cooling

• Jan 15th 2006, 06:11 AM
jacs
Newton's Cooling
Hi, I have tried this many times and just can't seem to figure it out.

A body, initially at room temperature 20C, is heated so that its temperature would rise by 5C/min if no cooling took place. Cooling does occur in accordance with Newton's Law of Cooling and the maximum temperature the body could attain is 120C. How long would it take to reach a temperature of 100C?

using dt/dt = -k(T - P) where T is the current temp and P is the temp of the surrounding medium.

any help wouuld be appreciated, thanks

the answer provided is 32.2 mins
• Jan 15th 2006, 08:43 AM
Jameson
The general form for Newton's Law of Cooling is...

$\displaystyle T=(T_{0}-T_{s})e^{-kt}+T_{s}$

if my memory serves my correctly.
• Jan 15th 2006, 10:01 AM
CaptainBlack
Quote:

Originally Posted by jacs
Hi, I have tried this many times and just can't seem to figure it out.

A body, initially at room temperature 20C, is heated so that its temperature would rise by 5C/min if no cooling took place. Cooling does occur in accordance with Newton's Law of Cooling and the maximum temperature the body could attain is 120C. How long would it take to reach a temperature of 100C?

using dt/dt = -k(T - P) where T is the current temp and P is the temp of the surrounding medium.

any help wouuld be appreciated, thanks

the answer provided is 32.2 mins

The rate of change of temperature is the sum of the heating component and
the cooling component.

So in this case:

$\displaystyle \frac{dT}{dt}=5-k(T-20)$

The $\displaystyle 5$ on the RHS is the contribution to the rate of change of
temp from the heating, the rest is the contribution to the rate of change of
temp due to the cooling (the $\displaystyle 20$ is the temp of the surrounding
medium).

At a temp of $\displaystyle 120$ the rate of change of temp is zero, so:

$\displaystyle 0=5-k(120-20)\Rightarrow k=0.05$.

Hence:

$\displaystyle \frac{dT}{dt}=6-0.05T$,

which is a DE of variables separable type, so:

$\displaystyle \int_{20}^{T(t)} \frac{1}{6-0.05T}dT=\int_0^t dt$.

Hence:

$\displaystyle \left[ \frac{-\ln (6-0.05T)}{0.05} \right]_{20}^{T(t)}=t$,

so:

$\displaystyle \ln (6-0.05T)-\ln (5)=-0.05t$

again:

$\displaystyle \ln \left(\frac{6-0.05T}{5} \right) = -0.05t$

Putting $\displaystyle T=100$ into this will allow you to find the corresponding
$\displaystyle t$.

RonL
• Jan 15th 2006, 10:46 AM
ThePerfectHacker
Quote:

Originally Posted by Jameson
The general form for Newton's Law of Cooling is...

$\displaystyle T=(T_{0}-T_{s})e^{-kt}+T_{s}$

if my memory serves my correctly.

Yes it is correct. Let me come to that conclusion, this is similar to what CaptainBlack did.

Law of cooling:
$\displaystyle \frac{dT}{dt}=k(T-H)$ H is air-temperature.
Diffrencial equation is seperable,
$\displaystyle \frac{dT}{T-H}=kdt$
Integrating,
$\displaystyle \ln(T-H)=kt+C_1$
Under base e,
$\displaystyle T-H=e^{kt+C_1}=e^{kt}e^{C_1}=Ce^{kt}$
Thus,
$\displaystyle T=H+Ce^{kt}$
when $\displaystyle t=0$ then $\displaystyle T=T_o$
thus,
$\displaystyle T=H+(T_0-H)e^{kt}$
• Jan 15th 2006, 02:39 PM
jacs
Thanks!!!! :)

Captain Black, you are a legend, it was the 5 that was really throwing me and just didn't know where to put it in, and that works out just nicely.

my gratitude

jacs