The Calculus of smoking a roast

Isaac Newton is a pretty bright coal in the fire pit of the scientific world. Okay, maybe he had his personal problems, but he was a brilliant thinker. One of his strongest assets was being able to find the simplicity and identify the essence of a complicated problem. His basic observations of natural phenomena became translated into elegant mathematical expressions.

One such example is Newton’s Law of Cooling and Heating. Our everyday experience tells us that ice cubes placed in hot water will melt faster than ice cubes placed in cold water. Newton’s observation was that any object’s temperature changes more quickly the larger the difference in temperature between that object and its surroundings is. A more mathematical-sounding expression would be:

(1) The rate of change in temperature of an object is proportional to the difference in temperature between the object and its (assumed) constant environment.

Well, mathematics is a wonderful language… we can express mathematical ideas both in words, as I just did in equation (1) above, and we can also express the same ideas, usually more succinctly, in symbols, as I will do in equation (2) below. Both equation (1) and equation (2) say and mean exactly the same thing!

(2) dT/dt = k(T – Te)

Here, we’re letting the symbol T represent the temperature at any time, t. We might also write this as T(t), but that gets cumbersome. T is fine. The little d means “a very small change.” So, the first part of equation (2) means, a very small change in T during some very small interval of time, dt. This is also called the “instantaneous change” of T, or the derivative of T with respect to time, t. From equation (1), it is associated with the part saying the rate of change in temperature.

We also see in equation (2) the letter k, which is often used to express constant values. In our case, it is representing what we call a “constant of proportionality,” which is a fancy way of saying “scaling factor.” It’s a mathematician or physicist’s way of hedging their bets. We know that the change in temperature is somehow related to the difference in temperatures, but we’re not ready to commit to a single numeric value. Besides, different materials may have different ways they respond to temperature, and the goal is to keep things as generalized as possible. From equation (1) it is associated with the part saying is proportional to.

The last part of equation (2) is the difference, or subtraction, of Te from T. We’re letting Te represent the assumed constant temperature of the surrounding environment. Why do we assume it is constant, and is that a valid assumption? Well, we assume it is constant because it makes the math easier! So much of physics is finding ways to simplify the complex world so that we can better understand it, even if slightly imperfectly. Besides, in practice, in an experiment, we can control the surrounding temperature by keeping our object in a water bath, or keeping it in a refrigeration unit. But is it a valid assumption to make in nature? Try this thought experiment: put an ice cube into the ocean. The ice cube will melt – it is obviously changing temperature. How much cooler is the ocean because it gave some of its heat to melt the ice cube? The change in temperature is immeasurably small! There’s so much more mass of water in the ocean compared to the mass of a single ice cube. We consider “the environment” to be something like the ocean – a large enough mass of material (water, air, whatever) so that any heat that is transferred into or out of it will not measurably change its temperature. Therefore, from equation (1), it is associated with  the difference in temperature between the object and its (assumed) constant environment.

So, how do we solve this equation? For that, we need integral calculus. But first, we need to rearrange equation (2) into a more useful form (this doesn’t change any of the truth of the equation, just makes it easier to work with):

(2a) dT / (T – Te) = k dt

All I did was swap the dt and the (T – Teparts around. No blood, no foul. Now, we have each side of the equation with its own d term. On the left is dT, a very small change in temperature, and on the right is dt, a very small change in time. When each side of an equation has its own “very small ____________” then we can integrate the entire thing. Integration is the process of undoing all the little d terms… turning them from “very small change in” expressions into regular old expressions. So, if we integrate dt all by itself, we just get t. If we integrate dT all by itself, we just get T. The trick is that the left side of equation (2a) doesn’t just have dT; it also has a T on the bottom there, and that complicates it a bit.

The complication need not stop our progress, though. There are entire, large, print volumes dedicated to cataloging all the integration rules. There are also many available on the intertubes. As it turns out, when we have a dT divided by a T (substitute any variable, as you like), then the result isn’t just T, it is the natural logarithm of T.

So, what the heck is a natural logarithm? What’s a regular logarithm, forget “natural”! When we write something like 102 = 100, there are hoighty-toighty mathematical labels for each of the numbers. The 10 is the base, the 2 is the exponent or power. So, in a regular sentence, we might say, “Ten raised to the power 2 is one hundred.” Great. But we were talking about logarithms… Well, what if we were asked, “What power do we need to raise 10 by in order to get 100?” or equivalently, “100 is 10 raised to what power?” That’s where logarithms (aka, logs) come in. The base-10 log of 100 is 2. And, wait now… 2 is also the power you need to raise 10 by to get 100! Let’s just put it out there:

(3) 102 = 100

(4) log10100 = 2

Just to confirm this… The base-10 log of 1,000 would be 3, because 103 = 1,000. Okay.

Natural logs, then, are just logs that have a specific base, different from base-10. For natural logs, the base is the number e = 2.71828-ish (the number keeps on going…). The number e has a fascinating history. The way we write “natural log” in symbols is ln. So, to say “the natural log of 4” we’d write ln(4). Logs (natural and otherwise) have a pretty cool set of rules, which I won’t really go into, except to say that:

(5) ln(10) – ln(3) = ln(10 / 3)

We can convert a difference of logs into a single log of the quotient of the same numbers! Freaky cool! Anyway, we’ll leverage that in our eventual answer, but I don’t want to bog down with more detail than I’ve already included.

Like I mentioned above, the integration of dT / (T – Teis going to involve a natural log. The integration of k dt is just going to be kt. Let’s write it all out.

(6) ln (T – Te) | To -> T = kt

Well, I’ve introduced a whole new thing here, not the least of which is a new variable, To , which is the starting temperature for our object. An ice cube might have To = 32 F. The rest of the new stuff just tells us to evaluate the natural log from the starting temperature to our variable temperature, T. It results in this:

(7) ln [ (T – Te) / (ToTe) ] = kt

Hey, we’re done! That’s what we need, is equation (7). Let’s recap what it all means, quickly. T is the current temperature of our object at a specific time, t. For our purposes, it’s the temperature of our roast. Te is the temperature of the environment. For our purposes, it’s either the temperature of the air (if we’re letting the roast sit following a stint in the fridge), or the temperature of our grill or smoker. To is the starting temperature of the object. For our purposes, it’s the starting temperature of our roast (about 40 F if you just pulled it out of the fridge).

Oh, but wait… we don’t know what k is! And we can’t know this ahead of time. We have to calculate it for our particular piece of meat. But it is not difficult. Let’s walk through an example. I did this very thing just the other day when I was trying to figure out how long to smoke a 15 pound rib eye roast.

I pulled the roast out of the fridge at 04:00. Using a temperature probe, I found it was 38F. It set it out on the table, on a rack, but otherwise in the open. The temperature in the room was 72F. By 06:30, the roast had risen in temperature to 45F. That’s all the information I needed to calculate my roast’s value of k!

In this case, T = 45, To = 38, and Te = 72. Also, = 2.5 hours. I used the scientific calculator on my computer because it had a natural log function. I calculated = -0.09220946. And that’s all I need, now, to find how long I have to cook this roast, for it to get to any temperature I want!

For this roast, I wanted to smoke it at 220F, and pull it off at an internal temperature of 115F, because I was then going to “reverse sear” it by running the Big Green Egg at 500F for 30 more minutes. Best estimates meant that this would bring it to 125F, and then it would rest for a time, rising another 5 to 10 degrees, for a finished internal temperature of 130-135F, a nice rare roast!

Now, T = 115, To = 45, and Te = 220. Plugging in the numbers, and using my value for k that I calculated, above, I found that I needed to smoke my 15 pounds rib eye roast for just over five hours. Perfect! Since I had to be serving at 1pm, I just backed off the smoke time and the “reverse sear” time and the rest time… and saw that I needed to get the roast on the Egg at 06:45. Brilliant! No guess work! And the roast came out perfectly done… rare and beautiful.

Now, you could argue that since I had a temperature probe, why not just use that, and pull the roast off when it hit 115F, and so on. Sure, that is great when you don’t have a time constraint… but I needed to serve this roast at precisely 1pm… there was no wiggle room! And when there’s no wiggle room, you can’t just guess; you have to use math and science.

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