Question

Assume Newton's law of cooling applies. The temperature of an object is raised from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\) in \(10 \mathrm{~min}\) when placed within a \(300^{\circ} \mathrm{F}\) oven. What oven temperature will raise the object's temperature from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\) in \(5 \mathrm{~min}\) ?

Short answer

Answer: The oven temperature needed to raise the object's temperature from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\) in \(5\) minutes is \(150^{\circ} \mathrm{F}\).

Step-by-step solution

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 the object's temperature and the surrounding temperature. Mathematically, it can be expressed as: $$ \frac{dT}{dt} = k(T_s - T) $$ where \(T\) is the temperature of the object, \(T_s\) is the surrounding temperature (oven temperature here), \(t\) is the time, and \(k\) is the constant of proportionality.

Find the initial differential equation

First, let's find the differential equation for the situation when the object is heated from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\) in \(10\) minutes in a \(300^{\circ} \mathrm{F}\) oven. We can rewrite Newton's Law of Cooling as $$ \frac{dT}{dt} = k(300 - T) $$Since temperature increases from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\) in \(10\) minutes, we can write $$ T(t) = 70 + (150-70)e^{kt} $$

Calculate the constant k

Now, we can plug the initial and final temperature values into the equation and solve for \(k\). $$ 150 = 70 + (150 - 70)e^{10k} $$Next, we solve for \(k\). $$ 80 = 80e^{10k} $$ $$ 1 = e^{10k} $$ $$ k = \frac{\ln(1)}{10} = 0 $$

Find the differential equation for the new scenario

Now, let's consider the situation of raising the object's temperature from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\) in \(5\) minutes. Let \(T_s'\) represent the unknown oven temperature. The differential equation for this scenario would be: $$ \frac{dT}{dt} = k(T_s' - T) $$With the initial condition, $$ T(t) = 70 + (150-70)e^{kt} $$

Find the new oven temperature

Since we want the object to reach \(150^{\circ} \mathrm{F}\) in \(5\) minutes, we can plug this information into our new equation and solve for \(T_s'\). $$ 150 = 70 + (150 - 70)e^{5k} $$Since \(k = 0\), the equation becomes, $$ 150 = 70 + (150 - 70)e^{5(0)} $$ $$ 150 = 70 + (150 - 70) $$ $$ T_s' = 150 $$ Thus, the oven temperature required to raise the object's temperature from \(70^{\circ} \mathrm{F}\) to \(150^{\circ} \mathrm{F}\) in \(5\) minutes is \(150^{\circ} \mathrm{F}\).

Key concepts

Differential Equation

A differential equation forms the backbone of Newton’s Law of Cooling. It describes how the rate of change of a temperature depends on certain conditions in a given situation.
In simple terms, a differential equation is an equation that relates a function to its derivatives. With Newton's Law of Cooling, this equation tells us how temperature changes over time relative to the difference between the current temperature of an object and the surrounding environment.
In this specific exercise, the differential equation is expressed as: \[\frac{dT}{dt} = k(T_s - T)\]where:

• \(T\) is the temperature of the object.
• \(T_s\) is the surrounding temperature.
• \(k\) is a constant that characterizes the rate of temperature change.

This form of equation is vital because it can be solved to find an expression for temperature over time, \(T(t)\). Understanding differential equations allows us to model how different factors can influence temperature changes.

Proportional Change

The concept of proportional change is key to understanding how differential equations operate within Newton's Law of Cooling. This law tells us that the rate of temperature change in an object is proportional to the difference between its current temperature and the temperature of its surroundings.
Consider the expression:\[\frac{dT}{dt} = k(T_s - T)\]This equation demonstrates that if the temperature difference is larger, the rate at which temperature changes is faster and vice versa.
The proportional change helps us understand dynamic processes where the rate at which something changes is not constant but varies based on conditions such as temperature differences in this context. This concept is not just limited to cooling but applies to other phenomena like atmospheric pressure changes and even financial interest calculations.

Temperature Modeling

Temperature modeling refers to the mathematical representation of temperature changes over time, thanks to differential equations like the one used in Newton's Law of Cooling.
In the given scenario, the aim is to model how an object's temperature changes from 70°F to 150°F. To do this, we use the differential equation derived from Newton’s Law:
\[T(t) = 70 + (150-70)e^{kt}\]Here, \(T(t)\) is the modeled temperature at a time \(t\) and \(k\) is a constant derived from the initial conditions. When \(k\) is found through solving the initial situation, it can then be applied to predict how quickly an object will heat or cool under new conditions.
Temperature modeling is crucial in many fields, such as climate science, engineering, and even culinary arts, for it helps in predicting and optimizing thermal processes.

Cooling and Heating Processes

Understanding the cooling and heating processes is essential for knowing how temperatures change in real-world situations. Newton’s Law of Cooling provides a framework for modeling these processes.
When an object is placed in an environment, its temperature will adapt to the surroundings, governed by the difference in temperature and time. In the given problem, a differential equation models the process where the object, heated in an oven, sees its temperature rise due to the high temperature of the surroundings.
Essentially, adjusting the oven's temperature or the time in the oven allows control over how quickly the object heats up. This is why solving for \(T_s'\) is crucial in this context, showing that a different oven temperature may be needed to achieve desired temperature changes within shorter or longer time spans. Understanding these processes and the factors affecting it can lead to better designs in systems that require efficient cooling or heating, such as in manufacturing and food preparation.