# Newton's cooling law [closed]

Can you explain the procedure to solve a differential equation like:

$\displaystyle \frac{\text{d}y}{\text{d}t} = \frac{y}{t+e^{t}}$

applying Newton's cooling law...

So the complete sentence is:

Supposing I have an environment with $20°C$ and I have an ice cube inside this environment, How do you apply Newton's cooling law?

• Also How do you model this situation?

• I will solve it later, but would like to know How to model this.

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## closed as off-topic by Rafflesia arnoldii, Ivo Terek, Antonio Vargas, Tunk-Fey, RecklessReckonerAug 9 at 4:48

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Did you make up the differential equation? I don't see how you can get that using Newton's law of cooling. Also, perhaps this is better suited to physics.SE. –  Aryabhata Jul 15 '11 at 21:49
COULD YOU PLEASE PROVIDE THE STEPS.... –  cMinor Jul 15 '11 at 22:17
That equation will not have a simple solution, because $\int dx/(x+e^x)$ does not. –  Henry Jul 15 '11 at 22:25
when I integrate $\int dx/(x+e^x)$, I can not solve it!!, Does it even exist??? –  cMinor Jul 15 '11 at 22:37
See here. Of course it exists, but it doesn't seem to have a closed form expression in terms of elementary functions. –  t.b. Jul 15 '11 at 22:40

The Newton Law Of Cooling says that if $y=y(t)$ is the temperature at time $t$ of a cooling warm object in an environment which is held at a constant $d$ degrees, then $$\frac{dy}{dt}=-k(y-d)$$ where $k$ is a constant. Informally, this says that the rate of change of the temperature is proportional to the difference between the current temperature of the object and the ambient temperature.

In your case, we do not really have cooling. The ice cube is not going to cool the room! Even if we imagine that the room is cooling, the cooling would not be proportional to the temperature difference, since as the ice cube melts its "strength of cooling" diminishes. The physics of the situation is very different from the physics of ordinary cooling. (By the way, the Newton Law of Cooling does not model actual cooling all that well.)

But back to equations. The Newton Law of Cooling DE is not difficult to solve. To me the easiest approach is to let $z=y-d$. Then we arrive at the very familiar $$\frac{dz}{dt}=-kz.$$ However, please remember that there is no reason to expect a connection of any kind between the Newton Law of Cooling and your problem.

Comment: By way of contrast, the differential equation of the post, which is not the Newton Law of Cooling, cannot be solved in terms of elementary functions. The difficulty lies with integrating $\dfrac{1}{t+e^t}$. Like all too many functions, it almost certainly does not have an elementary antiderivative. It does not seem to have an antiderivative even in terms of the exotic non-elementary functions used by Wolfram Alpha.

Comment: The rate of heat absorption of the melting ice cube is proportional to the surface area of the ice cube. You can take advantage of this fact to find out how fast the volume of the ice cube shrinks. The modeling here is fairly interesting, but to repeat, unrelated to the Newton Law of Cooling.

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And how does the differential equation $\frac{\text{d}y}{\text{d}t} = \frac{y}{t+e^{t}}$ enters here? so, I have to integrate $\frac{1}{t+e^t}$ could you explain why? –  cMinor Jul 15 '11 at 22:27
@cMinor: As mentioned by Theo Buehler, and also in my answer, $\dfrac{1}{e^t+t}$ probably does not have an elementary antiderivative. One would have to give a formal proof to be sure, but if there were one, Wolfram Alpha would likely have found it. If you really have a DE like the one you mention, you will need to use a numerical method to compute $y$ at a given $t$. (You would also need an initial condition.) Certainly Newton Law of Cooling does not give rise to your DE. –  André Nicolas Jul 16 '11 at 0:53