Question

A bottle of sun tea initially has a temperature of 78°F. It is left to cool in a refrigerator that has a temperature of 35°F. After 10 minutes the temperature of the sun tea is 62°F. Complete parts a through c. a. Use Newton's Law of Cooling, $T = C + (T_0 - C)e^{kt}$, to find a model for the temperature of the sun tea, T, after t minutes. T = \boxed{} + (\boxed{})e^{\boxed{}t} (Simplify your answer. Use integers or decimals for any numbers in the equation. Round to four decimal places as needed.)

          A bottle of sun tea initially has a temperature of 78°F. It is left to cool in a refrigerator that has a temperature of 35°F.
After 10 minutes the temperature of the sun tea is 62°F. Complete parts a through c.
a. Use Newton's Law of Cooling, $T = C + (T_0 - C)e^{kt}$, to find a model for the temperature of the sun tea, T, after t minutes.
T = \boxed{} + (\boxed{})e^{\boxed{}t}
(Simplify your answer. Use integers or decimals for any numbers in the equation. Round to four decimal places as needed.)
        
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A bottle of sun tea initially has a temperature of 78°F. It is left to cool in a refrigerator that has a temperature of 35°F.
After 10 minutes the temperature of the sun tea is 62°F. Complete parts a through c.
a. Use Newton's Law of Cooling, T = C + (T0 - C)e^kt, to find a model for the temperature of the sun tea, T, after t minutes.
T =  + ()e^t
(Simplify your answer. Use integers or decimals for any numbers in the equation. Round to four decimal places as needed.)

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Precalculus with Limits
Precalculus with Limits
Ron Larson 2nd Edition
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A bottle of sun tea initially has a temperature of 78deg F. It is left to cool in a refrigerator that has a temperature of 35deg F. After 10 minutes the temperature of the sun tea is 62deg F. Complete parts a through c. a. Use Newton's Law of Cooling, T=C+(T_(0)-C)e^(kt), to find a model for the temperature of the sun tea, T, after t minutes. T=+1|e| (Simplify your answer. Use integers or decimals for any numbers in the equation. Round to four decimal places as needed.) A bottle of sun tea initially has a temperature of 78'F. It is left to cool in a refrigerator that has a temperature of 35F. After 10 minutes the temperature of the sun tea is 62'F.Complete parts a through c. a. Use Newton's Law of Cooling, T = C+ (To - C) ekt, to find a model for the temperature of the sun tea, T, after t minutes. T=+( (Simplify your answer. Use integers or decimals for any numbers in the equation. Round to four decimal places as needed.)
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Newton's law of cooling states that the rate of change of temperature of an object is proportional to the difference between the temperature of the object and the ambient temperature. (a) Let u(t) be the temperature (in °C, degrees Celsius) of an object as a function of time t minutes, and let a be the ambient room temperature. Show that the differential equation governing the cooling is du/dt = -k(u - a) for some positive constant k. (b) Solve this differential equation. (c) To make tea, David puts a tea bag and boiling water (at exactly 100°C) into a cup and lets it sit for 5 minutes to brew in a room with ambient temperature 20°C. After 5 minutes, the tea has cooled to 60°C. Find the value of k. (d) David likes to have milk in his tea. When he adds milk to the tea, it decreases the temperature of the tea by exactly 5°C. He wants his tea as hot as possible. Should David add the milk before brewing the tea for 5 minutes, or after? Justify your answer mathematically.

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Newton's law of cooling states that the rate of change of temperature of an object is proportional to the difference between the temperature of the object and the ambient temperature. (a) Let u(t) be the temperature (in °C, degrees Celsius) of an object as a function of time t minutes, and let a be the ambient room temperature. Show that the differential equation governing the cooling is du/dt = -k(u - a) for some positive constant k. (b) Solve this differential equation. (c) To make tea, David puts a tea bag and boiling water (at exactly 100°C) into a cup and lets it sit for 5 minutes to brew in a room with ambient temperature 20°C. After 5 minutes, the tea has cooled to 60°C. Find the value of k. (d) David likes to have milk in his tea. When he adds milk to the tea, it decreases the temperature of the tea by exactly 5°C. He wants his tea as hot as possible. Should David add the milk before brewing the tea for 5 minutes, or after? Justify your answer mathematically.

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newton-law-of-cooling-states-that-the-rate-of-change-of-temperature-of-an-object-is-proportional-to-the-difference-between-the-temperature-of-the-object-and-the-ambient-temperature-let-ut-be-73932

Newton's law of cooling states that the rate of change of temperature of an object is proportional to the difference between the temperature of the object and the ambient temperature. (a) Let u(t) be the temperature (in °C, degrees Celsius) of an object as a function of time t minutes, and let a be the ambient room temperature. Show that the differential equation governing the cooling is du/dt = -k(u - a) for some positive constant k. (b) Solve this differential equation. (c) To make tea, David puts a tea bag and boiling water (at exactly 100°C) into a cup and lets it sit for 5 minutes to brew in a room with ambient temperature 20°C. After 5 minutes, the tea has cooled to 60°C. Find the value of k. (d) David likes to have milk in his tea. When he adds milk to the tea, it decreases the temperature of the tea by exactly 5°C. He wants his tea as hot as possible. Should David add the milk before brewing the tea for 5 minutes, or after? Justify your answer mathematically.

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Transcript

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00:01 Hi there, so for this problem we are given the following differential equation.
00:07 That is the derivative of u with respect to the times t is equal to minus a constant of proportionality and then this u minus a.
00:22 Okay, for um, so that's what part of this problem tells us that we need to show that the differential equation covering in the colon is this expression that we are given.
00:47 Now, in this case, we know that ewe represents the temperate war.
00:53 So, first of all, we know that the rate of change of ewe would respect the time.
00:57 First of all, we start by knowing that this is proportional to the difference between the tempore.
01:05 So that will be the deemperator at any given time, and this minus what we call in the case the ambient temperature that we label as a okay and with that set in order to change this proportionality to an equation we introduce a constant proportionality constant so then with this we can set the equality to just simply minus k times u minus a.
01:45 We need to include the minus sign because the temperate work is decreasing.
01:54 So that's the solution for part a of this problem.
01:59 Now for part b, the question for part b is to solve this differential equation.
02:08 So what we need to do in this case is to separate the variables.
02:11 So we will have the differential in u.
02:14 This divided by u.
02:15 Minus a, then this is minus the constant of proportionality times the differential in time.
02:25 Now we're going to integrate both sides of this.
02:32 For the left side of this, we obtain just simply then a variant logarithm of u minus a.
02:39 And for the right side, we obtained minus the constant of proportionality this times the time and this plus a constant of integration that we're going to label as c.
02:51 Now remember that in this case we need to solve for the temperature u.
02:55 So we are going to apply the exponential function in both sides of this in order to get rid of the neparian logarithm.
03:02 So we will have u minus a.
03:05 This is equal to the exponential of minus k times the time this plus c.
03:10 Now remember that when we have the exponential of a sum we can represent that as just simply the esponemential of a times the s penumtional of b.
03:23 Then in this case, that will be u minus a equals to the s penumptial of c, that we can just label a c, these times the s penumptial of minus the comes in proportionality times the time.
03:38 So finally solving for u, that will be then simply a plus c times the s penumption of minus the concept of proportionality times the time.
03:54 So that's a solution for part b of this problem.
03:59 Now for part c we are told that to make t, david puts a t back and boiling water at 100 celsius degrees.
04:10 So with this, we are given the initial temperature.
04:13 This is the temperature at the time equals to zero.
04:16 That will be 100.
04:19 And let's it sit for five minutes.
04:23 To brew in a room with an ambient temperature.
04:27 So for this, we are given the temperature a that is equal to 20 celsius degrees.
04:34 Now, after five minutes, the t has called to 60 degrees.
04:39 So we need to find the value of the concept of proportionality.
04:43 So the temperature after five minutes is equal to 60.
05:02 Then, with that said, what we need to do is to use these two conditions to determine the values of the constant of proportionality.
05:16 So with that said, let's use the first one at zero.
05:21 So we already know that a is 20.
05:25 So we will have 20 plus c and the exponential of zero is 1.
05:31 So we set this equal to 100.
05:34 So solving for c, that will be 80.
05:45 Then the temperate word for any given time now becomes 20, this plus c, which is 80, times the exponential of minus the constant of proportionality, times the time.
06:03 Now we use the other condition to determine the constant of proportionality.
06:07 So we set this equal to 60.
06:11 I will evaluate this at 5...
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