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Use Newton's Iaw for cooling (lecture 2) T(t) = T; + (To TH)e-k to solve the problem: Suppose that a cup of soup is placed in a Freezer where the temperature i...

Question

Use Newton's Iaw for cooling (lecture 2) T(t) = T; + (To TH)e-k to solve the problem: Suppose that a cup of soup is placed in a Freezer where the temperature is ~15"C, and it takes 5 minutes to cool down to 609 C How long would it take the soup to cool from 90'C to 35"C.12.2 min11.9 min15.5 min: 10.7 min11.03 min10.6 min:104 min:

Use Newton's Iaw for cooling (lecture 2) T(t) = T; + (To TH)e-k to solve the problem: Suppose that a cup of soup is placed in a Freezer where the temperature is ~15"C, and it takes 5 minutes to cool down to 609 C How long would it take the soup to cool from 90'C to 35"C. 12.2 min 11.9 min 15.5 min: 10.7 min 11.03 min 10.6 min: 104 min:



Answers

These exercises use Newton’s Law of Cooling. Cooling Soup A hot bowl of soup is served at a dinner party. It starts to cool according to Newton's Law of Cooling, so its temperature at time $t$ is given by $$ T(t)=65+145 e^{-0.05 t} $$ where $t$ is measured in minutes and $T$ is measured in $^{\circ} \mathrm{F}$. (a) What is the initial temperature of the soup? (b) What is the temperature after $10 \mathrm{min} ?$ (c) After how long will the temperature be $100^{\circ} \mathrm{F} ?$

Because the temperature of the room in this case is 20. The ts values here are gonna get replaced with twenties, so we'll go ahead and put those in and on minus 20. Now, the initial temperature of the soup was 90 degrees, So we'll put that in for our initial value and its final value that it cooled to with 60. So we'll put that in on the left side there. Okay, so those are our temperatures so far, and then we have that This took place in 10 minutes. So we hav e excuse me. This would be negative since its cooling and so would be negative. K times turn, and then we just want to solve and isolate. Okay, so from here, we could just say All right, left side is 40 of the right side. It's 70. Divide that 70 over, and then take the natural log of both sides. And then Ellen and e Well, of course, cancel. So this will just give us negative K times ton. These ones go away, okay. And then we're gonna divide both sides by 10. And so, Ellen, and even give it to the zeros. Ellen, 4/7 divided by 10 is going to give us our value. Oh, negative K. And this ends up becoming hey is approximately 0.0 five 96 This point. We are done with the prep work so we can jump in tow part day, which is one drink. We'll go ahead and just read it. How much longer would it take the soup to cool to 35 degrees. So assuming that the soup is now at 60 degrees, we want to say Okay, our final that we're interested is 35. The room is still 20. So minus pointing. We just said we're starting it off from you know what? We went out for it here. So in the short is a soon to be sexy. Subtract the temperature of the surrounding medium and then now we have this K value. So we could just say e of the negative k and we want to isolate, See? So we'll just go ahead and clean this up a little bit. The left side would give us 15. The right side would give us 40. So it's 15/40 after dividing that part over and then we'll take the natural log to get rid of the E there. So Alan on both sides is equal to negative K plans t divide out that negative, cater to the other side. Um, here will. How long it took tow. Cool toe. 35 degrees. This ends up being about 17.5, and the units are minutes. So, um, additional 17.5 minutes to cool down to 35. Part B is very similar to a except that the surrounding medium is different. So how long would it have taken if supposed, now that the temperature is the freezer, which is negative 15 degrees. So, as you can see, this is gonna change to a plus since minus and negative on the sides is assuming that waas from the beginning. And so the couple's 90 degrees minus a negative 15. Uh uh. These and then we have our you. So the negative Katie again, we know our k value and red over here. And so then we're just gonna isolate Tia's. We've been doing so on the left side. For instance, will get 50 on the right side, will get 105 will divide the one of five on both sides. we'll take the natural log, get rid of this e on both sides, aligned on both sides. That will leave us. What negative T dividing negative K over. We're gonna end up getting tea is approximately 13 point to sex. So just to make sure that this answer makes sense, it's much less time than it took otherwise. And so putting it in the freezer is going to reduce down to that temperature of 35 degrees much faster.

This problem. We're given the serving. Come between degrees pH. D director to be 90 and temperature up there. 10 minutes to be 60 real. These news is locally, and this is what it's a vocal and reads So let's first blood the values that we know in your mind this constant J doing that. We see that 60 miners 20 is equal to 90 minus 20 and then it's exponential minus time. Okay, so here we have 40 six months for me, and we have seven here. So then if you take them to live on both sides, we see that or seven will be with you to the time kay. It actually, uh, no financial black hole times. You don't see that, Casey? Negative off nation law. Off or over seven over 10. And that is equal to 0.56 All right. No, ain't for part a We don't wanna, uh, see the time to order structure to readies to 35 degrees someone less. Take h Not this time to be 60. So, Ryan country still the same. And the H is 35 again. Uses low. Be done. Have 35 minutes. 20 is defective. 60 minus 20 times exponential of 200.56 This is not king. They thought that time t on the left, us I'd be happy to, you know, like right inside the 40. So then he would have a 80 over 40 is equal to be negative. 0.56 turns t. So promise we see that tea is that equal to eventual negative a bitch. Walk three or eight 0.56 Something that is equal to about 17.5 minutes. So it would take about 18 more minutes for it to reach 35 p's in part B. We're going to assume that the, uh we're going to assume that the surrounding temperature is now. Thank you, Dean. So the ladies know is 90? Uh, h is 35 and it's around Cemeteries. Nothing. 15. So, a flying into locally BC that 35 fighters 30 Routine is 90 minus 15 and that is exponential. Negative 150.56 T from recipe. Have 50 or fortify disease. Exponential. A zero. Uh, they didn't point there about six times. T So now you take natural look of both times We see that tea is that equal to minus national off 10 or 21 you have 5.56 and that is it. About 31 3 minutes. So it would take about 14 minutes or to go for 90 degrees. Very fast agrees in a surrounding Ah, what native? 15 degrees?

Section 64 Problem number 30 ones who were dealing with Newton's Law of cooling. Ah, they give us a scenario here where you've got a cup of soup. This soup starts out at 90 degrees Celsius, and then it is 60 degrees Celsius 10 minutes later, and it's sitting in a room that is 20 degrees. According to Newton's Law of Cooling. We know that the temperature, uh, minus um, the initial term should be the steady state room temperature can be expressed as, um T's of zero minus T c. S you to the minus. Katie. So it follows exponential. So there's an exponential model that fall that follows the cooling of this subject. So in this, this is going to be our temperature, Uh, time little tee tee's of s, that is our steady state or our room temperature or the ambient temperature and then t subzero. That is our initial temperature. So in this case, we know that the initial temperature was that 90 degrees. Okay, we know that the ambient room temperature is going to be this 20 degrees. Okay, so that we get a substitute in a couple of places. Okay? Now what? We're asked to find. First of all, is to find how much longer until the soup cools to, um, soup cools to 30 degrees. So let's take a look at that and see what we can come up with. Um, so let's just take a look. Our formula is T minus. Tisa, Bess, This sequel to t subzero Mind his teeth of S E to the minus. Katie. Now, in this formula, we know that our steady state room temperature is 20 degrees. We know that the initial temperature that soup started out at 90 degrees. Okay? And we know that it was equal to 60 degrees when the tea was equal to 10 minutes. Okay, so let's substitute all of that into, um, our formula. So we're gonna have 60 minus 20 is equal to 90 minus 20 you to the minus K. And this was 10 minutes later. So this is going to allow us to solve for the value of K. And then once we have that, we can answer how much longer it would take for that temperature to get the 35 degrees. So let's go and do this. We're gonna have 40 is equal to 70 you to the minus 10 K. Yeah, so we get divide both sides by 70. You get 4/7 equal E to the minus 10 K. Then we take the natural log on both sides. The natural log of 4th 7th is equal to minus 10 K. Therefore, K is equal to minus 1/10 natural log of 4/7. It's okay. Equal minus 1/10 natural log for seventh. So you'll substitute this into your calculator. U K is equal to 0.55 96 So what do we know in our formula t minus? And we knew that the steady state temperature was 20. It's Our formula is T minus 20 is equal to. And then we had tea subzero minus to suggest that was 70 e to the minus 0.0 5596 t. So for this scenario, the soup This is Newton's law of Cooley. So they wanted to ask us How long will it take for that temperature to get the 35 degrees? So when that happens, you're going tohave 35 minus 20 equal 70 e to the minus 0.55 96 t So 15 equal 70 you to the minus 0.55 96 t divide both sides by 70 and you get 15. Divided by 70 equals E to the minus 0.55 9 60 Then I'm gonna take the natural log of both sides of this equation. So I get, um the natural log of 15 over 70 is equal to minus point No. 55 96 t. Therefore, t is equal to minus 1.55 96 natural log of 15 over 70. So on your calculator when you substitute this and you're gonna come up with T is approximately 27 0.5 to 8. So it's gonna take about 27.5 to 8 minutes for this super cool from 90 down, too. 35. Now what they ask, the specific question was, how much longer would it take to get the 35 degrees? So we were told it took 10 minutes, um, to get down to 60 degrees. So if it took 10 minutes in that original problem to go from 90 to 60 and I just found out it's 27.5 to 8. 27.5 to 8. That's how long it takes to cool from 90 2 35 So it's only an additional 17.5 to 8 minutes. And that's how long it takes to cool from 60 two. 35. Okay, so an additional 17 point 5 to 8 minutes to make that cooling happen. Now, let's move on. We've got part B of this problem. If instead of being left to stand at room temperature, that cup of 90 degrees super put into a freezer at negative 15 degrees, how long would it take to cool from 92 35? So I'm gonna start in 90 degrees is gonna be my initial temperature, and then I want my final temperature apartment will be back out. No, um, and it's instead of being a room, temperature is put into a room and negative 15. So negative 15 degrees. That's gonna be my steady state. So that is now the room Ambient temperature. Um, that I'm plugging in. And I wanted to be my final temperature. I wanted to cool down to 35 degrees and there we should be good to go. So remember our formula WAAS T minus tease of s is equal to t subzero minus tease of S E to the minus K. But in this case, we know what Kay was. We solved it a moment ago. 0.0, 55 96 t. Okay, so I wanted to reach 35 degrees. So 35 minus negative 15. It's equal to that. Initial temperature was 90 and then that steady state temperature is negative. 15 Youth, a minus 0.55 96 t. So this gives me 50 is equal to 105 you to the minus 0.0, 5596 t. So 50 divided by 105 is equal to E to the minus 0.5 96 Now I'm gonna take the natural log of both sides of ah, this equation. So this is 50 over 105 That's gonna be by the both of those by five. That's 10 over 21. So I might have the natural log of 10 over 21 is equal to minus 0.0, 5596 t. Therefore, T is equal to minus one over point 05596 times the natural log of 10 over 21 and if you substitute this in your calculator, you come up with about 13.26 so it's gonna take about 13.26 minutes to go from 90 degrees, 2 35 degrees in a negative 15 degree. That it was a freezer Come again, pretty long problem here, but we all we're doing is Newton's law of cooling started out here is the law of cooling. We use the temperatures they gave us to figure out how long it would take to cool something toe 35 degrees. And then we saw for value of K and R T and then in part B. They said if we wanted to stick it into a freezer at negative 15 degrees, the steady state temperatures change, then how would that affect everything going forward? And we found out that it would have to be about 13 minutes to take this

We already know that the mortal function off Newton laws of cooling this do you off is equal to be It bless D not e bowler, my s k b So we are PS is around in temperature The noticed our difference between the initial later than does start learning temperature is the time taken for the cooling door open. So in the given question, given information is that only them But it's 64 degrees father here. So he s is 65 degrees father here That difference just a minute. So let's simplify this. It will become 35 his equal dough 1 45 Ebo minus zero point. You know, for your people So that will be e over minus 0.5 p is equal to 35 by 1 45 So if we apply alone on boats, it's the EPA will go. So we will apply long on both sides. That will give us minus, you know, points. You know, five d is equal to lawn that he forgave by 1 45 so that just be easy. Quarto minus one by 0.5 Indo Lawn 35 by 1 45 So after the simplify it is, he's equal. Do 28 point fourth to the minutes so that their nature 100. If these five need Dr, it'll be just 100 degrees track after to 28.4 30 minutes.


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