Question

Food, initially at a temperature of \(40^{\circ} \mathrm{F}\), was placed in an oven preheated to \(350^{\circ} \mathrm{F}\). After \(10 \mathrm{~min}\) in the oven, the food had warmed to \(120^{\circ} \mathrm{F}\). After \(20 \mathrm{~min}\), the food was removed from the oven and allowed to cool at room temperature \(\left(72^{\circ} \mathrm{F}\right)\). If the ideal serving temperature of the food is \(110^{\circ} \mathrm{F}\), when should the food be served?

Short answer

Answer: The food should be served when its temperature reaches \(110^{\circ} \mathrm{F}\), which can be found using the temperature function outside the oven: \(T(t) = 72 + (T(10) - 72)e^{k(t-10)}\). Solve this equation for \(t\) to find the appropriate time to serve the food.

Step-by-step solution

Analyze Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change in the temperature of an object is proportional to the difference between the object's temperature and the temperature of the environment. The equation is given by: $$ \frac{dT}{dt} = k(T - T_{env}) $$ where \(T\) is the temperature of the food, \(T_{env}\) is the temperature of the environment, and \(k\) is the cooling constant.

Determine the temperature function inside the oven

We know the initial temperature of the food is \(40^{\circ} \mathrm{F}\), and the oven is preheated to \(350^{\circ} \mathrm{F}\). After \(10 \mathrm{~ min}\), the food temperature rises to \(120^{\circ} \mathrm{F}\). We can use this information to find the temperature function of the food in the oven: 1. Rearrange Newton's Law of Cooling to isolate the cooling constant \(k\) and find its value when the food is in the oven: $$ k = \frac{\frac{dT}{dt}}{T - T_{env}} $$ 2. Here, \(T_{env} = 350^{\circ} \mathrm{F}\). After \(10 \mathrm{~ min}\), \(T = 120^{\circ} \mathrm{F}\) and \(\frac{dT}{dt} = \frac{120 - 40}{10} = 8^{\circ} \mathrm{F} \text{ per min}\) 3. Substituting the values, we get $$ k = \frac{8}{120 - 350} = \frac{8}{-230} $$ 4. Once we have the cooling constant \(k\), we can integrate the equation of Newton's Law of Cooling: $$ dT = k(T - 350) dt $$ $$ \int \frac{dT}{T - 350} = \int k dt $$ 5. Let \(t_0 = 0\), and apply the initial condition \(T(t_0) = 40^{\circ} \mathrm{F}\) to find the constant of integration. We obtain the temperature function of the food in the oven: $$ T(t) = 40 + (350 - 40)e^{kt} $$

Determine the temperature function outside the oven

After \(20 \mathrm{~ min}\), the food reaches a temperature at which it is removed from the oven. We need to find that temperature and then determine the function that models the cooling outside the oven. 1. Plug \(t = 10\) into the temperature function: $$ T(10) = 40 + (350 - 40)e^{k * 10} $$ 2. The environment temperature outside the oven is \(72^{\circ} \mathrm{F}\). Using Newton's Law of Cooling, we obtain a new equation for the cooling outside the oven: $$ \frac{dT}{dt} = k(T - 72) $$ 3. Integrate the equation and apply the initial condition \(T(t_0) = T(10)\) to find the constant of integration. We obtain the temperature function of the food outside the oven: $$ T(t) = 72 + (T(10) - 72)e^{k(t-10)} $$

Find the time at which the food reaches the ideal serving temperature

Now we need to find when the food reaches the ideal serving temperature of \(110^{\circ} \mathrm{F}\) using the temperature function outside the oven. 1. Set the temperature function equal to \(110^{\circ} \mathrm{F}\): $$ 110 = 72 + (T(10) - 72)e^{k(t-10)} $$ 2. Solve the equation for \(t\) to find the time at which the food should be served. Note that this equation might not have a closed-form solution, and numerical methods may need to be used. When the food reaches a temperature of \(110^{\circ} \mathrm{F}\), it should be served.

Key concepts

Differential Equations

Differential equations play a pivotal role in understanding how systems change over time. They provide a mathematical framework that describes the relationship between a function and its derivatives. In the context of Newton's Law of Cooling, the differential equation is expressed as \( \frac{dT}{dt} = k(T - T_{env}) \). Here, \( \frac{dT}{dt} \) represents the rate of change of temperature with respect to time, while \( k \) is the cooling constant, and \( (T - T_{env}) \) indicates the temperature difference between the object and the environment.

This formula shows that the temperature change rate is proportional to how much hotter the object is compared to its surroundings. Essentially, it captures how quickly an object cools or heats towards the ambient temperature, depending on the sign and value of the constant \( k \). Solving this differential equation involves integrating both sides, which helps to find the temperature function \( T(t) \) over time.

Understanding differential equations can be challenging, but they are vital for mapping real-world processes. They allow us to see changes dynamically and can be applied in various fields such as physics, engineering, and biology.

Temperature Modeling

Creating a model for temperature changes, like in our cooling and heating scenario, is essential for predicting the behavior of systems over time. Temperature modeling uses mathematical equations, often differential equations, to simulate the variation of temperature under different conditions.

In the given problem, temperature modeling is done using Newton's Law of Cooling. This law helps us create a function that describes how the food's temperature changes in the oven and also outside.

• The first function is for when the food is heating in the oven. This uses the oven's constant temperature of \( 350^{\circ} \mathrm{F} \) as the environmental temperature.
• The second function models the cooling process once the food is removed, using \( 72^{\circ} \mathrm{F} \) as the room temperature.

By plugging in initial conditions and solving the differential equation, these models can predict the food's temperature at any given time.

Temperature modeling is powerful as it gives insights into how quickly or slowly a process like cooling occurs, helping determine precise timings, like when the food is just right for serving."},{