The Piradian Null Device Company

Newton's Law of Cooling

Newton's Law of Cooling is given by:

T - T_m = ce^kt

...where T is the temperature of a cooling object, T_m is the ambient ("room") temperature, t is time, and k is a coefficient of cooling particular to an object. (A ball of aluminum foil will cool faster than a cast-iron skillet; this is what k describes.) It is derived here (.gif image or .xml).

Example 1: The engine in Tiger Woods' Buick has reached a temperature of 190°. He turns off the car. Ten minutes later, the engine temperature is 130°. If the ambient temperature is 65°, how long will it be before the motor cools to 90°?

Well, we're talking about the relationship between temperature and time, and we've already got some valuable information from this. Our problem starts when the car is turned off and begins cooling. So the initial temperature at the inital time is 190°, which we denote mathematically as an ordered pair (time, temperature):

(0, 190)

Second, we are told that after ten minutes, the engine has cooled to 130°, and so we have another time/temperature pair:

(10, 130)

Now we're asked to find out the time t at which the engine will be 90°, so our third ordered pair has an unknown in it. We know we're looking for 90°, but we don't (yet) know the time:

(x, 90)

Lastly, we also know the ambient temperature:

T_m = 65°.

Let's start with our first ordered pair, and see what we can glean of our many unknown variables:

T-T_m = ce^kt

Using (0, 190), our first time/temp combo, we can rewrite this as:

190 - 65 = ce^k(0)

With e's exponent of zero, c gives itself up very quickly:

190 - 65 = c(1)
c = 125

That's useful to know, and now we can use our second ordered pair (10, 130) to figure out what k is. (Remember that after ten minutes, the engine had cooled to 130.) Begin again with Newton's law, and we'll plug in our second ordered pair (10 minutes, 130 degrees):

T - T_m = ce^kt
130 - 65 = 125e^k(10)
65/125 = e^k(10)
ln (13/25) = 10k
k = (ln 13 - ln 25) / 10 {the exact answer}
k ≈ -0.06539 {the answer we can actually USE!}

So now we have all the variables in Newton's equation, except the one we're asked to solve for. The question in the problem asks how long it will be before temperature cools to 90°. So now our ordered pair is (x, 90). We already found c and k, so the only thing left to solve for is time t:

T - T_m = ce^kt
90 - 65 = 125e^(-0.06539)t
1/5 = e^(-0.06539)t
ln 1 - ln 5 = -0.06539t {recall than ln 1 = 0}
-ln 5 / -0.06539 = t
t = 24.61 minutes

That's the answer! It's really three problems in one. In the first problem, we use the first ordered pair to solve for c. In the second, we use the second ordered pair and our newly discovered c value to solve for k. And in the third, we solve for the unknown time t.

Example 2: We could have just as easily solved for an unknown temperature at a known third time. Consider if we kept the first two ordered pairs (since we've already done the work there), but changed the third pair to (24.61 minutes, x temperature). We're simply solving for T instead of t:

T - T_m = ce^kt
T - 65 = 125e^{(-0.06539)(24.61)}
T = 90.0048 ≈ 90°

Precision errors are to be expected. We could've carried out k to an infinite number of digits, but who has the time?

Example 3: Why Newton's Law of Cooling is disturbing:

Consider a cake coming out of a 300° oven to a room whose ambient temperature (T_m) is 75°. After ten minutes, the cake is 170°. When will the cake reach ambient temperature?

Again, we have three ordered pairs:

(0, 300°) Right out of the oven, time zero, 300°
(10, 170°) Cake is 170° after 10 minutes
(x, 75°) Cake will be 75° (same as ambient "room" temperature in X minutes)

Start with Newton's formula to describe cooling:

T - T_m = ce^kt

Using the first ordered pair, we plug in what we know and solve for c.

300 - 75 = ce^k(0)
c = 225

Using the second ordered pair, we plug in what we know and solve for k:

T - T_m = ce^kt
170 - 75 = 225e^k(10)
19/45 = e^k(10)
10k = ln 19 - ln 45
k = (ln 19 - ln 45) / 10
k ≈ -0.08622235106 {I want good precision here, to help illustrate the horror}

Finally, using the third ordered pair, we discover something awful:

T - T_m = ce^kt
65 - 65 = 225e^{(-0.08622235106)t}0 = e^{(-0.08622235106)t}

Well now we have a problem. In order to solve for t we have to take the natural log of zero, and that's not defined. We can cheat a little bit, though, and just choose a very small, non-zero number to see where this is headed. Let's cheat:

0.00000000001 = e^{(-0.08622235106)t}
ln 0.00000000001 = -0.08622235106t

In this case, the cake reaches nearly room temperature (1x10e-11 degrees above it) in 294 minutes. We can use smaller values for T in order to get a more accurate picture of how long it'll take, but according to Newton, it will NEVER reach ambient temperature. So there!