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When a cold drink is taken from a refrigerator, its temperature is $ 5^o C. $ After 25 minutes in a $ 20^o C $ room its temperature has increased to $ 10^o C. $ (a) What is the temperature of the drink after 50 minutes? (b) When will its temperature be $ 15^o C? $

a) $13.3^{\circ} \mathrm{C}$b) $67.74 \mathrm{min}$

03:42

Wen Z.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 8

Exponential Growth and Decay

Derivatives

Differentiation

Missouri State University

University of Nottingham

Idaho State University

Lectures

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Involve Newton's Law …

we have a Newton's law of cooling problem and when we read the section in the book, we see that the Newton's law of cooling problems fall into the same category as the exponential growth and decay. And we could make our model a little bit more specific. We can replace the why with t minus t sub s, where the tea is the final temperature of the object and the T's of S is the temperature of the surroundings and we can replace why not with t not minus tisa Best t not is the initial temperature of the object and Tisa best is still the temperature of the surroundings. So for this problem, we know that the room temperature or the surroundings temperature is 20 degrees Celsius. We know that the cold drink started at a temperature of five degrees Celsius. So that's our t not and then we have one more piece of information and we're going to use that to help us find the value of K in our model. So it substitutes some numbers in, so the temperature is 10 degrees. We're going to use that for tea and we have our TS, which is 20. So 10 minus 20 equals And then for tea, not minus t s. We have five minus 20 and then we have e raised to the K T. And the time is 25 minutes. So we have 25 times K. So we're going to use this to solve for K. So let's simplify it. We have negative 10 equals negative 15 times each of the 25 K divide both sides by negative 15 and we have 2/3 equals each of the 25 k. Then we can take the natural log of both sides and we have natural log of 2/3 equals 25 k, and we divide both sides by 25 we get our value of K natural log 2/3 over 25. So we substitute that into our equation and we have t minus 20 equals t not, which is going to be five degrees for the duration of the problems. So I'll just go ahead and have that five minus 20. That negative 15 there times e to the natural log of 2/3 over 25 times time. So there's our model that we're going to use to find the answers to part A and part B. All right, so here's part. A. We're going to find the temperature when the time is 50 minutes. So let's substitute 50 in for tea, little tea time and sell for big T temperature. So we substitute the 50 in there, and then we compute the right side of the equation with the calculator and we get approximately six negative 6.7. So we have t minus 20 is approximately negative. 6.7, add 20 to both sides, and we have the temperature is approximately 13.3 degrees Celsius when 50 minutes have elapsed. Okay, Now we'll move on to part B, and in this part, we're finding the time that it would take for the temperature of the drink to get to 15 degrees Celsius. So we're going to substitute 15 degrees Celsius in for capital T, and we're going to solve for lower case T. Okay, let's go ahead and subtract 20 from 15. We get negative five, then we're going to divide both sides by negative 15. We get 1/3 then we're going to take the natural log on both sides now to get t by itself. We want to multiply both sides of the equation by the reciprocal of this fraction. So we're going to end up with T equals 25 times a natural log of 1/3 over the natural log of 2/3. And we put that into a calculator and approximated, and we get about 67.74 minutes.

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