# Thread: Newton's Law of Cooling

1. ## 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.

2. 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.

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.

3. $
\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?