Problem 40 - College Algebra

Question

For the following exercises, use this scenario: A pot of boiling soup with an internal temperature of \(100^{\circ}\) Fahrenheit was taken off he 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 \( T(t) = 69 + 31e^{-0.0113t} \).

Step 1: Understand Newton's Law of Cooling

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

Step 2: Define the Variables

Identify the values from the problem: the initial temperature of the soup \( T_0 \) is \(100^{\circ} \mathrm{F}\), the ambient temperature \( T_a \) is \(69^{\circ} \mathrm{F}\), and after 15 minutes, the temperature \( T(15) \) is \(95^{\circ} \mathrm{F}\).

Step 3: Set Up the Differential Equation

The differential equation using Newton's Law is \( \frac{dT}{dt} = -k(T - 69) \).

Step 4: Solve the Differential Equation

This is a separable differential equation. Rearrange to get \( \int \frac{1}{T - 69} \, dT = -k \int dt \). Integrating both sides, we get \( \ln |T - 69| = -kt + C \).

Step 5: Solve for the Integration Constant

Exponentiate both sides to remove the natural logarithm: \( |T - 69| = e^{-kt+C} = Ce^{-kt} \). Since temperature cannot be negative, \( T - 69 = Ce^{-kt} \).

Step 6: Apply Initial Conditions to Find C

Use initial conditions where \( t = 0 \) and \( T(0) = 100 \). Substitute these into the equation: \( 100 - 69 = Ce^{0} \). Thus, \( C = 31 \).

Step 7: Use Given Data to Find k

Use the data at 15 minutes where \( T(15) = 95 \). Substitute into the equation: \( 95 - 69 = 31e^{-15k} \). Solve for \( k \): \( 26/31 = e^{-15k} \). Taking the natural logarithm gives \( -15k = \, \ln \left( \frac{26}{31} \right) \). Solve for \( k \): \( k = -\frac{1}{15}\ln \left( \frac{26}{31} \right) \approx 0.0113 \).

Step 8: Write the Final Formula

Substitute \( C \) and \( k \) back into the temperature equation: \( T(t) = 69 + 31e^{-0.0113t} \). This is the model for the cooling of the soup using Newton's Law of Cooling.

Key Concepts

Differential Equations

Differential equations are mathematical tools used to relate the rate of change of a variable to other variables. In the context of Newton's Law of Cooling, the differential equation helps us understand how temperature changes over time.
In this situation, the differential equation is given by \( \frac{dT}{dt} = -k(T - T_a) \). Here,

• \( \frac{dT}{dt} \) represents the rate of change of the temperature of the soup with respect to time.
• \( T \) is the temperature of the soup at any given time.
• \( T_a \) is the ambient temperature.
• \( k \) is a positive constant known as the cooling rate constant, which is influenced by factors like the properties of the material and the surroundings.

The form of this equation, known as a separable differential equation, indicates that the rate of cooling is proportional to the difference in temperature between the soup and the room. The process of integrating such an equation yields a solution that describes the temperature as a function of time.

Temperature Modeling

Temperature modeling involves creating a mathematical representation of how temperature changes within a system. Newton's Law of Cooling provides a model for this by establishing the relationship between an object's temperature and the surrounding environment.
In the given problem, our model can be formulated from the solved differential equation:\( T(t) = 69 + 31e^{-0.0113t} \), where:

• \( T(t) \) is the temperature of the soup at time \( t \).
• The term \( 69 \) denotes the ambient temperature, which is a constant since the room temperature is steady.
• \( 31 \) is the initial temperature difference between the soup and the room.
• The exponential factor \( e^{-0.0113t} \) depicts how fast the soup approaches the room temperature over time.

This model helps predict the temperature of the soup at any given time, which is useful in various real-life applications, such as culinary science and industrial processes.

Initial Conditions

Initial conditions are essential for solving differential equations as they provide specific information to find particular solutions. In Newton's Law of Cooling, we apply these conditions to determine constants like \( C \) in our model equation.
For our problem:

• We begin with the initial condition \( T(0) = 100 \), indicating the soup's temperature right off the stove is 100°F.
• By using this information, we find \( C = 31 \), representing the initial head start above room temperature.
• Another key point, \( T(15) = 95 \), gives us information about the temperature after 15 minutes, which is crucial to determine the cooling rate constant \( k \).

The initial conditions allow us to transition from a general solution to one that accurately reflects our specific situation, enabling precise predictions for the soup's cooling over time.