Question

Assume Newton's law of cooling applies. A chef removed an apple pie from the oven and allowed it to cool at room temperature \(\left(72^{\circ} \mathrm{F}\right)\). The pie had a temperature of \(350^{\circ} \mathrm{F}\) when removed from the oven; \(10 \mathrm{~min}\) later, the pie had cooled to \(290^{\circ} \mathrm{F}\). How long will it take for the pie to cool to \(120^{\circ} \mathrm{F}\) ?

Short answer

Answer: It takes approximately \(21.33\) minutes for the pie to cool to \(120^{\circ} \mathrm{F}\).

Step-by-step solution

Write the equation for Newton's Law of Cooling

We can represent Newton's law of cooling by a differential equation: \(\frac{dT}{dt} = -k(T - T_{room})\) where, - \(T\): temperature of the object (in this case, the pie) - \(t\): time - \(k\): proportionality constant - \(T_{room}\): ambient temperature

Solve the differential equation

In order to solve the differential equation, we can separate the variables and integrate: \(\int\frac{dT}{T - T_{room}} = -k \int dt\) Let \(u = T - T_{room}\), then the above equation becomes: \(\int\frac{du}{u} = -k \int dt\) \(\ln|u| = -kt + C\), where \(C\) is the integration constant Since \(u = T - T_{room}\), we have: \(\ln|T - T_{room}| = -kt + C\)

Find the proportionality constant (\(k\))

We are given that the pie has a temperature of \(350^{\circ} \mathrm{F}\) when removed from the oven (\(t = 0\)) and cools to \(290^{\circ} \mathrm{F}\) after \(10\) minutes (\(t = 10\) min). We can use this information to solve for \(k\): When \(t = 0\), we have: \(\ln|350 - 72| = -k(0) + C\) \(\ln|278| = C\) When \(t = 10\) min, we have: \(\ln|290 - 72| = -10k + C\) \(\ln|218| = -10k + \ln|278|\) Now we can solve for \(k\): \(-10k = \ln|218| - \ln|278|\) \(k = \frac{\ln(|278|) - \ln(|218|)}{10}\)

Find the time needed for the pie to cool to \(120^{\circ} \mathrm{F}\)

Now that we have \(k\), we can plug this value into the equation and solve for \(t\) when the pie cools to \(120^{\circ} \mathrm{F}\) : \(\ln|120 - 72| = -kt + \ln|278|\) Now plug in the value of \(k\): \(\ln|48| = -\frac{\ln(|278|) - \ln(|218|)}{10}t + \ln|278|\) Solve for \(t\): \(t = \frac{\ln(|278|) - \ln(|218|)}{\ln(|278|) - \ln(|48|)}\times 10\) Calculate \(t\) using the given values: \(t \approx 21.33\) minutes So, it will take approximately \(21.33\) minutes for the pie to cool to \(120^{\circ} \mathrm{F}\).

Key concepts

Differential Equations

Differential equations are mathematical expressions that relate a function with its derivatives. They show how things change over time or in relation to other variables. In the context of Newton's Law of Cooling, the differential equation is used to model how the temperature of an object, like an apple pie, changes as it cools down in a room. Newton's law describes the cooling process using the equation:

• \( \frac{dT}{dt} = -k(T - T_{room}) \)

Here, \( \frac{dT}{dt} \) represents the rate of change of temperature with respect to time, \( T \) is the current temperature of the object, and \( T_{room} \) is the ambient temperature, or room temperature. The change of temperature over time is proportionate to the difference between the object's temperature and the room temperature, symbolized by the constant \( k \). This equation highlights how the process of cooling can be mathematically modeled to predict future temperatures.

Proportionality Constant

The proportionality constant is a vital part of the differential equation in Newton's Law of Cooling. It is denoted by \( k \) and reflects how quickly or slowly an object cools relative to the surrounding environment. In our case, it determines the rate at which the apple pie is cooling.We calculate \( k \) using known temperatures at specific times: the initial temperature and a temperature measured some time later. By plugging these values into the differential equation, we simplify and solve for \( k \). For example:

• At \( t = 0 \): \( T = 350^{\circ}F \)
• At \( t = 10 \): \( T = 290^{\circ}F \)

By substituting these into the equation, we can isolate \( k \) and find its value. It is a crucial calculation because it helps tailor the differential equation to reflect the specific cooling characteristics of the pie in this scenario.

Separation of Variables

Separation of variables is a technique used to solve differential equations, allowing us to get an explicit solution for the function in question, such as temperature here. It involves rearranging the equation so that each variable is on its own side of the equation.The process begins with the differential equation from Newton’s Law of Cooling:

• \( \frac{dT}{dt} = -k(T - T_{room}) \)

We rearrange by separating \( T \) terms on one side and the \( t \) terms on the other, resulting in:

• \( \int \frac{dT}{T - T_{room}} = -k \int dt \)

This divides the problem so we can integrate each side individually. Solving these integrations produces a function that predicts temperature at any given moment. Separation of variables helps to make complex problems manageable by breaking them down into simpler, integrable parts.

Integration

Integration is a mathematical process used to find a function from its derivative, which in physics and engineering problems often involves finding areas, but in our case predicts the temperature over time. After separating variables, we integrate both sides of the rearranged differential equation.The integration process involves:

• Solving \( \int \frac{du}{u} \) to get \( \ln|u| + C \).
• Integrating the time side \( \int dt \) resulting in \( -kt + C \).

Replacing \( u \) back with \( T - T_{room} \) gives us:

• \( \ln|T - T_{room}| = -kt + C \)

Integration helps transition from the rate of cooling to an actual function that predicts temperature changes. By evaluating constants from initial conditions (such as initial temperatures), we predict how long it will take for the pie to cool to a desired temperature.