Problem 541 - Precalculus

Question

For the following exercises, use this scenario: A soup with an internal temperature of \(350^{\circ}\) Fahrenheit was taken off the stove to cool in a \(71^{\circ} \mathrm{F}\) room. After fifteen minutes, the internal temperature of the soup was \(175^{\circ} \mathrm{F}\) . Use Newton’s Law of Cooling to write a formula that models this situation.

Step-by-Step Solution

Answer

The formula is \( T(t) = 71 + 279e^{-0.0677t} \).

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. Mathematically, this is modeled as: \( \frac{dT}{dt} = -k(T - T_{room}) \), where \( T \) is the temperature of the object, \( T_{room} \) is the ambient temperature, and \( k \) is a positive constant.

Step 2: Set Up the Differential Equation

Using Newton's Law of Cooling, we set up the equation for this situation. The cooling process is described by the differential equation \( \frac{dT}{dt} = -k(T - 71) \) since the room temperature is \(71^{\circ}F\).

Step 3: Solve the Differential Equation

Separate variables and integrate: \( \int \frac{1}{T-71} dT = -k \int dt \). Solving this gives \( \ln|T-71| = -kt + C \). Exponentiating both sides, we have \( T - 71 = Ce^{-kt} \), where \( C \) is a constant determined by initial conditions.

Step 4: Apply Initial Condition

The initial condition given is that the soup's temperature \( T \) at time \( t=0 \) is \( 350\). Substitute these into the equation: \( 350 - 71 = Ce^{0} \). Thus, \( C = 279 \). So \( T(t) = 71 + 279e^{-kt} \).

Step 5: Use Additional Condition to Find k

After 15 minutes, the temperature of the soup is \( 175^{\circ}F\). Substitute \( T=175 \) and \( t=15 \) into the equation: \( 175 = 71 + 279e^{-15k} \). Thus, \( 104 = 279e^{-15k} \). Solve for \( k \) by dividing, \( e^{-15k} = \frac{104}{279} \). Take the natural log: \( -15k = \ln\left(\frac{104}{279}\right) \). Solve for \( k \), yielding \( k \approx 0.0677 \).

Step 6: Write the Final Temperature Formula

Replace \( k \) in the formula: \( T(t) = 71 + 279e^{-0.0677t} \). This formula models the cooling of the soup over time.

Key Concepts

Differential Equation

A differential equation is a mathematical model where the rate of change of a variable is expressed in terms of the variable itself and its derivatives. In the context of Newton's Law of Cooling, the differential equation describes how the temperature of an object changes over time in relation to the ambient temperature. The equation takes the form \(\frac{dT}{dt} = -k(T - T_{room})\). Here, \(\frac{dT}{dt}\) denotes the rate of change of temperature, \(T\) is the object's temperature, \(T_{room}\) is the constant room temperature, and \(k\) is a positive proportionality constant. Differential equations like this are crucial for modeling processes where change is continuous, helping us predict future conditions.

Initial Conditions

When solving a differential equation, initial conditions are vital as they help determine the specific solution to a general equation. These conditions are the known values of the variable at a particular point in time. In this exercise, the initial condition is that the soup's temperature is initially \(350^{\circ}F\) when taken off the stove. Using this information, we set \(T=350\) when \(t=0\). This condition allows us to find the constant \(C\) in the equation \(T - 71 = Ce^{-kt}\), leading to a specific solution tailored to our scenario. Initial conditions ensure that our mathematical model perfectly reflects the real situation being analyzed.

Temperature Modeling

Temperature modeling refers to the process of creating mathematical representations to predict how the temperature of an object changes over time. In this scenario, Newton's Law of Cooling provides a framework for modeling the soup's temperature as it cools down. The derived formula \(T(t) = 71 + 279e^{-0.0677t}\) arises from the initial differential equation and captures the soup's temperature \(T\) at any time \(t\). With this model, one can not only understand but also visualize the cooling process, anticipating future temperatures of the soup as it equilibrates with room temperature. This practical application is vital in fields like engineering and environmental science, where temperature control is critical.

Integration

Integration is an essential calculus technique used to solve differential equations, enabling us to find original functions given their derivatives. In the solution for Newton's Law of Cooling, integration helps transition from the rate of temperature change to an explicit formula describing the temperature at any given moment. We start by separating variables in the equation \(\frac{1}{T-71} dT = -k dt\). By integrating both sides, we obtain \(\ln|T-71| = -kt + C\), which can be further solved to express \(T\) directly. This step showcases how integration transforms differential relationships into practical, understandable models that describe real-world phenomena, like cooling.