# Newton's Law of Cooling

• February 8th 2009, 01:02 PM
Konidias
Newton's Law of Cooling
Newton's Law of Cooling can be written as the DE $\frac{dT}{dt}=k(T-T_{a})$ for some value of k.

If $T-T_{a} > 0$ is k positive or negative? Briefly explain why.

Find the GS of the DE $\frac{dT}{dt}=k(T-T_{a})$. You may assume $T-T_{a} > 0$ and the final solution should give $T$ explicitly in terms of $k, t, T_{a}$ and a constant of integration.

Could someone give me a helping hand with this? I'm new to differential equations and this question was just thrown at me unexpectedly.
• February 8th 2009, 04:44 PM
HallsofIvy
Quote:

Originally Posted by Konidias
Newton's Law of Cooling can be written as the DE $\frac{dT}{dt}=k(T-T_{a})$ for some value of k.

If $T-T_{a} > 0$ is k positive or negative? Briefly explain why.

What do "T" and " $T_a$" mean? If T is the variable temperature and $T_a$ is a constant temperature, then $T- T_a> 0$ means temperature T is greater than the the temperature $T_a$ and T will decrease to $T_a$. So dT/dt must be negative and that means that k must be negative.
By the way, although it wasn't asked, if T< $T_a$, $T- T_a< 0$ so T increases to $T_a$. dT/dt is positive now which means that, since T- $T_a$ is negative, k must still be negative. The "k" in Newton's law of cooling is always negative.

Quote:

Find the GS of the DE $\frac{dT}{dt}=k(T-T_{a})$. You may assume $T-T_{a} > 0$ and the final solution should give $T$ explicitly in terms of $k, t, T_{a}$ and a constant of integration.

Could someone give me a helping hand with this? I'm new to differential equations and this question was just thrown at me unexpectedly.
"Separate" it: $\frac{dT}{T- T_a}= kdt$. Integrate both sides. You will get a logarithm on the left so solving for T will give an exponential of kt.
• February 8th 2009, 07:30 PM
Konidias
$
\frac{dT}{dt}=k(T-T_{a})
$

$\int \frac{dT}{T-T_{a}}=\int kdt$

$
ln|T-T_{a}|+C=kt
$

$
T-T_{a}+C=e^{kt}
$

$
T=e^{kt}+T_{a}+C
$

Is this correct?