Cooling Time of a Pizza Pan A pizza pan is removed at 5: 00 PM from an oven whose temperature is

Cooling Time of a Pizza Pan A pizza pan is removed at 5: 00...

Key Concepts

Asymptotic Behavior

As time goes on, the temperature of the cooling object approaches the ambient temperature, meaning that although the temperature continuously decreases (or increases if coming from a lower value), it never actually reaches the ambient temperature in finite time. This concept highlights the long-term behavior of systems described by exponential decay.

Differential Equation Modeling

The cooling process can be modeled using a first-order differential equation, typically written as dT/dt = k(T - T_ambient), where k is a constant. Solving this differential equation leads to an explicit formula for the temperature, which is essential for predicting the cooling time at various temperatures.

Exponential Decay

Exponential decay describes a process where the quantity—here, the temperature difference between the object and the ambient temperature—decreases at a rate proportional to its current value. In the context of cooling, the temperature drop follows an exponential curve as time progresses.

Newton's Law of Cooling

This principle states that the rate at which an object cools (or warms) is proportional to the difference between its own temperature and the ambient temperature. It forms the basis for modeling cooling processes and leads to an exponential temperature decay towards the environment’s temperature.

Transcript

00:01 They want us to use newton's law of cooling to help us answer a couple of questions about this pizza pan.

00:09 So they first tell us that it's removed at 5 p .m.

00:13 And it's starting at a temperature of 450 degrees.

00:17 And in the room that we're leaving this pan, it's 70 degrees.

00:22 So that's going to be our surrounding temperature of t is equal to 70.

00:27 And our u -n is going to be the initial temperature of our object, which is the feet of the pan, which is 450 degrees.

00:39 They also tell us that after five minutes, the pan becomes 350 degrees fahrenheit.

00:44 So we'll use that last part in a little bit.

00:49 But for now, let's go ahead and first see what we can do for this problem.

00:56 So they tell us that we want to find the time the pen is 135 degrees fahrenheit.

01:04 So before we actually use anything with that, let's plug everything into newton's law of cooling, which is this equation up here in the top left, and then see what we end up with.

01:16 So u of t is going to be our final, so we won't change anything there.

01:21 Our initial temperature is 70, or our surrounding temperature is 70, and then plus u -not, which is our initial temperature of the pizza pan, or 50, and then minus the surrounding temperature again, 70.

01:38 And then we have e to the kt.

01:41 And we don't know what k is yet.

01:46 So we can simplify this down though to give us 70 plus 380, e to the kt.

01:57 Now, if we want to solve for time in both a and b, it may be a good idea to go ahead and try to solve for t to start.

02:10 And then after that, we can just plug in our values.

02:14 But it looks like we'll have to solve for k as well.

02:20 So let's go ahead and first just solve for k or t.

02:26 And then we can just plug in things as we need it.

02:30 So let's see.

02:31 Well, the first thing we would need to do to get k or t by itself is to subtract 70.

02:37 So i'm going to do that.

02:39 So you'd get u of t minus 70 is equal to 380, e to the kt.

02:47 And then we would need to divide each side by 380.

02:52 So we have u of t minus 70.

02:57 Over 380 is equal to e to the kt.

03:01 And then lastly, to get rid of that natural log, remember, or to get rid of that base e, we want to take the natural log on each side.

03:10 So i'll just go ahead and actually write that here.

03:12 Natural log like this, those cancel out, and then we're just left with kt over here.

03:19 So now, depending on if we're going to solve for k or t, then we can just plug stuff in.

03:25 And remember, we first need to figure out what our, k is, so let's go ahead and solve for that.

03:30 So we would divide by little t.

03:32 So actually let me put a little star by this because we'll probably need to come back to this equation in a second.

03:40 And also let me go ahead and mark this piece off here.

03:46 All right.

03:46 So let's plug everything in that we know to try to solve for k.

03:52 So we're told that after five minutes our temperature should be 300.

04:00 So that's coming from right here.

04:02 So let's go ahead and plug that in.

04:05 So we'd have natural log of 300 minus 70 over 380.

04:13 And this is going to be equal to, so it was 5 minutes, so 5k.

04:18 And then we would just need to divide by 5.

04:21 And that's going to give us.

04:23 So this implies that k is equal to.

04:26 So if we divide by 5 over, it would be 1 5th, natural log, of so in the numerator here their 300 minus 70 would give us 230 and then 230 divided by 380 well we can simplify that by dividing each by 10 so that would give us 23 over 28 so that would be our value for k so we have that done and now if we want to figure out what is our time that the pen is going to be 135 degrees fahrenheit? well, we can go ahead and now plug in what our k is.

05:17 Let's do that really quickly.

05:18 So we're going to take this fact along with this fact and combine them.

05:27 And in doing that, that's going to give us.

05:30 So let me scroll down a little bit.

05:32 First, the equation just ln of u .t.

05:36 Minus 70 over 380 is equal to, so k we found was 1 5th natural log of 23 over 28 t...

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