0
$\begingroup$

In Newton's law of cooling, we see that the rate of loss of heat is directly proportional to temperature. Therefore we formulate the differential equation like this, $$-\dfrac{dQ}{dt}=k(T_{system}-T_{environment})$$

And we solve for $T(t)$.

But when I think about Newton's second law I cannot see a negative sign, i.e., essentially Newton is talking about the rate of decrease of momentum is directly proportional to the acceleration. So, shouldn't we have something like this? $$-\dfrac{dp}{dt}=ma$$

But after solving we get $p=-mv$. So, what am I not understanding?

$\endgroup$

closed as unclear what you're asking by Kyle Kanos, Yashas, honeste_vivere, Jon Custer, Emilio Pisanty Jul 25 '17 at 17:45

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Why do you think there is an equivalence between heat and momentum equations? $\endgroup$ – Kyle Kanos Jul 25 '17 at 10:01
  • $\begingroup$ Because both increases by the decrease of a quantity $\endgroup$ – Jyotishraj Thoudam Jul 25 '17 at 10:03
  • 2
    $\begingroup$ If I accelerate for a moment, then my momentum has increased & if I decelerate, then my momentum has decreased, so your assertion there is false. $\endgroup$ – Kyle Kanos Jul 25 '17 at 10:05
  • $\begingroup$ I get the gap now. I found out that the negative sign would be valid if we have a source of the momentum given to the particle. Because the source's momentum would decrease as the acceleration of the particle increases $\endgroup$ – Jyotishraj Thoudam Jul 25 '17 at 10:08
  • $\begingroup$ Momentum is a property of an object in motion, there's no 'source' for it. $\endgroup$ – Kyle Kanos Jul 25 '17 at 10:10
3
$\begingroup$

These are two completely unrelated laws that both happen to share their name with an old dead British guy. You cannot derive Newton's Second Law of Motion from Newton's Law of Cooling.

$\endgroup$
  • $\begingroup$ I'm not trying to derive the second law from cooling. I was asking a fact of science and how they are using calculus $\endgroup$ – Jyotishraj Thoudam Jul 25 '17 at 2:52
  • $\begingroup$ Oh well, I thought I'd have a physical meaning to the conflict? $\endgroup$ – Jyotishraj Thoudam Jul 25 '17 at 3:02
  • 2
    $\begingroup$ @jyotishrajthoudam What conflict? That two different laws are different? $\endgroup$ – Emilio Pisanty Jul 25 '17 at 17:45
0
$\begingroup$

I found the fundamental gap between my understanding of the use of calculus for the second law and the cooling problem.

I realized that we are not connecting two systems in one equation. We cannot fit two systems in one equation. So, when I formulated the question, I thought that in order for the law of cooling to work, given the changing heat source, the heat will flow in the direction of decreasing temperature and thus the negative sign. So, I miscalculated the fact that if there is a negative sign here why isn't there a negative sign in Newton's second law of motion, because, the particle accelerates as the applied momentum decreases, so, the decreasing momentum of the momentum applying object is equal to the increasing acceleration. So, the decreasing heat will increase the temperature difference. Therefore I thought there will be a negative sign in the second law.

But I found out from analyzing the differential equation that there was consideration of a momentum providing body to the particle. The equation meant, if there is a positive/negative change in momentum of the particle(from anywhere), there will be and increase/decrease in the acceleration, thereby cancelling the negative signs.

From this I was able to deduce that it was saying something about a particle's motion and not concerned about the source of its motion.

Therefore, I find the law of cooling more satisfying once I understood that he didn't say about two systems, he wanted to study the object itself and its consequences without considering the source of its consequences. Hence, a part of a block cooling means, heat flows in with decreasing temperature.

Suppose if $$\dfrac{dQ}{dt}=-ve$$ it means heat is decreasing with increasing time. The slope is the temperature difference multiplied by a constant. $$\dfrac{dQ}{dt}=-k(T-T_{initial})$$

But also, the case of $-ve$ sign infront of the second law would perhaps (I don't know) be valid if we consider two separate objects addressed by a single equation i.e., the decreasing momentum of the momentum source leads to the increasing acceleration.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.