Question

A cake is removed from the oven after baking thoroughly, and the temperature of the oven is \(450^{\circ} \mathrm{F}\). The temperature of the kitchen is \(70^{\circ} \mathrm{F}\), and after 10 minutes the temperature of the cake is \(430^{\circ} \mathrm{F}\). a. Write the appropriate initial-value problem to describe this situation. b. Solve the initial-value problem for \(T(t)\). c. How long will it take until the temperature of the cake is within \(5^{\circ} \mathrm{F}\) of room temperature?

Short answer

It takes approximately 470 minutes for the cake to cool within 5°F of room temperature.

Step-by-step solution

Understand the Problem

We are dealing with Newton's Law of Cooling which states that the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the ambient temperature. We need to set up a differential equation with given initial conditions.

Set up the Differential Equation

Newton's Law of Cooling can be expressed as \( \frac{dT}{dt} = -k(T - T_{room}) \), where \( T \) is the temperature of the cake at time \( t \), \( T_{room} = 70^{\circ} \mathrm{F} \), and \( k \) is the proportionality constant to be determined.

Apply Initial Conditions

Initially, when \( t = 0 \), the temperature of the cake \( T = 450^{\circ} \mathrm{F} \). After 10 minutes, \( T = 430^{\circ} \mathrm{F} \). These conditions will help us find the value of \( k \).

Solve the Differential Equation

We integrate the differential equation \( \frac{dT}{dt} = -k(T - 70) \) to get the equation of temperature over time, \( T(t) = T_0 e^{-kt} + 70 \), where \( T_0 \) is the initial temperature difference: \( 450 - 70 = 380 \).

Determine the Constant \( k \)

Use the condition \( T(10) = 430 \) to find \( k \). Substitute into the equation: \( 430 = 70 + 380 e^{-10k} \). Solve for \( k \): \( e^{-10k} = \frac{360}{380} \), which gives \( k \approx 0.00526 \).

Express the Solution for Temperature Over Time

Substitute \( k \) into \( T(t) = 70 + 380 e^{-0.00526t} \). This equation gives the temperature of the cake over time.

Calculate Time for Cake to Cool to Within \(5^{\circ} \mathrm{F}\) of Room Temperature

We need to solve \( T(t) = 75 \) (as \( 70 + 5 \)). Replace in the temperature equation: \( 75 = 70 + 380 e^{-0.00526t} \), and solve for \( t \). Simplify to get \( e^{-0.00526t} = \frac{5}{380} \) and find \( t \approx 470 \) minutes.

Key concepts

Differential Equation

In the scenario of cooling cake, we employ a differential equation to describe how the temperature changes over time. A differential equation involves a function and its derivatives—essentially, it's an equation that relates a quantity to its rate of change. In this exercise, we're using Newton's Law of Cooling, which provides a perfect example. It's succinctly expressed as \[\frac{dT}{dt} = -k(T - T_{room})\]where:

• \(T\) is the temperature of the object at time \(t\).
• \(T_{room}\) is the ambient room temperature.
• \(k\) is a constant proportional to the rate of cooling.

The equation tells us that the rate of temperature change of an object is directly proportional to the difference between the object's current temperature and the surrounding temperature. Understanding this lays the groundwork for solving temperature dynamics in physics and engineering applications.

Initial-Value Problem

An initial-value problem is a type of differential equation that also includes conditions at the start of the observation. These initial conditions are crucial as they allow us to find a unique solution to the differential equation by determining unknown constants. In our example, the initial conditions are:

• At time \(t=0\), the temperature of the cake is \(450^{\circ} \mathrm{F}\).
• After 10 minutes, the temperature is \(430^{\circ} \mathrm{F}\).

These conditions help to calibrate the equation, allowing us to find the constant \(k\) and express the decay of temperature over time. By incorporating these specifics into the equation, we tailor the mathematical model specifically to our situation. This is essential because every situation, governed by similar laws, might have different initial starting points.

Temperature Decay

Temperature decay refers to how the temperature of an object decreases over time relative to its surroundings, especially when influenced by Newton's Law of Cooling. In our problem, the cake cools from a higher temperature towards the ambient temperature, which is cooler.
Utilizing the formula derived from our differential equation, \[T(t) = T_{room} + (T_0 - T_{room}) e^{-kt}\]we can track the temperature decay. Here, \(T_{room}\) is the ambient temperature and \(T_0\) is the initial temperature. The term \( e^{-kt} \) models the exponential decay based on the constant \(k\) found using initial conditions.
This naturally diminishing process means that the rate of change in temperature is rapid initially but slows as it approaches the ambient temperature. Calculating the time for the cake to come within \(5^{\circ} \mathrm{F}\) of room temperature involves solving the equation for \(t\) when the temperature is close to \(75^{\circ} \mathrm{F}\). This highlights how mathematical models predict real-life changes over time.