Question

You have a cup of coffee at temperature \(70^{\circ} \mathrm{C}\) and the ambient temperature in the room is \(20^{\circ} \mathrm{C}\). Assuming a cooling rate \(k\) of \(0.125\), write and solve the differential equation to describe the temperature of the coffee with respect to time.

Short answer

The coffee's temperature over time is \( T(t) = 20 + 50e^{-0.125t} \).

Step-by-step solution

Understanding 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 own temperature and the ambient room temperature. The differential form of the law is given by \( \frac{dT}{dt} = -k(T - T_{ambient}) \), where \( T \) is the temperature of the object, \( T_{ambient} \) is the ambient temperature, and \( k \) is the cooling constant.

Establish the Differential Equation

For this problem, substitute the given values into Newton's Law of Cooling. Use \( T = T(t) \) for the coffee's temperature. Hence, the equation becomes \( \frac{dT}{dt} = -0.125(T - 20) \).

Solve the Differential Equation

The equation \( \frac{dT}{dt} = -0.125(T - 20) \) is a separable differential equation. We can rewrite it as \( \frac{dT}{T - 20} = -0.125 \, dt \) and integrate both sides. First, integrate the left side with respect to \( T \) and the right side with respect to \( t \).

Integrate Both Sides

The integration results in \( \ln|T - 20| = -0.125t + C \), where \( C \) is the constant of integration. Exponentiating both sides to remove the natural log gives \( |T - 20| = e^{-0.125t + C} = Ce^{-0.125t} \). Hence, \( T - 20 = Ce^{-0.125t} \).

Solve for Constant Using Initial Condition

Use the initial condition \( T(0) = 70 \) to find \( C \). Substitute \( T = 70 \) and \( t = 0 \) into the equation: \( 70 - 20 = Ce^{0} \) means \( 50 = C \). Now the equation becomes \( T(t) = 20 + 50e^{-0.125t} \).

Write the General Solution

The general solution describing the temperature of the coffee over time is \( T(t) = 20 + 50e^{-0.125t} \). This equation can be used to find the coffee's temperature at any given time \( t \).

Key concepts

Differential Equations

Differential equations are crucial for modeling real-world behaviors that evolve over time, such as temperature changes in our coffee example. They are equations that involve functions and their derivatives; in our case, it describes how the rate of change of temperature \( \frac{dT}{dt} \) is related to the temperature itself.

In the context of Newton's Law of Cooling, the differential equation \( \frac{dT}{dt} = -k(T - T_{ambient}) \) tells us that the rate of change of the temperature of the coffee is proportional to how much hotter it is than its surroundings.

• \( T \) is the temperature of the coffee;
• \( T_{ambient} \) is the ambient temperature of the room;
• \( k \) is a cooling constant that characterizes how quickly the coffee cools.

Solving this equation gives us the function \( T(t) \), which can predict future temperatures based on current conditions.

Exponential Decay

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. In our coffee cooling example, as the coffee cools, its rate of cooling decreases over time because the temperature difference with the ambient environment reduces.

The temperature function \( T(t) = 20 + 50e^{-0.125t} \) that we derived is an example of exponential decay. Here:

• The term \( 20 \) represents the ambient temperature, below which the coffee doesn't cool;
• The coefficient \( 50 \) reflects the initial temperature difference;
• The expression \( e^{-0.125t} \) shows how the temperature approaches the ambient temperature over time.

This shows that the coffee's temperature never quite reaches zero but asymptotically approaches the ambient room temperature.

Initial Condition

An initial condition is a crucial piece of information needed to solve differential equations. It is a specific value that the unknown function satisfies at a certain point, helping to define a unique solution.

For our coffee example, the initial condition was that the coffee was \( 70^{\circ} \) at time \( t = 0 \).

• We use this initial condition to find the constant \( C \) in the solution of our differential equation.
• Substituting \( T = 70 \) and \( t = 0 \) gives \( 50 = C \).

With the initial condition, we determine the complete temperature function \( T(t) = 20 + 50e^{-0.125t} \), which accurately models how the coffee temperature changes over time.

Separation of Variables

Separation of variables is a common technique for solving differential equations, used here to solve the cooling equation. It involves rearranging the equation so that all terms involving one variable are on one side and all terms involving another variable (and its derivative) are on the other side.

In our problem, we started with \( \frac{dT}{dt} = -0.125(T - 20) \). We separated the variables to obtain \( \frac{dT}{T - 20} = -0.125 \, dt \).

• This step allows us to integrate both sides independently.
• The integration yields the logarithmic expression \( \ln|T - 20| = -0.125t + C \), leading to our exponential solution after exponentiation.

Separation of variables is a powerful method when the equation can be expressed in a separable form, allowing for a straightforward integration process to find solutions.