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Aggignment We Kaxt to think about how hot cup of coffee (or tea or any hot liquid) E888= "s make some predictions: What would be reasonable starting temperatur...

Question

Aggignment We Kaxt to think about how hot cup of coffee (or tea or any hot liquid) E888= "s make some predictions: What would be reasonable starting temperature for cup of hot coffee? After about how long would it be cold? About what temperature is it when it'$ cold? (Why?) If we were t0 oxrape the coolin? %rocqdene" with time on the X Kheis and ieepevouzes on y axis general Would it be a straight line? Make a Skezqe _ %f whete EeRergtapk would look like_ Describe the end "be

Aggignment We Kaxt to think about how hot cup of coffee (or tea or any hot liquid) E888= "s make some predictions: What would be reasonable starting temperature for cup of hot coffee? After about how long would it be cold? About what temperature is it when it'$ cold? (Why?) If we were t0 oxrape the coolin? %rocqdene" with time on the X Kheis and ieepevouzes on y axis general Would it be a straight line? Make a Skezqe _ %f whete EeRergtapk would look like_ Describe the end "behavioe oerthe of the graph. Katah the video for Aoignpedkctiond and look at actual data for a sample of a cookinaitect Were your correct? "Giquidyhat factors can you think of that will the rate of cooling of a



Answers

Suppose a hot cup of coffee is left in a room for 2 hours. Sketch a reasonable graph of what the temperature would look like as a function of time. Then sketch a graph of what the rate of change of the temperature would look like.

For part A. The coffee cools down most quickly in the beginning, when it is taken off the heat source, and as time goes on, the rate of cooling will slow down as it approaches. Room temperature are what we will be referring it thio as the ambient temperature heartbeat. Newton's all of cooling states at the rate of cooling oven object is proportional to the difference between the temperature oven object and the ambient temperature To write that in mathematical terms going to say the rate of change in temperature in respect of time should be proportional to the temperature oven object in the ambient temperature. Gonna put a negative in front there to say that it is cooling. Do you not like that negative With the K, we can always rewrite it like this. The only thing we have left to do now is plug in the ambient temperature. We're also asked what the initial temperature is. Just 95 degrees Celsius. We can express this just like this. Part C. We're gonna do a rough sketch of a graph 20 95 going to draw a dotted line just to make sure I don't go past my ambient temperature. We're going to say yes. The model in the answers support each other because as the temperature approaches the ambient temperature of 20 the rate of change in temperature with respect to time is approaching zero. Thank you so much.

There's a 42 being, uh, yeah. No, no. Just meaning 60. Yeah. All right. Nice. But look great when you think you're what's it's okay. Nature works trying to prison. This is thinks that somebody no Cherries on the makes those Yeah, injuries. And then, you know something that's a major real. And then names. Okay, this would lighter where your injury is and no much for essentially that. Maybe, you know, And on their second case, prove it's just Thank you. Do you ever sit down, extrapolate? Let's sit down and think all the different iterations shouldn't is a very strange from, um this No way. Yeah, it's no more. So you can unto those questions, right? Okay. For the five. Okay, guys, let's take a look at 42 as pulling all the data into decimal. You can get a C and A So that's movie one for a or a f t e code To see Time's 18 poverty equal you 93 which is dusty, comes a zero point my a approximately to the poverty. So with pocket pc, it's asking, This is our model for part B is one hour, one hour ridges, 60 minutes, 60 minutes minus 06 So, yeah, t you this 60 minutes equal to 93 times the 90 i Quincy to the power 60 equal to put them into calculator. You got this answer?

So this problem is a Newton's law of cooling problem, and we want to find out and party when this is when the temperature is decreasing most quickly. When do we think the coffee cools most quickly? And this would be at the beginning, when the temperatures dropping most drastically towards room temperature and so we could go ahead and just jump right to part. Um, see here. So it starts off. We know at 95 degrees the room itself is 20 degrees. That's going to be the lowest that it gets. And that's our horizontal Assam to it there and then a big decrease and then followed by a much smaller decreases as we go. So what happens to the rate? Basically, the rate, um, decreases as time goes on, Or I guess, another way to say that is it kind of increases towards the tends towards more positive values is another way. Toe say that as well. And then finally, if we write, rewrite the oh spell bad getting and then for part B. If we right at this differential equation here, it's proportional, which means it's gonna have a K value. It's negative, since it's a decreasing, but you could call it positive or negative, I think, and then to the difference between whatever the temperature of the object is and the surrounding temperature. So this is the directly proportional to the difference between these two.

In this video, we're going to be looking at Newton's law of cooling. So Newton's law of cooling states at the rate of change of temperature of an object is proportional to the difference between its temperature and the temperature of its surroundings. So the example will look at is a cup of coffee that is 95°C cooling down in a room that is 20°C.. So, um if we first, just intuitively think about what Newton's law of cooling is saying, we know that the temperature of our hot object in our cool room is going to decrease and um that temperature is going to um temp film. It's going to decrease the fastest when the difference between it and surroundings. Yes, greatest. And that makes sense if we put um two different cups of coffee in the same 20° room with one of them already at 50°C.. Um if we work to measure the temperature a few seconds later, in both cups, the hotter cup would have cooled off more. Um Just intuitively, that makes sense to us. So let's go ahead and try and write a different equation modeling this. So we know the rate of change of the coffee's temperature is proportional to the temperature difference. So right of change sets off derivative bills in our heads right away. So we know that D. T. D. T. Big T. Being 10th and little t being time going to be equal to. Um Let's write that term that it's proportional to first uh T -20° in this case. And these are not equal their proportional who needed proportionality constant. And we know that our temperatures decreasing. So this needs to be negative. So that's our differential equation. You think about what the initial value would be if we want to make this an initial value problem. Um Our temperature T at time zero is 95 degrees Celsius, add that on there. So if we were to make a quick sketch of the graph of this function, this is a very rough sketch, but just so you get an idea, um There's 95°, there's 20 degrees. So our function would start at 95. So we know there will be a point there and as time goes on approach 20 and we already said that it's going to decrease the fastest when this difference is the greatest. So that means our slope will be much larger, larger in magnitude closer to T equals zero. This is T big T. So let's make a quick sketch here. What was kind of rough? There we go. So starting at two equals zero will be at 95° and then the temperature will decrease exponentially.


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