“Do the hotplates heat the food through properly?”

“Oh yes, they come out of the oven at 200 degrees and the temperature drops by a degree every minute.”

To @dragondodo’s credit, she did not launch into a lecture on Newton cooling. But she did grumble about it to me - and it got me wondering.

### All models are wrong

Clearly, dropping by a degree a minute is not sustainable, or else three hours later, the hotplate would make a functional icepack.

The usual model for objects cooling or heating - under sensible conditions - is that the rate of cooling is proportional to the difference between the current temperature ($T$) and the ambient temperature ($T_a$).

That gives a differential equation: $\diff T t = -\alpha (T - T_a)$, where $\alpha > 0$ is assumed to be a constant, and $T_a$ is the ambient temperature.

Anyone who’s been through the second year of A-level maths should recognise that’s separable:

$\int \frac{1}{T - T_a} \diff T t \d t = -\int \alpha \dt$

So $\ln(T - T_a) = -\alpha t + C$

… which gives $T = T_a + Ae^{-\alpha t}$.

If you know the initial temperature (say, $T_0$), you can use that fact to work out $A$: $T = T_a + (T_0 - T_a)e^{-\alpha t}$.

### On the hotplate

In the steakhouse, we can probably assume that $T_a$ is about 20 degrees, and we’re told that $T_0$ is 200, so:

$T = 20 + 180 e^{-\alpha t}$, for a value of $\alpha$ we can work out.

The server didn’t give us a mathematically precise set-up, so let’s surmise that they meant the temperature is initially dropping at 1 degree per minute. We can go back to the initial differential equation:

$\diff T t = -\alpha (T- T_a)$

So, if we measure $t$ in minutes, we have:

$-1 = -\alpha (200 - 20)$, so $\alpha = \frac{1}{180}$.

So, the natural question for me is: how does “one degree per minute” hold, at least approximately. Or rather: for how long is the rate of change between a full degree and (let’s be generous) half a degree?

### A matter of degree

We know the rate of change is given by $\diff T t = -\frac{1}{180} (T- 20)$, and that $T$ itself is $20 + 180 e^{-\alpha t}$.

That makes the rate of change, after a bit of cancelling, $\diff T t = - e^{\alpha t}$.

That’s equal to $-\frac{1}{2}$ when $e^{-\alpha t} = \frac{1}{2}$, or $\frac{1}{180} t = \ln(2)$

Now, $180\ln(2)$ is… hello sensei!

whoosh!

Their eyes roll. “It’s about 18 sevens, or 126. You can lose a percent if you want. 125, a shade over two hours. Can I take this hot plate? I have an idea for some… high steaks testing.”

whoosh

So, for a good two hours, the rate of change rounds to one degree.

Over the… I have no idea how long it takes to cook a steak on a hot plate. 20 minutes? If you cooked your steak for 20 minutes like this, assuming the model still holds, its temperature would end up at $20 + 180e^{-\frac{1}{9}}$, which is…

whoosh

“Honestly, can’t anyone do basic arithmetic? $e^{-\frac{1}{9}}$ is about $\frac{8}{9}$. So… 180 degrees.”

“You know, sensei, I never thought I’d hear you utter that phrase.”

A furrowed brow. A glance at the nearby knives.

The whoosh was mine this time. As I left, I noted that the server’s assessment may have been morally wrong - but it wasn’t at all far off.

## Anecdote

As a vegetarian of 20-odd years, I don’t often - read ever - frequent steakhouses. I’m reminded of an anecdote, possibly @gavinesler’s, about a steakhouse in Yellowstone Park. One of the producers asked what the server would recommend for a vegetarian; she sighed, put a pencil behind her ear, and said “Frankly, I would recommend leaving the state of Wyoming.”

* Edited 2019-10-21 to fix some broken code. Thanks to the LaTeX Ninja, @christianp, for pointing it out.