# Thread: Newton's law of cooling vs heating URGENT!

1. ## Newton's law of cooling vs heating URGENT!

Hello all,

I thought I had this nailed when a little confusion reared its ugly head. When dealing with objects that are cooling (coffee etc), I want to use dT/dt = -k(T-A) where A is ambient temp, T is temp of e.g. coffee.

Now the dead body in the sauna problem, of a thing heating up. What equation do I use here? dT/dt = k(T-A)? dT/dt = k(A-T) (Which is effectively dT/dt = -k(T-A)?

Any help appreciated, but please bear in mind I have an exam on this tomorrow!

2. NVM - turns out prof wrote their equation down wrong. I was sure that, for heating, it should be dT/dt = k(T-A), and that's what it turns out to be (unless somebody wants to contradict that?)

Cheers anyway.

3. Originally Posted by DangerousDave
NVM - turns out prof wrote their equation down wrong. I was sure that, for heating, it should be dT/dt = k(T-A), and that's what it turns out to be (unless somebody wants to contradict that?)

Cheers anyway.
It turns out that it doesn't actually matter,

newtowns heating law can be expressed as

$\displaystyle \frac{dT}{dt} = k(A-T)$

as well as what you have up there. It's a good exercise to prove this, just follow through with the seperation in both cases, and compare the results.

4. Your expression equates to dT/dt = -k(T-A) for a scenario of warming. I was told that it matters that you use a negative k for cooling and a positive k for heating (at least by one source) - something like one produces a converging exponential and one produces a diverging exponential... unless I'm getting muddled with formulae for compound interest - which could well be the case.

5. As AllanCuz said whether you have $\displaystyle \frac{dT}{dt} = k(T-A)$ or $\displaystyle \frac{dT}{dt} = -k(T-A)$ it doesn't matter. The IC and extra condition will give you the correct sign of $\displaystyle k$ (for heating or cooling).

6. Reeto - well that ship has sailed now anyway. The exam was this morning. There was a differential equation question about sprouts, but I think I got it right. Most of the rest of the exam was, however, a complete bugger. I ultimately got stumped by a question on so-called 'homogeneous differential equations', which appear to involve some kind of substitution of some function of x and y for z. But since it wasn't the basic stuff I've seen before on that topic, it was all way too much for my tired mind (no sleep).