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Acup has been flled with coffee with initial temperature 210* F at 7.00 _ The coffee $ temperature was 1809 F at 7:15 a.m a.In _ Room the coffee'$ temperature ...

Question

Acup has been flled with coffee with initial temperature 210* F at 7.00 _ The coffee $ temperature was 1809 F at 7:15 a.m a.In _ Room the coffee'$ temperature is 709 F. Find temperature at 7.20 a.m Hint: use Newton'$ cooling law

Acup has been flled with coffee with initial temperature 210* F at 7.00 _ The coffee $ temperature was 1809 F at 7:15 a.m a.In _ Room the coffee'$ temperature is 709 F. Find temperature at 7.20 a.m Hint: use Newton'$ cooling law



Answers

GENERAL: Temperature A covered mug of coffee originally at 200 degrees Fahrenheit, if left for $t$ hours in a room whose temperature is 70 degrees, will cool to a temperature of $70+130 e^{-1.8 t}$ degrees. Find the temperature of the coffee after: a. 15 minutes. b. half an hour.

Okay. We know they've given us an equation. And we know now that if capital T is the temperature of a coffee after time teeth, then we know using Newton's law of cooling, which is listed in this chapter. We know we have duty over TFT minus 20 and we know the integral of this is equivalent to and to roll of negative K. Did you? Which gives us cheat to the T minus 20. Even C E to the negative. Katie could remember natural log times e we end up lifting up the exponents. Therefore, we have capital. Chief Chief is 20 co c e to the negative. Kate. Cheap number K is your point. You're too and t is zero there for capital to use 95 see if 75 therefore capital t of tea, the equation would be 20 with 75 e to the negative 0.2 times. T, this is part A. Now we're moving on to part B of this question. We know we're gonna be using the average value Formula one over B minus. Eso won over 30 minutes. You're from A to B from 0 30 capital G of tea. Do you see? We know what our function is. We just figured this out in part, right? Snowy, canned. Hello again. I'm plugging in the function divide by the constant care. When we take the integral from 0 to 30 this is equivalent to 76.4 degrees Celsius upon plugging in.

Cool states that the rate of change of temperature of a body is equal to or is proportional to the difference in the temperature of the body and the surrounding temperatures. T m, where m is the medium. So the water or air or gravy, whatever the object is in. So the temperature of it. Okay, we're given an initial condition. Uh, what happens five seconds later and the temperature of the surrounding medium? We need all that to find the constant. And then when we can answer the question, when will the temperature be 50? Okay, so first thing, this separable equation So we have d t over t minus. Let's go ahead and put the 21 in 40 m equals K d. T in a great natural log of T minus 21. Because Katie plus c all right, we could use it like this, but we would really like a better for the salt for t. So I'm gonna use those both sides as exponents for E because that's the only way to get things out of the logger rhythm. T minus 21. I'm gonna go ahead and just change those to print this. He's in absolute value because I'm going to assume that the temperature of the coffee is more than the temperature of the room. E in natural log are inverse functions. They cancel each other out. So you get T minus 21 equals e to the K t times E to the C. Alright, see, was just a constant. So eat to the sea is just want a constant So let's just call it. See, it's a different see than this, but we don't care. That was just temporary. So we'll call it Big C if you like, eat to the K T plus 21. So now we saw the differential equation for tea, but we have to constant C and K, and we have to find both of those before we can answer the question. So first we know that at time zero the temperature was 95. So 95 equals C E to the zero plus 21 eat to the zero is one. So C is 95 minus 21 which is 74. So now our equation is T equals 74. Eat to the K T plus 21. Okay, Now we have to find out what K is. That's what the second condition was for. When the time is five. The temperature is, uh, 80. Alright. 90 s five temperatures. 80. So 80 equals 74. Eat the K times five plus 21. Okay, subtract 21 from 80. I did it wrong before, so I checked my calculator. 59 equals 74. Eat to the five k, divided by 74. All right, we're trying to solve for Kay. The only way to get the k down from the exponents take the log rhythm of both sides. I'm gonna do natural log, because then the well go away. So we get five k over here. Last step, divide by five. Okay. So you could either leave Kay like this or go ahead and change it to a decimal. If you leave it like this, you're more likely to get the exact or closer to the answer, exact the answer. I'm gonna go ahead and change it to a dismal though 59 divided by 74. Take the natural log of that and then divide by five. And it's negative because K is a cooling constant. The temperature is going down. So negative point. 045306 I'm going to keep all those decimal places since I rounded off right there. Okay, so now our equation from up above is T equals 74 e to the negative point. 04530 60 plus 21. All right, now we can answer the question. And the question is, when? When will the temperature be 50? There. You could see them when I did it long before. All right. 50 equals 74 e to the minus 0.4530 60 plus 21. Subtract 21 from both sides. 50 minus 20. Books 50 minus 21. 29 equals 74 e to the minus 740.4530 60. Divided by 74. Just again. Three. Only way to get something down from the next. Parents take the log rhythm. So l n 29/74 equals the Ellen of this. Those air inverse functions. They cancel each other out. Last step. Divide. All right, Let's see what we get. 29 to 5 by 74. Take the natural log. Divided by 0.45 306 and I get t equals 20.6765 or about 21. I think it was seconds.

Okay. In this question, a formula of temperature of a cup of coffee is given that its capital T equals to 70 plus 100. He raised to the power negative 0.446 small T. Where this T. Is the time in minutes after the coffee is poured in the cup. Okay. And now there are two parts of discussions apart. A what was the temperature of the coffee when it was cold? Or we can say at equals to zero. What will the temperature? Okay, so temperature will be 70 plus 100 years to department negative 0.446 And the time is zero. So he raised to the power negative zero. It will be one. So the temperature will be 70 plus 100 it will be 1 70 F. So this will be the answer of part of discussion. And now part B. That is what when will the coffee be cool enough to drink? Say 1 20 degree for a night. Okay. So the temperature is given 1 20 we have to find out that time. Okay, so now we will put equals temperature records to 1 20 here and we will find out the time T. So uh in the formula 1 20 equals to 70 plus 100 years to the power negative 0.446 small T. That is time. Okay, so here we have to find out this small T. And now, first of all, we will subtract both sides of the equation by 70. So 1 20 minus 70. That is 5200 years to department negative 0.44 60. Now we will divide both sides of the equation by 100. So here 50 divided 100 that can be written. Okay, first of all, 50 divided 100 equals two years to the power negative 0.44 60. And now 5500, that is one by two. Or we can say 0.5 equals two years to the power minus 0.44 60. Okay. And now we can say it will be Yeah, he raised to the power minus 0.44 60 equals to 0.5. Now we will take natural logarithms in the both side of the question and it will be L N E raised to the power negative 0.44 60 equals two Ln 0.5. And now we we know the property of the logarithms that is LNM raised to the power and can be written S n L N M. So this exponent will come before the lock and it will be minus 0.446 T L n E equals two Ln 0.5. And we know the value of L N. Is one because natural logarithms has the base E. And L N. E will be one. So negative 0.0446 T equals two Ln 0.5. Okay. And now we will divide both of the side of the equation by This negative 0.0446. Okay? So that we can find out the value of time t. And it will be Ellen 0.5 divided by negative 0.446 And when we solve it will be approximate 15.54 minutes. Okay. And this will be the answer of part B of our question. Okay. I'm writing down the both the answer in one place. Okay. That is the answer of part A. It will be 1 70 F. Okay. And the answer of part b. That is 15.54 minutes. Okay. Thank you.

Hi And this problem, we're gonna look at an application of an exponential function up in the upper left hand corner. You'll see our function in equation form function. TFT on the right side. You'll see a graph of the same function. We also have a table will be using. And in the bottom left hand corner we're gonna give a description of the problem. What is this about? So our situation here is we're gonna grab a cup of coffee and uh we bring that cup of coffee back to our desk. We set it down on the desk. Uh And what do we start to notice that the temperature of our coffee starts to decrease? So our output in this function or T. F. T. Will be the temperature of our coffee and degrees Fahrenheit, That temp will be affected by our input. T or the number of minutes that coffee is sitting on our desk. So let's come over to our graph and we're gonna look at are y intercept to start or initial value. All right. And we'll see that we get 0180. Right? Uh so this is communicating when you filled your cup of coffee from the pot. Uh it was at 180°F. All right. That's our initial temperature. And if you trace the graph as you carry that coffee back to your desk and let it sit on your desk, the temperature starts to decrease. Right? And you're gonna notice it starts to look like it approaches some value here. Right? The value that it's going to be approaching. Uh Is this 65 here? Uh 65 is representing the temperature of the room. It's pretty cool room 65 degrees. Um And you know, it's that our temperature in the coffee will keep decreasing, but it's only going to decrease to the temperature of the room. Right? If the room is 65 degrees, The coldest, our coffee can get a 65°, it's not going to get cooler than that. Just sitting there on the desk. All right. So, there's kind of the general idea of what's happening with this function. We're going to ask you some questions around that now. All right. So, first question we'll ask you is you fill your cup of coffee? It's 180° when you get it. What temperature will it be? After 10 minutes? I'm gonna punch that into my table. Gives me an answer. I could also come over to the graph here and click on that coordinate When x equals 10. If you wanted to do this by hand, Algebraic Lee. Or with your calculator, you could input 10 for tea. And then use your calculator to find the value of the function. Lots of different approaches there. What is this Coordinate or what is this information communicating to us? The most important part that after 10 minutes of sitting the coffee sitting on your desk It cools to a temperature of 140.56°. We've dropped almost 40° in 10 minutes there. All right. Second question. We're going to switch it around and ask you, when will your coffee at what time or how many minutes will it take for your coffee To get to a temperature of 100°. All right. So in this case we've given you the output of this function. We told you that we know the temp That was 100°. We're asking at what time would that happen? So graphically we can do this by looking for the point of intersection. Notice when a graph y equals 100. End up creating a horizontal line here. That I can then find the point of intersection and see that uh a little after 28 minutes, 28.3 minutes after I set my cup of coffee down on the table, that coffee will have a temperature of 100°. If you are going to try to solve this algebraic lee, You would replace TFT your output with 100 and then use inverse operations. And specifically allow algorithm to help you isolate the variable team, find your time. All right. I personally find the graphing method much quicker. Put in your function uh more more user friendly. But you can definitely also do this by hand


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