solve-the-given-problems-by-integrationa-hot-metal-rod-with-an-initial-temperature-of-425circ-mathrm

Name: solve the given problems by integrationa hot metal rod with an initial temperature of 425circ mathrm
Uploaded: 2020-06-29
Duration: 4 min 12 s
Channel: Numerade
Description: Solve the given problems by integration.A hot metal rod with an initial temperature of $425^{\circ} \mathrm{C}$ is placed in a room with temperature $20^{\circ} \mathrm{C}$. The time $t$ (in min) required for the temperature $T$ of the rod to cool to $175^{\circ} \mathrm{C}$ is given by $$t=8.72 \int_{175}^{425} \frac{1}{T-20.0} d T$$.Find this time.

Question

Solve the given problems by integration.A hot metal rod with an initial temperature of $425^{\circ} \mathrm{C}$ is placed in a room with temperature $20^{\circ} \mathrm{C}$. The time $t$ (in min) required for the temperature $T$ of the rod to cool to $175^{\circ} \mathrm{C}$ is given by $$t=8.72 \int_{175}^{425} \frac{1}{T-20.0} d T$$.Find this time.

Stephen Hobbs · Accepted Answer

in this Newton&#x27;s law of cooling problem, We start off with the initial value and we&#x27;re waiting to find out how long it will take to cool to a certain temperature. Now, a big part of this one is we&#x27;re already given the K value, which is what we&#x27;ve had this all foreign previous problems. So that will save us some steps. But we&#x27;ll start off with the original equation of the temperature as a function of time is going to equal be ah constant you and that is gonna cool but exponentially decreasing rate. So e to the negative k t bless the ambient temperature. And right away we could go ahead and pull again some initial values. So let&#x27;s go ahead and pull again. 900 degrees is the temperature which will substitute on the left side. Vehicles Oh, are constant times the exponential raised to zero. Since we have a time of zero, it is the beat of the zero is the same thing as one and then plus R and B, a temperature of the room which is 30 degrees. This gives us our see that constant value, which we could see is gonna be 870 degrees civil Scroll down a little here, So that gives us see is 800 and 70. Next, we will use this in conjunction with the rest of the problem to set the whole thing equal to 100 would just use black this time. And we want to find out how long it will take before this rod reaches a temperature of 100. So we&#x27;ll replace the left side with that people to again, we&#x27;re gonna basically use this form of the equation. Um, so it is 8 70 times e. So the negative K, which is negative 0.2 times t which we do not know. So we could just go ahead and leave Battisti and then plus that ambient temperature of the room of 30 degrees tool sulfur t Here, um, we&#x27;ll subtract the 30 over so little give us 70 and then from there we&#x27;re gonna divide both sides by 8 70 or just combining a stuff or two. And that&#x27;ll leave us in the end. On the right side, that&#x27;s just e to the negative 0.0 to t now to solve for T, we would have to get rid of the exponential. So this is where we&#x27;re gonna take the natural log of both sides. Now, one of the left equals a line of the right. And if we do take the natural log of of the exponential there in verses and so Bill cancel. And so we&#x27;ll end up with Ln of 70 over 80 8 70 Divided by this is what&#x27;s left over on the right side. So we&#x27;ll divide both sides by mm. Negative. 0.2 The best by negative zero to. And I don&#x27;t get finally just the time. And as we do, So, um, we&#x27;re going to find out that this gives us about 126 minutes, so it will take about 126 minutes or a rod to cool down to that value of 100 degrees

Katelyn Chen · Answer

for this question were given. T is equal to 85 life, three times a square e of 25 mice. T Where&#x27;s there is lost in your equal to see which is laughter equal to 25. And so first of all were asked to find the room temperature. When T is equal to zero when T is equal to 16 in T is equal to 25. So we first trying t zero just equal to 85. My street 25 minus zero, which is equal to 85 minus three, has five, which we&#x27;ll get US 70. Next we have t of 16 which is equal to 85. My three times 25 months 16 which is equal to 85 minus three. Hers nine, which is equal to 85 minus 27. What? Which will gives 85 miles. 27. Which is equal to Let&#x27;s see. So I&#x27;m not 85 miles 27 85 minus the silverfish. Every three. So 85. My sign which will get us 76. What is Fahrenheit? Lastly, we are Tier 25. She was any fire minus treatments were 25 lives 25 is just equal to 85 F, so for per P were asked to find the rooms average temperature. So the first thing we can do is we can try to integrate area underneath the curve to find our total temperature and then divide 25 so we&#x27;ll get are integral from 0 to 25 85 money, three routes. Did you find my misty, which is equal to our integral from 0 to 25 of 85 minus? You can factor out our three just square it are starting three times into girl. Well, it&#x27;s from 0 to 25 of the square It the 25 minus x the X which we&#x27;re gonna see five x from jurors 25 minus three turns into girl 25 0 Neither roots you, Dean, you when we apply, use institution, which will get us 2125. I see. But you&#x27;ll get us minus three times 2/3 you to 3/2. Gerald&#x27;s 25 which will get US 250. Her story 2125 miles 250 which will get US 2875. We know that there are 25 minutes so we can say that&#x27;s you are average temperature people to account 725 divided by 25 which will guess 75 degrees for a night.

Kurt Kleinberg · Answer

Hi. We&#x27;re here to answer another Newton&#x27;s law of cooling question way. Have a metal that currently is narrative. You re Celsius in a room has an ambient temperature of 30 degrees Celsius. Nicely were given that the proportional constant is pointed to. Normally. We need another initial condition to figure that out, but they&#x27;ve swapped out instead of giving us the initial condition. A 2nd 1 and us finding K. They kind of just gave us Kay. Um, we want to know when is the metal going to be a 100 resources? Remember that Newton&#x27;s law of cooling can nicely be summarized with this formula, your temperature in terms of or the specter time you will have some constant C times e to the negative Katie, plus your ambient temperature. And you should recognize this is a, well, a exponential decay with an asthma toe horizontal as minto at a. So we have something like this going on. There&#x27;s a restaurant. Oh, a In fact, I could do a little better. Here&#x27;s our asthma, took a and are metals cooling, exponential decay and approaching. Uh, in this case, 30 degree Celsius. So we can do is since we&#x27;re very nice. With a lot of info. We can start going and right at finding C someone replace TFT with 900. It&#x27;s gonna be equal to see times e to the zero with because you&#x27;re gonna have negative que No, we don&#x27;t. We know K is pointed to, but I&#x27;m using the initial conditions at time zero. So we&#x27;re just gonna get even. Zero plus her ambient temperature of 30. That leaves us with 870 degrees Celsius equals r constancy. So because but the Celsius see and our constancy So now we know that our temperature pardon me, nor that our temperature. So it&#x27;s recap Europe. We have a even metal 900 degrees currently Celsius in a room of 30 degrees Celsius. We want to know what is gonna be able to want to re Celsius. We started off with our general form or are nice condensed form of new love cooling on and use our initial conditions to figure out our see value. So we do now know is that our temperature with respect to time okay is equal to 8 70 times e to the negative 0.0 to t plus 30 degrees based upon our info. So the question now is when we&#x27;re gonna have a temperature 100 degrees Celsius. So that&#x27;s when Hey, when is that 100? Essentially, let&#x27;s solve 30. It would be 70 equals 8 70 e to the negative point to T 70/8. 70 reduces to 7 87th course. It&#x27;s negative. Pointed to t. I want to natural law both sides to cancel the exponential we now have the natural log of 70/87 equals negative point Go to tea. So one last step is that vision and the natural log of seven divided by 87 that divided by negative 1.2 gets a value of roughly 126 minutes like 1 25.9998 so forth. So it&#x27;s gonna be roughly about 126 minutes for this piece of metal to cool 200 degree Celsius.

Ramzi Deek · Answer

I would got integral of tea cubed signed t duty. So let&#x27;s start picking. Ah, you and Devi. So you will be t cubed So right from beginning I know that we&#x27;re gonna do integration by parts another two times. So do you. Is three t squared. The problem will be longer than previous problems. So Devi is sign t in that case, V is minus co sign t. All right, so integral. A few devi is equal to u V minus integral v Do you you ve in this case is t cubed times minus co sign t So it is going to put minus here minus two cubed co sign T minus into grow off three t squared times minus co sign t So I&#x27;m just gonna take the negative out. So plus co sign t. All right, let&#x27;s take the three out again. No, we can move on to the next Ah, integration by parts. So this will be our new you And this will be our new devi. So you is t squared and D v is well scientist e do you is to t and V is equal to scientific in this case who signed t No. I remember our first part was minus t Cute plus co sign T or Times Co sign T se Long problem. So make sure you all right, follow through and make sure you go back and check your copying things, right? Sometimes this error might happen so plus three times So you&#x27;re gonna keep three out you ve So that&#x27;s t squared times signed t minus into go off V do you? So that&#x27;s two t sign teeth. It&#x27;s now again, we can take the two out. So that&#x27;s two here. Now, this will be our final, um, integration by parts. So T scored 70 minus two t sign t Can&#x27;t really integrate that except by parts. So you is t do you use one? Ah Devi is Scient E and we&#x27;ve done that before. Devious scientist. So v is minus co sign t And now let&#x27;s start plugging everything in so minus t cubed co sign t That&#x27;s way from beginning. Plus three into so and t squared sine t minus two. Now you got open the currencies again so minus two into UV. So tee times minus co scientist. So that&#x27;s minus t co sign t minus Integral. Uh ah v d u So that&#x27;s minus co sign teens. I just put plus go sign t and then closed the bracket here, so we know that integral of co sign t is signed t So now let&#x27;s copy everything you don&#x27;t need to expand. So t cubed co sign T plus three and to t squared scientist E minus two into minus t co signed t plus sign t close, close again. And finally add your constant C. It&#x27;s so this is your final equation here.

Robert Daugherty · Answer

Section 67 Problem number one Dealing with a lot of problems here to do with the integration by parts. So our formula for the integration by parts is the integral of you. Devi is equal to you ve minus the integral VD you and this just evolves out of the product rule. So what you want to do is pick something for you and pick something for a DV. And if you could do that So I&#x27;m looking at this interval. Make a substitution for you and one for TV. What you want for you is something that when you differentiate, it becomes simpler. What you want for Devi is something that you can integrate, uh, rather easily. So when you see a polynomial like T, that is a logical choice for you. So if you is equal to tea, then do you is equal to d T? Then that leaves Devi is equal to e to the five t d t. Then V would be the integration of that which would just be 1/5 yay to the five t. So this integral which is in the form you devi now can be written as you ve so that&#x27;s gonna be t over five. He to the five T minus the integral of the D u. So that is 1/5 e to the five t d t. And so this turns into t over five. You to the five. T minus 1 25th e to the five. See, plus a constant of integration. You can simplify this just by factoring out that, um, 1/5. So you&#x27;re gonna have one fifth, he to the five t, and that leaves you with one minus. Um, excuse me. It was t minus 1/5 okay?

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Solve the given problems by integration.In determ…

Problem 45 Easy Difficulty

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