Problem 29 - Algebra and Trigonometry

Question

Cooling Soup When a bowl of hot soup is left in a room, the soup eventually cools down to room temperature. The temperature \(T\) of the soup is a function of time \(t\) . The table below gives the temperature (in "F) of a bowl of soup \(t\) minutes after it was set on the table. Find the average rate of change of the temperature of the soup over the first 20 minutes and over the next 20 minutes. During which interval did the soup cool off more quickly? $$ \begin{array}{|c|c|c|c|}\hline t(\min ) & {T(\mathbf{F})} & {t(\min )} & {T(\mathbf{F})} \\ \hline 0 & {200} & {35} & {94} \\ {5} & {172} & {40} & {89} \\ {10} & {150} & {50} & {81} \\ {15} & {133} & {60} & {77} \\ {20} & {119} & {90} & {72} \\ {25} & {108} & {120} & {70} \\ {30} & {100} & {150} & {70} \\ \hline\end{array} $$

Step-by-Step Solution

Answer

The soup cooled more quickly during the first 20 minutes with an average rate of change of -4.05°F/minute.

Step 1: Identify the Temperature Values

The soup temperature at each time point given in the table is needed for the calculation of the average rate of change. For the first 20 minutes, the temperature values are at 0 minutes (200°F) and 20 minutes (119°F). For the next 20 minutes, they are at 20 minutes (119°F) and 40 minutes (89°F).

Step 2: Calculate the Change in Temperature for the First 20 Minutes

Subtract the final temperature from the initial temperature for the first 20 minutes. \[\Delta T = T_{20} - T_{0} = 119^\circ F - 200^\circ F = -81^\circ F\]

Step 3: Calculate the Change in Temperature for the Next 20 Minutes

Subtract the final temperature from the initial temperature for the next 20 minutes. \[\Delta T = T_{40} - T_{20} = 89^\circ F - 119^\circ F = -30^\circ F\]

Step 4: Calculate the Average Rate of Change for the First 20 Minutes

The average rate of change is calculated as the change in temperature divided by the change in time. For the first 20 minutes:\[\text{Average Rate of Change} = \frac{\Delta T}{\Delta t} = \frac{-81^\circ F}{20 \, \text{minutes}} = -4.05^\circ F/\text{minute}\]

Step 5: Calculate the Average Rate of Change for the Next 20 Minutes

The average rate of change for the next 20 minutes is calculated similarly. \[\text{Average Rate of Change} = \frac{\Delta T}{\Delta t} = \frac{-30^\circ F}{20 \, \text{minutes}} = -1.5^\circ F/\text{minute}\]

Step 6: Compare the Average Rates of Change

Compare the two average rates of change to determine during which interval the soup cooled more quickly. The rate of \(-4.05^\circ F/\text{minute}\) in the first 20 minutes is greater in magnitude than \(-1.5^\circ F/\text{minute}\) in the next 20 minutes, indicating that the soup cooled off more quickly during the first interval.

Key Concepts

Understanding Temperature Change

Temperature change refers to how the temperature of an object, such as soup, varies over time. In our exercise, the temperature of the soup starts at a high point when it is initially set out and gradually decreases as time progresses. This phenomenon is common and can be observed in many everyday situations.

When calculating temperature change, it's important to note the temperature at specific points in time. This helps us understand how quickly or slowly the temperature is changing. In our example of the cooling soup, the temperature decreases as it approaches room temperature. The challenge is to measure and quantify that change over specified intervals.

Exploring the Cooling Process

The cooling process describes the manner in which an object loses heat over time, gradually reducing its temperature until it stabilizes with the surrounding environment. For a bowl of soup, this means it will cool down until it reaches room temperature.

At the start, the soup is at 200°F. The natural process of cooling involves heat transferring from the soup into the surrounding air. Several factors can affect this process, such as the size of the room, air temperature, and even whether there is a breeze.

In our exercise, we see that the soup cools faster at the beginning. This is typical because the temperature difference between the soup and the air is largest initially. As the temperature difference decreases, the rate of cooling slows down.

Using Mathematical Modeling

Mathematical modeling helps us understand and predict the behavior of real-world processes. In this case, we use mathematical models to represent the cooling of soup.

One common model for cooling is Newton's Law of Cooling, which states that the rate of change of temperature is proportional to the difference between the object's temperature and the ambient temperature. Though we don't explicitly use this law in our step-by-step solution, it helps guide our understanding of how objects cool.

In our task, we calculate the average rate of cooling over specific time intervals. By doing this, we create a simple mathematical model that helps quantify the process. This involves calculating the overall temperature change and relating it to the duration over which the change occurs.

The Importance of Time Interval Analysis

Time interval analysis involves examining how a process changes over specific periods. In our soup-cooling exercise, we analyze two intervals: the first 20 minutes and the subsequent 20 minutes.

This type of analysis is important because it reveals how changes occur over time and allows us to identify where, within a process, the most significant changes are happening.

In our example, by comparing these two intervals, we determine that the soup cools more rapidly during the first interval. This finding is significant because it highlights the non-linear nature of cooling - the process is faster when the temperature difference is greater. Analyzing time intervals thus provides a deeper understanding of dynamic processes and their behaviors.