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Newton's law of cooling says that a hot object cools rapidly when the difference between its temperature and that of the surrounding air is large, but it cools more slowly when the object nears room temperature. Suppose a piece of aluminum is removed from an oven and left to cool. The following table gives the temperature \(A=A(t)\), in degrees Fahrenheit, of the aluminum \(t\) minutes after it is removed from the oven. $$ \begin{array}{|c|c|} \hline t=\text { Minutes } & A=\text { Temperature } \\ \hline 0 & 302 \\ \hline 30 & 152 \\ \hline 60 & 100 \\ \hline 90 & 81 \\ \hline 120 & 75 \\ \hline 150 & 73 \\ \hline 180 & 72 \\ \hline 210 & 72 \\ \hline \end{array} $$ a. Explain the meaning of \(A(75)\) and estimate its value. b. Find the average decrease per minute of temperature during the first half- hour of cooling. c. Find the average decrease per minute of temperature during the first half of the second hour of cooling. d. Explain how parts b and c support Newton's law of cooling. e. Use functional notation to express the temperature of the aluminum after 1 hour and 13 minutes. Estimate the temperature at that time. (Note: Your work in part c should be helpful.) f. What is the temperature of the oven? Express your answer using functional notation, and give its value. g. Explain why you would expect the function \(A\) to have a limiting value. h. What is room temperature? Explain your reasoning.

Step by step solution

01

Understanding the meaning of A(75)

The notation \( A(75) \) represents the temperature of the aluminum 75 minutes after it has been removed from the oven. Since 75 minutes is halfway between 60 minutes (with a temperature of 100°F) and 90 minutes (with a temperature of 81°F), we can estimate the temperature at 75 minutes by averaging these two values: \( \frac{100 + 81}{2} = 90.5 \) degrees Fahrenheit.

02

Calculating average decrease per minute during first half-hour

To find the average rate of temperature decrease during the first 30 minutes, we calculate the difference in temperature from \( t = 0 \) to \( t = 30 \) and divide by the time elapsed. The temperature changes from 302°F to 152°F, so the decrease is \( 302 - 152 = 150 \)°. Over 30 minutes, the average decrease per minute is \( \frac{150}{30} = 5 \) degrees per minute.

03

Calculating average decrease per minute during first half of second hour

In the first half of the second hour (from \( t = 60 \) minutes to \( t = 90 \) minutes), the temperature decreases from 100°F to 81°F. Thus, the change in temperature is \( 100 - 81 = 19 \)°. Since this time span is 30 minutes, the average decrease per minute is \( \frac{19}{30} \approx 0.633 \) degrees per minute.

04

Supporting Newton's law of cooling

Parts b and c show that the rate of cooling decreases over time, which supports Newton's law. During the first 30 minutes, temperature decreases at 5 degrees per minute, while the first half of the second hour shows a much slower rate at approximately 0.633 degrees per minute. This aligns with the law, as cooling rate decreases as the temperature difference with the surrounding air lessens.

05

Expressing temperature after 73 minutes and estimating

To express the temperature 73 minutes after removal in functional notation, we use \( A(73) \). Since 73 minutes is closer to 60 minutes (100°F) than to 90 (81°F), we can use interpolation: Estimate \( A(73) \) as \( A(73) = A(60) + \frac{73 - 60}{90 - 60} \times (A(90) - A(60)) \approx 100 + \frac{13}{30} \times (81 - 100) \approx 93.8 \)°F.

06

Determining the oven's temperature

The oven temperature is represented as \( A(0) \), which is the initial temperature of the aluminum at time \( t = 0 \). From the table, this is 302°F.

07

Explaining the limiting value of the function A

Newton's law of cooling suggests that as time progresses, the temperature of the aluminum will approach the surrounding air temperature (room temperature) but will never quite reach it, creating a limiting value. This behavior forms a property of exponential decay toward an asymptote, which is the room temperature.

08

Identifying room temperature and reasoning

The room temperature can be observed as the limiting value that \( A(t) \) approaches over time. As \( t \) increases after 180 minutes, the temperature stabilizes around 72°F, indicating the room temperature to be approximately 72°F.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Temperature Change

Understanding how temperature changes is key to comprehending Newton's Law of Cooling. When you initially remove the aluminum from the oven, the temperature difference between the aluminum and the surrounding air is maximal. This significant difference causes the aluminum to lose heat rapidly at first. As the aluminum continues to cool, and its temperature nears the surrounding room temperature, the rate at which it loses heat slows down.

This change is evident from the exercise, where you first observe temperatures dropping from 302°F to 152°F in just the first 30 minutes. As time progresses, the rate of temperature change decreases, supporting the principle that larger temperature differences result in quicker heat transfer.

Recognizing these patterns is essential to predict how objects will cool down in real-world settings.

Exponential Decay

Exponential decay describes processes where the value of a quantity decreases at a rate proportional to its current value. In Newton's Law of Cooling, this concept explains why the object's cooling rate is faster at higher temperature differences and slower as it approaches equilibrium (room temperature).

This behavior is formally represented using the equation:\[A(t) = T_r + (A_0 - T_r) e^{-kt}\]where:

• \( A(t) \) is the temperature at time \( t \)
• \( T_r \) is the room temperature
• \( A_0 \) is the initial temperature of the object
• \( k \) is the cooling constant

The negative exponent—\(-kt\)—is what enables the decay, reflecting the gradual slowing down of temperature change as equilibrium nears. This results in a curve that appears to flatten out over time.

Cooling Rate

The cooling rate refers to how quickly the aluminum loses heat. Initially, this rate is relatively high because of the larger temperature difference between the aluminum and the room. As you progress through the exercise, you note that the aluminum's cooling rate is very fast at the start, at 5 degrees per minute within the first half-hour.

As time goes on, the difference in temperature between the aluminum and its surrounding air becomes smaller, reducing the cooling rate to about 0.633 degrees per minute by the first half of the second hour. This slower rate of cooling indicates the object's temperature is reaching closer to that of the room, exemplifying the concept of decreasing cooling rates over time as explained by Newton's Law of Cooling.

Through careful observation and calculation, you can identify these cooling rates and encapsulate an essential part of thermal interaction, governed by the surrounding environment.

Functional Notation

Functional notation provides a concise and clear way to represent relationships between variables—in this case, time and temperature for the cooling aluminum. Using notation such as \( A(t) \), you can denote that the temperature of the aluminum depends on the elapsed time \( t \) since removal from the oven.

In the problem, functional notation is also used to estimate unknown values. For instance, estimating \( A(75) \) means finding the temperature at 75 minutes by interpolating between known values at 60 and 90 minutes. This form of notation simplifies the communication of mathematical relationships and makes it straightforward to apply calculations like interpolations or use formulas such as those expressing exponential decay.

It empowers you to quickly represent and work with sophisticated concepts without delving into lengthy descriptions, facilitating a deeper understanding of how variables interact in Newton's Law of Cooling.