# Newton's Law of Cooling

• Feb 8th 2009, 12:02 PM
Konidias
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.
• Feb 8th 2009, 03:44 PM
HallsofIvy
Quote:

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.

Quote:

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.
• Feb 8th 2009, 06:30 PM
Konidias
$\displaystyle \frac{dT}{dt}=k(T-T_{a})$

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

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

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

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

Is this correct?