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# In Exercise 14 in Section 9.1 we considered a $95^oC$ cup of coffee in a $20^oC$ room. Suppose it is known that the coffee cools at a rate of $1^oC$ per minute when its temperature is $70^oC.$(a) What does the differential equation become in this case?(b) Sketch a direction field and use it to sketch the solution curve for the initial-value problem. What is the limiting value of the temperature?(c) Use Euler's method with step size $h = 2$ minutes to estimate the temperature of the coffee after 10 minutes.

## a) $\frac{d T}{d t}=-0.02(T-20)$b)Equilibrium Temperature is $20^{\circ} \mathrm{C}$T varies with Time as $T(t)=20+75 e^{-t / 70}$c)Temperature after 10 mimutes is $\approx 81.153$

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Differential Equations

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##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

Hello Today we're looking at a differential equation problem, and in this problem we have a cup of coffee sitting at room temperature of 20 degrees Celsius and it is The cup of coffee is 95 degrees Celsius and we know that it calls at a rate of one degrees when it's 70 degrees. The cup of coffee Sony periods Celsius. So first we want to write a differential equation for this problem. So in order to do this, we need to know that the rate of cooling of ah hot body is proportional to the temperature difference off its surroundings. So that can be rewritten as d t d t, which is the rate of cooling that then we have okay, which is proportional to the temperature difference with the surroundings so t minus the temperature difference. And we know that it starts off with a temperature. The room temperatures 20 degrees. So we know that TM is 20. Okay, so now we have to solve for K to finish this equation. So in order to do that, we need to know that it we need to look at this right here, which tells us that it cools at a rate of one degrees Celsius permanent at seven degrees Celsius. So we know at this time T t T T is negative one. It's negative one because it's cooling. And we know that at this point in Time T equals 70. So it's called Unplugged. Those two values since we have native one equals k time 70 minus 20 to have negative one equals 50 k swift k equals negative zero 0.2 So when we re are the equation, we get DT t t equals negative 0.2 times T minus 20. And that is the answer for part of his problem. Okay, so then, for part B of this problem, we've been asked to sketch a soap filled and then sketches solution curve through that salt field. So when we're looking at our differential equation, you notice that there's Onley the y value, which is teeth. So we can, uh, go ahead and create a little table for ourselves. We have are Tee And then what is our d t p. T? So let's just go ahead and start at 0 100. So win t, there's 100. We can plug that into our calculator and we get 100 minus 20 times a negative 0.2 Do we get DT? DT is negative 1.6 um, so we know that a t t we were sketching are still fields that a value of negative one has a soap of negative 45 degree angle, and a value of zero is flat. And then so we can look at negative 1.6 and know that it's a little bit more than a 45 degree angle. And that will be true across the X values, because this equation is not affected by the X values. Okay, so then we can go ahead and go move on to 90. So 90 and we plug that into our equation. 90 minus 20 times negative. 0.2 We get 90 equals negative, 1.4. So again, not too much of a difference. Pretty. It's just a little bit less than at 100. Okay, let's move on the 80 and we get I'm negative 1.2 still more than 45 degrees, but not as much. And these don't have to be exact, so let's just keep working and we have 70 equals. Negative one point. Oh, and that makes sense because, um since we know that it's going at a rate of one degree when it's 70 degrees. So this is going to be right at that 45 degree angle. Alright, then 60 is at negative. 0.8 was a little bit less than that 45 degree angle. 50 40 30 20. So we noticed that it's just going down Bye point to each time before we get to 0 20 so we can just go out and graft thes points. We're just gonna keep getting smaller and flatter as we go down until we reach 20 which is flat. So I'm just going to go ahead and finish this up. Okay, so we've had this finished slope field and we just need to draw solution, curve. Starting at our point, T f zero is equal 95 so that would be right here on ours. And let's go ahead and draw a line through here. So have a curve that will look something like this and it will plateau at the At T equals 20. So we know that the limiting value off. The temperature is going to be 20 degrees. And this makes sense because the room temperature is 20 degrees, so it be you couldn't get below 20 degrees. Otherwise you the cop couldn't keep below 20 degrees because it would stop cooling. At that point, it would be at equilibrium. So that's how you do problem be. So we're gonna move on to problems, you know? Okay, So, for part C, we need to use jewelers method to estimate the temperature after 10 minutes. So we have our starting value off X zero. Yeah. And why is 95? Why Miss Cases Teeth. Okay. And we know that h is too. So the step size in between each of these will be the change in H. So this is 2468 10. Okay. And so we're going to need to do this a couple of times. So to find d t d t, we plug the 5,000,095 in, and when we do that, we have 95 minus 20 time, the negative 0.2 and we get negative 1.5, and then we multiplying 1.5 times that, uh, changing X, which is the same as H. This is lots of the H value. Okay, so negative 1.5 times to is negative three. Okay. And then we subtract three from the Y value to get our next y vice. We have 95 minus three, which is 90 to, So you just continue to fill out this table. So when you do that, you get a table look like something like this, As you can see, the typical a little bit wonky because we needed more space. But now we have our estimate for 10 minutes, that is at T 10. The temperature in the room will be approximately 80 in the cups are in a through B 81.1 five Thio, let's go in around four decimal places. So we have 153 zero degrees Celsius and that would be your final answer for part

University of Washington

#### Topics

Differential Equations

##### Heather Z.

Oregon State University

##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

Lectures

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