Problem 13 - A Graphical Approach to College Algebra

Question

Newton's law of cooling says that the rate at which an object cools is proportional to the difference \(C\) in temperature between the object and the environment around it. The temperature \(f(t)\) of the object at time t in appropriate units after being introduced into an environment with a constant temperature \(T_{0}\) is $$f(t)=T_{0}+C e^{-k t}$$ where \(C\) and \(k\) are constants. Use this result. A pot of coffee with a temperature of \(100^{\circ} \mathrm{C}\) is set down in a room with a temperature of \(20^{\circ} \mathrm{C}\). The coffee cools to \(60^{\circ} \mathrm{C}\) after 1 hour. (a) Write an equation to model the data. (b) Estimate the temperature after a half hour. (c) About how long will it take for the coffee to cool to \(50^{\circ} \mathrm{C} ?\) Support your answer graphically.

Step-by-Step Solution

Answer

(a) \( f(t) = 20 + 80 e^{-\ln(2)t} \); (b) \( \approx 71.55^\circ C \); (c) \( \approx 1.678 \) hours.

Step 1: Establish Equation for Temperature Model

Given Newton's law of cooling, the temperature of the coffee as a function of time is modeled by the equation \( f(t) = T_0 + C e^{-kt} \). In this case, \( T_0 = 20^{\circ} \mathrm{C} \) is the ambient temperature, and initially, \( f(0) = 100^{\circ} \mathrm{C} \). So, at \( t = 0 \), we have:\[ f(0) = 20 + C = 100 \]Solving this gives us \( C = 80 \).

Step 2: Find Constant 'k' Using Given Data

We know that after 1 hour (\( t = 1 \)), the temperature of the coffee is \( 60^{\circ} \mathrm{C} \). We use this to find \( k \):\[ f(1) = 20 + 80e^{-k} = 60 \]Solving for \( k \):\[ 80e^{-k} = 40 \]\[ e^{-k} = \frac{1}{2} \]Taking the natural logarithm of both sides, we get:\[ -k = \ln\left(\frac{1}{2}\right) \]\[ k = \ln(2) \].

Step 3: Model Equation for the Scenario

Substituting back the values of \( C \) and \( k \), the temperature model is:\[ f(t) = 20 + 80 e^{-\ln(2) t} \].

Step 4: Estimate Temperature After Half Hour

To find the temperature after half an hour (\( t = 0.5 \)), substitute \( t = 0.5 \) into the model:\[ f(0.5) = 20 + 80 e^{-\ln(2) \cdot 0.5} \]\[ f(0.5) = 20 + 80 \cdot \sqrt{\frac{1}{2}} \]\[ f(0.5) = 20 + 80 \cdot \frac{1}{\sqrt{2}} \]Calculating this gives approximately \( f(0.5) \approx 71.55^{\circ} \mathrm{C} \).

Step 5: Time for Temperature to Reach 50°C

We need to find \( t \) such that \( f(t) = 50 \):\[ 20 + 80 e^{-\ln(2) t} = 50 \]\[ 80 e^{-\ln(2) t} = 30 \]\[ e^{-\ln(2) t} = \frac{3}{8} \]Take the natural logarithm:\[ -\ln(2) t = \ln\left(\frac{3}{8}\right) \]\[ t = -\frac{\ln\left(\frac{3}{8}\right)}{\ln(2)} \]Calculating this gives approximately \( t \approx 1.678 \) hours.

Key Concepts

Temperature Model Equation

Newton's law of cooling provides a mathematical description of how the temperature of an object changes when exposed to a constant surrounding environment. The equation used to model this cooling process, typically represented by \( f(t) = T_0 + C e^{-kt} \), is crucial for predicting temperature changes over time. In this equation:

• \( T_0 \) is the ambient or room temperature, which stays constant.
• \( C \) represents the difference between the initial temperature of the object and the ambient temperature.
• \( e \) is the base of the natural logarithm, important in capturing the pattern of decay.

Here, \( f(t) \) denotes the temperature of the object at time \( t \), and \( k \), a positive constant, indicates the rate of cooling. In our example with the pot of coffee, initially, the temperature was \( 100^{\circ} \mathrm{C} \) while the ambient temperature was \( 20^{\circ} \mathrm{C} \), giving us \( C = 80 \). This formula is a cornerstone for understanding how objects respond thermally to their environments.

Rate of Cooling

Understanding the rate of cooling is essential in analyzing how quickly the temperature of an object changes. This rate is directly related to the constant \( k \) in Newton's law of cooling equation. The bigger the value of \( k \), the faster the cooling occurs. In the coffee cooling example, the rate of decrease in temperature was significant during the first hour, since the coffee temperature dropped from \( 100^{\circ} \mathrm{C} \) to \( 60^{\circ} \mathrm{C} \). To calculate \( k \), we used the information that after 1 hour, the coffee's temperature was \( 60^{\circ} \mathrm{C} \). By inserting \( t = 1 \) into the temperature model equation, we calculated \( k = \ln(2) \), telling us about the rate at which our coffee cooled. This concept not only helps in practical applications, like estimating when your coffee might reach drinking temperature, but is also widely applicable in physics and engineering contexts.

Exponential Decay

The term exponential decay aptly describes the process of cooling as per Newton's law of cooling. This type of decay is characterized by the fact that the rate of change of the object’s temperature is proportional to its current temperature relative to the surrounding environment. With exponential decay, the rate at which the temperature changes gradually decreases over time. Initially, the temperature drops rapidly, and as time goes on, the rate becomes slower. This happens because the temperature difference between the object and the environment decreases. The factor \( e^{-kt} \) within the temperature model equation captures this behavior, systematically reducing the temperature difference \( C \) over time, hence leading to a stabilizing effect as the object reaches the ambient temperature. Exponential decay processes like this are important not only in understanding cooling phenomena but also find applications in fields like finance, population studies, and radioactive decay. That’s why mastering this concept is vital for students across various scientific disciplines.