Problem 58 - Calculus Early Transcendentals

Question

Suppose that at time \(t=0\) an object with temperature \(T_{0}\) is placed in a room with constant temperature \(T_{a} .\) If \(T_{0}T_{a},\) then the temperature will decrease. Assuming that Newton's Law of Cooling applies, show that in both cases the temperature \(T(t)\) at time \(t\) is given by $$ T(t)=T_{a}+\left(T_{0}-T_{a}\right) e^{-k t} $$ where \(k\) is a positive constant.

Step-by-Step Solution

Answer

The temperature function is \( T(t) = T_a + (T_0 - T_a)e^{-kt} \).

Step 1: Understand Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature. Mathematically, it is expressed as: \( \frac{dT}{dt} = -k(T - T_a) \), where \( k \) is a positive constant.

Step 2: Derive the Differential Equation

The differential equation \( \frac{dT}{dt} = -k(T - T_a) \) tells us how the temperature changes over time. The negative sign indicates cooling (or heating, depending on the initial condition towards \( T_a \)). Our goal is to find \( T(t) \).

Step 3: Solve the Differential Equation

To solve \( \frac{dT}{dt} = -k(T - T_a) \), we separate variables: \( \frac{dT}{(T - T_a)} = -k \, dt \). Integrate both sides: \( \int \frac{dT}{(T - T_a)} = \int -k \, dt \).

Step 4: Integrate

The integral of the left side yields \( \ln |T - T_a| \), and the right side yields \( -kt + C \), where \( C \) is the integration constant. Hence, \( \ln |T - T_a| = -kt + C \).

Step 5: Exponentiate to Solve for Temperature

Exponentiating both sides of \( \ln |T - T_a| = -kt + C \) gives: \( |T - T_a| = e^{C}e^{-kt} \). Let \( e^{C} = C_1 \), as it's just a constant, thus \( T - T_a = C_1 e^{-kt} \).

Step 6: Solve for the Constant Using Initial Conditions

At \( t = 0 \), \( T = T_0 \). Substitute into the equation: \( T_0 - T_a = C_1 \times 1 \). Therefore, \( C_1 = T_0 - T_a \).

Step 7: Substitute Back the Constant

Substitute \( C_1 = T_0 - T_a \) into \( T - T_a = C_1 e^{-kt} \), giving \( T(t) = T_a + (T_0 - T_a)e^{-kt} \). This is the solution for \( T(t) \).

Key Concepts

Differential Equations

Differential equations play a crucial role in Newton's Law of Cooling, as they allow us to mathematically describe how the temperature of an object changes over time. The fundamental differential equation in this context is given by:
\[ \frac{dT}{dt} = -k(T - T_a) \]Understanding this equation is key because it effectively conveys how the rate of temperature change \( \frac{dT}{dt} \) is proportional to the difference between the object's temperature \( T \) and the ambient temperature \( T_a \).
This proportionality is pivotal in showing how the temperature of an object deviates from its surroundings, either cooling down or warming up based on initial conditions.

• The negative sign accounts for a reduction in temperature when \( T > T_a \), aiming towards equilibrium.
• The constant \( k \), which is always positive, controls the rate of adjustment towards the ambient temperature.

Differential equations like this one are often solved through a process called separation of variables, enabling us to isolate and integrate each part of the equation. This technique helps with solving for the function \( T(t) \), which provides the temperature at any given time \( t \).

Temperature Change

The concept of temperature change is at the heart of Newton's Law of Cooling, which models how an object's temperature converges to its environment over time. When \( T_0 < T_a \), the object's temperature increases since the room is warmer. Conversely, when \( T_0 > T_a \), the object will lose heat to the room, cooling down.
This change in temperature is described by the equation:
\[ T(t)=T_{a}+(T_{0}-T_{a}) e^{-kt} \]Here, \( T(t) \) represents the temperature of the object at a specific time \( t \).
The rate at which the temperature changes relies heavily upon the parameter \( k \) and the initial temperature difference \( T_0 - T_a \).

• As time passes, the term \( e^{-kt} \) decreases exponentially.
• The object's temperature approaches \( T_a \), the ambient temperature, as \( t \to \infty \).

This mathematical model showcases the gradual nature of temperature change, emphasizing how external conditions influence the object's final state.

Exponential Functions

Exponential functions are a critical component in understanding the predicted temperature behavior over time according to Newton's Law of Cooling. The function \( e^{-kt} \) characterizes the rate at which an object moves toward the ambient temperature, affecting the magnitude of the temperature difference.
The exponential term results in a rapid initial change in temperature that slows over time, which is a common characteristic of cooling processes in nature.

• At \( t=0 \), the value of \( e^{-kt} \) is 1, maximizing the initial difference \( T_0 - T_a \).
• As \( t \) increases, \( e^{-kt} \) approaches zero, causing \( T(t) \) to approach \( T_a \).

Exponential functions are helpful in describing real-world phenomena where processes initially occur rapidly and then gradually slow down. This behavior is well-suited for scenarios where temperature differences minimize over time, illustrating how exponential decay leads to stabilization.