Newton's Law of Cooling

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February 8th 2009, 12:02 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, 03:44 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, 06:30 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?