Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Solve the differential equation for Newton's Law of Cooling for an arbitrary \(T_{0}, T_{1}\), and \(k\), assuming that \(T_{0}>T_{1}\). Show that \(\lim _{t \rightarrow \infty} T(t)=T_{1}\).

Step-by-Step Solution

Answer

As \(t \rightarrow \infty\), \(T(t) \rightarrow T_1\), so the temperature approaches the ambient temperature.

Step 1: Understand Newton's Law of Cooling Equation

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference in temperature between the object and its environment. This can be written as: \(\frac{dT}{dt} = -k(T - T_1)\), where \(T(t)\) is the temperature of the object at time \(t\), \(T_1\) is the ambient temperature, and \(k\) is a positive constant.

Step 2: Separate Variables

To solve the differential equation \(\frac{dT}{dt} = -k(T - T_1)\), we separate the variables: \(\frac{1}{T - T_1}dT = -kdt\). This allows us to integrate both sides individually.

Step 3: Integrate Both Sides

Integrate the left side with respect to \(T\) and the right side with respect to \(t\). The integral of \(\frac{1}{T - T_1}dT\) is \(\ln|T - T_1|\), and the integral of \(-kdt\) is \(-kt + C\), where \(C\) is an integration constant: \(\ln|T - T_1| = -kt + C\).

Step 4: Solve for T(t)

Exponentiate both sides to remove the natural logarithm: \(T - T_1 = Ce^{-kt}\), where \(C\) is a constant. Solve for \(T(t)\): \(T(t) = T_1 + Ce^{-kt}\).

Step 5: Substitute Initial Condition

Apply initial condition \(T(0) = T_0\). Substitute \(t = 0\) and \(T_0\) into the equation: \(T_0 = T_1 + C\cdot 1\). Simplify to find \(C = T_0 - T_1\). Thus, the solution becomes \(T(t) = T_1 + (T_0 - T_1)e^{-kt}\).

Step 6: Show Limit as t Approaches Infinity

To find \(\lim_{t \rightarrow \infty} T(t)\), substitute \(T(t) = T_1 + (T_0 - T_1)e^{-kt}\). As \(t \rightarrow \infty\), \(e^{-kt} \rightarrow 0\), so \(T(t) \rightarrow T_1\). This shows that as time approaches infinity, the temperature of the object approaches the ambient temperature \(T_1\).

Key Concepts

Differential Equations

Differential equations are powerful mathematical tools that help us understand how things change. They are equations that describe the relationship between a function and its derivatives, representing how a particular quantity changes over time or space.

Newton's Law of Cooling is a classic example of a first-order linear differential equation. In this context, the differential equation is given by:

• \(\frac{dT}{dt} = -k(T - T_1)\)

Where:

• \(T(t)\) is the temperature of the object at time \(t\).
• \(T_1\) is the ambient or surrounding temperature.
• \(k\) is a positive constant that depends on the characteristics of the object and its environment.

This equation implies that the rate of change of the temperature, \(\frac{dT}{dt}\), is proportional to the difference between the object's temperature and the ambient temperature.

Solving this differential equation involves techniques such as separation of variables and integration, which we will explore further.

Cooling Process

The cooling process, as described by Newton's Law of Cooling, models an object losing or gaining heat in relation to the ambient temperature around it. The law states that the rate at which an object cools (or heats up) is proportional to the temperature difference between the object and the environment.

In simpler terms:

• The bigger the temperature difference, the faster the cooling occurs.
• As the object's temperature gets closer to the ambient temperature, the rate of temperature change decreases.

This effect can be seen in various real-life scenarios. For example, a hot cup of tea will cool more quickly if the room is significantly cooler, and the rate of cooling will slow down as the tea approaches the room temperature.

This cooling model is widely used in science and engineering to predict the cooling rates of different materials and objects.

Temperature Change

Temperature change in the context of Newton's Law of Cooling refers to the adjustment of an object's temperature over time as it exchanges heat with its surroundings. By solving the differential equation for this law, we find that the temperature at any time \(t\) is given by:

• \(T(t) = T_1 + (T_0 - T_1)e^{-kt}\)

Where:

• \(T_0\) is the initial temperature of the object at time \(t = 0\).
• \(e^{-kt}\) is an exponential decay term demonstrating how the temperature difference decreases over time.

The equation shows that temperature change follows an exponential pattern, gradually slowing down as the temperature difference with the environment reduces.

In long-term scenarios, as \(t\) approaches infinity, the exponential term \(e^{-kt}\) becomes negligibly small, and the object's temperature effectively stabilizes at \(T_1\), the ambient temperature.

Integration Techniques

Integration techniques allow us to solve differential equations like the one seen in Newton's Law of Cooling. For this equation, the approach involves a few key steps.

Firstly, we separate variables in the equation \(\frac{dT}{dt} = -k(T - T_1)\) to get:

• \(\frac{1}{T - T_1}dT = -kdt\)

This makes it easier to integrate each side with respect to their respective variables. Now, we integrate:

• The left side with respect to \(T\), resulting in \(\ln|T - T_1|\).
• The right side with respect to \(t\), resulting in \(-kt + C\), where \(C\) is the constant of integration.

Exponentiating both sides eliminates the logarithm and yields:

• \(T - T_1 = C e^{-kt}\)

Here, \(e^{-kt}\) represents the natural decay factor, and \(C\) is determined using the initial condition. These integration steps show how mathematical techniques guide us in finding comprehensive solutions to dynamic problems like temperature change.