# Thread: Newton's Law of Cooling

1. ## Newton's Law of Cooling

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

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

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

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

Find the GS of the DE $\displaystyle \frac{dT}{dt}=k(T-T_{a})$. You may assume $\displaystyle T-T_{a} > 0$ and the final solution should give $\displaystyle T$ explicitly in terms of $\displaystyle k, t, T_{a}$ 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: $\displaystyle \frac{dT}{T- T_a}= kdt$. Integrate both sides. You will get a logarithm on the left so solving for T will give an exponential of kt.

3. $\displaystyle \frac{dT}{dt}=k(T-T_{a})$

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

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

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

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

Is this correct?