a-glass-of-water-whose-temperature-is-50circ-mathrmf-is-taken-outside-at-noon-on-a-day-whose-tempera

Question

A glass of water whose temperature is $50^{\circ} \mathrm{F}$ is taken outside at noon on a day whose temperature is constant at $70^{\circ} \mathrm{F}$. If the water's temperature is $55^{\circ} \mathrm{F}$ at $2 \mathrm{p} . \mathrm{m} .,$ do you expect the water's temperature to reach $60^{\circ} \mathrm{F}$ before $4 \mathrm{p} . \mathrm{m}$. or after $4 \mathrm{p} . \mathrm{m} . ?$ Use Newton's law of cooling to explain your answer.

Michael Jacobsen · Accepted Answer

in this problem. We&#x27;re dealing with a glass of water and t at lower case T indicates the temperature of that water after noon. 12 PM were given that tia zero so directly at 12 PM the water starts out as 50 degrees. Two hours later, the temperatures increased to 55 degrees and the surrounding temperature is a nice 70 degrees itself. What we like to do is determine the solution and its form for this differential equation. They want to term decide whether the temperature will reach 60 degrees before or after four o&#x27;clock. So let&#x27;s start by writing down the solution to this differential equation, we have t at Time T is equal to t sebum the surrounding temperature, plus a constant see times e to the power of negative Katie. First, we&#x27;re given that T sebum is 70 degrees. So start by doing this is a direct substitution here. So the next form of our solution to the differential equation is that temperature of Time T is equal to 70 degrees plus the constant C times Either the naked of Katie. The next step in this process will be determined. The value of that constant C Let&#x27;s look at the information we&#x27;ve used so far. We&#x27;ve used that the surrounding temperature of 70 degrees and we have to initial conditions. Let&#x27;s start with this initial conditions here it says T s zero is 55 degrees, but we sub substitute zero into the differential equation we have so far we would obtain 70 degrees, plus the constant C times e to the power of negative K time. Zero. Then another Substitution tells us that temperature at time zero is 50 degrees itself. So we have the equation. 50 degrees is equal to 70 degrees, plus the constant C, since each of the power of zero invokes a cancellation immediately, then subtracting 70 degrees from both sides we know have negative 20 degrees. Is that constant? C? Let&#x27;s input that back into our solution on the left. Now the equation reads temperature at Time T is 70 degrees minus 20 degrees times E to the power of negative Katie. The next step is to determine the value of that constant lower case K, and to do that will use the final initial condition that wasn&#x27;t used temperature at time to is 55 degrees. Let&#x27;s write that out on the right hand side, so temperature at time two hours from noon would be equal to 70 degrees minus 20 degrees times E to the negative K times two. Source substitution happens at the value for T. Now what we&#x27;re left with is an equation that if we saw four K itself will be able to make a full substitution and know exactly what the solution would look like. So let&#x27;s work on next temperature at time to is 55 degrees, which is equal to 70 degrees minus 20 degrees times each of the power of negative two K. So the next step in solving for the variable K will be to subtract 70 from both sides. The next equation is negative. 15 degrees equals negative 20 degrees each of the power of negative two K free divide bull sites. By negative 20 we would get positive. 3/4 equals e to the power of negative two K. Now let&#x27;s take the natural log of both sides of this equation to obtain Natural log of 3/4 is the natural lock of the Basie exponential, but that gives us a cancellation to say the right hand site is now negative. Two k Once we divide both sides by negative, too. We obtained that that constant K is negative 1/2 times a natural log of 3/4. Let&#x27;s substitute that back into our original solution to that differential equation Temperature at time T. After all, this work is equal to 70 degrees minus 20 degrees. Let&#x27;s substitute next in for K, where we already have one negative sign to account for. So this will be eats the power of positive 1/2 natural log of 3/4 times T. So this is our full format for the solution to that differential equation. In a calculator, we can now check that temperature at time. Four. It&#x27;s 70 degrees minus 20 degrees times E to the power of 1/2 times the natural log of 3/4 times. For if we can put all this information to a calculator, we would get 58 0.75 degrees four hours after the water has been left outside. Likewise, temperature at time five would be equal to 60.2 degrees after we make the substitution and then put the value to the calculator. And so this work tells us finally that the temperature will reach 60 degrees Fahrenheit after 4 p.m. Since we just discovered that at four PM were close to 60 degrees and we obtained just a little bit more than 60 degrees at five PM itself.

Kush Khamesra · Answer

All right, so we&#x27;re giving a variety of information. Let&#x27;s start by converting the minutes and hours to 24 is equal to zero point four hours 24 minutes 0.4 hours. So now we know 50 because we&#x27;re given after 24 minutes. Reserve point for hours be 50 degrees Celsius. So 50 years ago, 200 what we started with e que and then t is your 14 All right? Now we just saw for ah que So we&#x27;ll divide by 100 on both sides and they&#x27;ll take natural logs just to bring this in. The regular question outside of the exponents will get not your log of zero point Viber 1/2 is equal to K time 0.4. Now we divide both sides by serving four. We&#x27;ll get Ln 0.5. Divide adviser four equals K, which happens to be on right up here. So I remember the next portions question and that happens to be Alan zero point. So negative 1.73 again. You can round however far you feel comfortable. I&#x27;m just gonna leave it here. Doesn&#x27;t just the further you just know the further you around the more precise and exact your answer will be. But But at the end of day, it it will be about the same. I want a results. I don&#x27;t pay any. Um, true anyway, now that we know. Okay, all we gotta do is after 96 minutes ago, 96 men is equal to 1.6 hours. The hit was telling us covered these minutes into hours. So I go, Why equals 100? Our original temperature E r k. Is negative. 1.73 times 1.6 hours from appear equals. Just put that into a calculator and you&#x27;ll get around six point Ah, 6.107 e. You&#x27;ll get around 6.2 28 degrees. So, self see, some pretty sure it was, uh, again, depending on how far you rounded your answer. My very But anywhere in this six point to something range

Ian Shi · Answer

All right. So this question is also pretty straightforward plugging what you know and then use that information to solve for not invariable and this cancer trying to solve for the time it takes for, um, something to freeze over. So let&#x27;s get started. So we have this equation laid out with all the admission were given. So we&#x27;ll start by subtracting by from their size. This gives us 27 equal, will combine what&#x27;s inside the premises, and this gives us for it to you to negative 0.0. Want to itch analysis by both sides by 42. This leaves us with 0.64 equals e to the negative. 0.12 pitch. I&#x27;m old announced Baltic Unnatural lock. Besides, to give her to the exponential. So we have natural law give 0.64 equals negative 0.12 each and we&#x27;ll try to stop. Creates so much about both sides. By negative 0.12 It supplies the H. He caused natural lock of zero point 64 over negative 0.12 police In your calculator, you get about 36.2 takes that amount of time for the temperature of 47 degrees, expose an environment of five degrees to reach freezing.

Ankit Gupta · Answer

so Hello, Stone. No. Where would a next on this question on this is about Newton&#x27;s law? Really? So why in water at 100 degrees, integrate is blessed at Fletcher at the relentless integrate the temperature off water each ridiculous indicate opportunity for two minutes. Approximate the temperature off water Upper 96 minutes. Let us see. This is f&#x27;d equal toe d, no less sea of our minus. Oh, my escape. Where be not is temperature environment environment? Uh, B is time efty is temperature off that time The on C is temperature difference between objects on a environment not in SC Here it is given a boiling water 100 C list Indigenous Listen to your temperature differences See which is dead minus Jiro Degutis integrate but just 100 bigness in together. Do you know what temperature it in one environment Temperature zero degrees integrate and efty temperatures after. Don&#x27;t be up to 24 minutes. Did you get the syndicate and time that is given 20 partners Now let us find gate. So Agent ARN T f t equals toe Be not Let&#x27;s see it aboard Minus Katie Putting this identity This is f t equals two, uh, implanted Jiro Lester. Bloody difference. See is 100. And it bade minus okay in 2 24 24 minutes. So now we divide with hundreds of 50 by 100 people Eat four minus Kate times before So 50 Buddha This one by two paper y minus k Interpret before no digging. Ellen Bull say Ellen half the quarter Ln minus K in tow. 24. So this is a hobby. Gina. Point 69315 equal. This is minus. This is also whiteness minus 24. Okay. Minus minus. A concert and K quarto geo 0.6 91 5 upon before. So this is the veil off on stand K. This is different for different material. No, uh, again. FD. It is good. Not plus C. He wore minus K p. Now they&#x27;re asking them Richard upper 96 minutes. So that would be anyone. Deal then put a difference by the same 100. Hey, this and eat about minus gay Geo point 629315 up one before and time. Now they&#x27;re giving 96 minute. So 24 14 96. So this is for 200 in about minus 2.7 it. So ah per solving, you&#x27;ll find this is 6.2 degrees centigrade, so temperatures after R 96 minutes will be 6.2 degrees integrate.

Carson Merrill · Answer

So first we want Thio, identify what Newton&#x27;s law of cooling states. And that is, um the temperature as a function of time is equal to the temperature of the surrounding environment. So piece of s plus cannot minus key s times B to the K T. Where K is a constant that&#x27;s different for each object and t is time right hours, that is So what we see here is, um, we have our initial temperature or our temperature of the surrounding area, which is he&#x27;s a past at 70. And we know that, um, it&#x27;s poured out at 50 F. So what we have is 50 minus 70 and as a result of that, that&#x27;s gonna be negative. 20. So, so far, we have negative 20 each of the K T plus 70 and now all that&#x27;s left to find is the value. Okay, um, to find K, this is just going to be some constant, Um and we know that based on the temperature change that we have, the K value is going to ultimately end up Equalling 0.7. So now it will have is it will be than the natural log of 0.7 key over to. So now what we&#x27;ll have is Yeah, the natural log of 0.7 t over to and this right here would be our final equation. If we look at the graph, we see that it starts at 70 degrees and then, um, after two minutes so we could go to t equals two a t equals two. We see that the temperature is down to about 46 point 336 degrees. Um, and that&#x27;s just based on the drink itself. If the temperature of the room starts off here, we see that the temperature of the drink will wind up there. But this is gonna be the final equation that we end up getting for our function.

Heather Zimmers · Answer

all right, Here&#x27;s the equation. Were given to use Newton&#x27;s law of cooling and figure out wth e value of K the constant if the temperature is 90 degrees when the time is three minutes. So let&#x27;s go ahead and substitute those values into our equation. We have 94 capital t the temperature, and we have 34 lower case t the time. Just offer Kay. Let&#x27;s start by subtracting 60 from both sides. Now we have 30 equals 40 e to the three K. Then we&#x27;ll divide both sides by 40 and now we have 3/4 equals e to the three K. Then we&#x27;ll take the natural log of both sides so we have the natural log of 3/4 equals three K, and then we&#x27;ll divide both sides by three. So we get natural log of 3/4 divided by three for K. That&#x27;s the exact value. If you want the approximate value, you can put that into your calculator. Be real careful with parentheses with logarithms to make sure you get the order of operations. Correct. Now we get a K value of negative 0.959

Problem

On a cold winter day $\left(10^{\circ} \mathrm{F}…

Problem 9 Medium Difficulty

Answer

after $4 \mathrm{pm}$

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after $4 \mathrm{pm}$

Differential Equations and Linear Algebra

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Stephen W. Goode, Scott A. AnninDifferential Equations and Linear Algebra

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Stephen W. Goode, Scott A. Annin

Differential Equations and Linear Algebra