Problem 40 - College Algebra with Corequisite Support

Question

For the following exercises, use this scenario: A pot of warm soup with an internal temperature of \(100^{\circ}\) Fahrenheit was taken off the stove to cool in a \(69^{\circ} \mathrm{F}\) room. After fifteen minutes, the internal temperature of the soup was \(95^{\circ} \mathrm{F}\). Use Newton's Law of Cooling to write a formula that models this situation.

Step-by-Step Solution

Answer

The formula is: \( T(t) = 69 + 31e^{-\left(-\frac{1}{15}\ln\left(\frac{26}{31}\right) \right)t} \).

Step 1: Understand Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature. The formula is: \[ \frac{dT}{dt} = -k (T(t) - T_{a}) \]where \( T(t) \) is the temperature of the object, \( T_a \) is the ambient temperature, and \( k \) is a positive constant.

Step 2: Integrate the Differential Equation

To find the formula, we need to integrate the differential equation. Separating variables, we have:\[ \frac{dT}{T(t) - T_a} = -k \, dt \]Integrating both sides, we get:\[ \ln|T(t) - T_a| = -kt + C \]where \( C \) is the integration constant.

Step 3: Solve for the Temperature Function

Exponentiating both sides gives:\[ |T(t) - T_a| = e^{C}e^{-kt} \]Let \( C' = e^{C} \), hence:\[ T(t) = T_a + C'e^{-kt} \]This is the general solution of the temperature as a function of time.

Step 4: Use Initial Conditions to Find Constants

We have the initial condition \( T(0) = 100^{\circ} \). Setting \( t = 0 \) in the function:\[ 100 = 69 + C' \cdot e^{0} \Rightarrow C' = 31 \]Now for the condition at 15 minutes: \( T(15) = 95^{\circ} \):\[ 95 = 69 + 31e^{-15k} \]Solving for \( k \):\[ 26 = 31e^{-15k} \Rightarrow e^{-15k} = \frac{26}{31} \Rightarrow -15k = \ln\left(\frac{26}{31}\right) \]\[ k = -\frac{1}{15}\ln\left(\frac{26}{31}\right) \]

Step 5: Write the Final Formula

Substitute the value of \( k \) and \( C' \) back into the general solution:\[ T(t) = 69 + 31e^{-\left(-\frac{1}{15}\ln\left(\frac{26}{31}\right) \right)t} \]

Key Concepts

Differential Equations

Differential equations play a crucial role in understanding how quantities change over time. They can express complex real-world problems in mathematical terms, describing how a particular quantity evolves. For example, in the case of Newton's Law of Cooling, the equation \( \frac{dT}{dt} = -k (T(t) - T_{a}) \) models the rate of cooling of an object. Here,

• \( \frac{dT}{dt} \) represents the rate of change of temperature with respect to time.
• \( T(t) \) is the temperature of the object at a specific time \( t \).
• \( T_a \) is the constant ambient temperature.
• \( k \) is a positive constant that needs to be determined.

This differential equation implies that the rate at which the temperature changes is proportional to the difference between the object's temperature and the room temperature. This is why differential equations are powerful tools in modeling dynamic systems.
By solving these equations, you can predict the future behavior of the system, which is precisely what Newton's Law of Cooling enables us to do.

Temperature Modeling

Temperature modeling involves creating a mathematical representation of temperature changes over time. The purpose is to simulate how a temperature, like that of the soup, decreases when placed in a cooler environment. In Newton's Law of Cooling, the model is constructed using the differential equation derived above. This allows us to form an explicit temperature function, \( T(t) \), which describes how temperature evolves.In the context of the cooling soup, the initial conditions state that:

• The internal soup temperature is initially \( 100^{\circ}\)F.
• The surrounding room temperature is \( 69^{\circ}\)F.

By utilizing these initial conditions, we can define specific parameters to closely represent the scenario. The equation models how quickly the soup's temperature approaches room temperature over time. Understanding temperature modeling is essential, particularly in fields like physics, engineering, and climate science, where predicting temperature variation is crucial.

Integration

Integration is a fundamental mathematical tool used to solve differential equations like the one in Newton's Law of Cooling. In the given problem, once the differential equation \( \frac{dT}{dt} = -k (T(t) - T_{a}) \) is set up, integration helps us to find \( T(t) \), the temperature as a function of time.The process begins with separating the variables to isolate terms involving \( T \) on one side and terms involving \( t \) on the other:

• \( \frac{dT}{T(t) - T_a} = -k \, dt \)

Integrating both sides:

• \( \ln|T(t) - T_a| = -kt + C \)

where \( C \) is the integration constant. Exponentiating both sides gives us \(|T(t) - T_a| = e^{C}e^{-kt}\), leading to a clear form of the equation \( T(t) = T_a + C'e^{-kt} \) once we solve for \( C' = e^{C} \). Integration transforms the abstract rate of change into a tangible function that describes temperature behavior over time, providing substantial insights into the cooling process.

Exponential Functions

Exponential functions frequently appear in temperature modeling and various natural phenomena due to their property of representing continuous growth or decay processes. When we solve the differential equation for Newton's Law of Cooling, we derive an exponential function: \( T(t) = T_a + C'e^{-kt} \). This form indicates:

• \( T(t) \) will approach the ambient temperature \( T_a \) over time.
• The rate of decay is influenced by the constant \( k \) and the initial temperature difference.
• As time \( t \) increases, the impact of \( e^{-kt} \) diminishes, causing \( T(t) \) to stabilize at \( T_a \).

In this exercise, the soup's decreasing temperature over time is captured by \( 31e^{-kt} \), a part of our solution that decays exponentially. The exponential function showcases how fast or slow this process occurs. Such functions are vital in modeling scenarios like radioactive decay, population growth, and cooling processes, making them indispensable tools in mathematics and science.