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Chapter 9: Problem 29

Solve the differential equation for Newton's Law of Cooling to find the temperature function in the following cases. Then answer any additional questions. A glass of milk is moved from a refrigerator with a temperature of \(5^{\circ} \mathrm{C}\) to a room with a temperature of \(20^{\circ} \mathrm{C}\). One minute later the milk has warmed to a temperature of \(7^{\circ} \mathrm{C}\). After how many minutes does the milk have a temperature that is \(90 \%\) of the ambient temperature?

Short Answer

Expert verified
Answer: The milk reaches 90% of the ambient temperature after \(\dfrac{\ln\left(\dfrac{15}{13}\right)}{\ln\left(\dfrac{13}{15}\right)}\) minutes.

Step by step solution

01

Differential equation for Newton's Law of Cooling

\(k(T(t)-T_a)=\dfrac{dT}{dt}\), where \(T(t)\) is the temperature of the object at time \(t\), \(T_a\) is the ambient temperature, and \(k\) is the proportionality constant. In this problem, the ambient temperature is \(20^{\circ} \mathrm{C}\), and the initial temperature of the milk is \(5^{\circ} \mathrm{C}\). The differential equation becomes:
02

Modified differential equation

\(k(T(t)-20)=\dfrac{dT}{dt}\) #Step 2: Solve the differential equation# To solve the differential equation, we can use separation of variables. Divide both sides by \((T(t)-20)\), and multiply both sides by \(dt\):
03

Separating the variables

\(\dfrac{1}{T(t)-20} \dfrac{dT}{dt} = kdt\) Now, integrate both sides with respect to their respective variables:
04

Integrating both sides

\(\int\dfrac{1}{T(t)-20} \dfrac{dT}{dt} dt = \int k dt\) This yields:
05

Obtaining the logarithmic function

\(\ln|T(t)-20| = kt+C\) To find the temperature function, exponentiate both sides:
06

Exponentiating both sides

\(T(t)=20+Ae^{kt}\), where \(A = e^C\) #Step 3: Determine the constants A and k# Using the initial condition, when \(t=0\), \(T(0)=5\):
07

Applying the initial condition

\(5 = 20+Ae^{0} \Rightarrow A=-15\) Now we need to find \(k\). After one minute, the milk has warmed to \(7^{\circ}\mathrm{C}\), so \(T(1)=7\):
08

Using the one-minute condition

\(7 = 20 - 15e^{k} \Rightarrow k=\ln(\dfrac{13}{15})\) The temperature function becomes:
09

Temperature function

\(T(t)=20-15e^{\ln(\frac{13}{15})t}\) #Step 4: Find the time when the milk is 90% of the ambient temperature# We want to find the time, \(t\), when the milk has warmed to 90% of the ambient temperature, i.e., \(0.9(20)=18^{\circ}\mathrm{C}\):
10

Setting T(t) to 90% of the ambient temperature

\(18=20-15e^{\ln(\frac{13}{15})t}\) Solve for \(t\):
11

Solving for t

\(2=15e^{\ln(\frac{13}{15})t}\Rightarrow t= \dfrac{\ln\left(\dfrac{15}{13}\right)}{\ln\left(\dfrac{13}{15}\right)}\) #Final Answer# The milk reaches 90% of the ambient temperature after \(\dfrac{\ln\left(\dfrac{15}{13}\right)}{\ln\left(\dfrac{13}{15}\right)}\) minutes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equations
Differential equations are mathematical equations that involve rates of change and the relationships between varying quantities. In the context of Newton's Law of Cooling, we use a differential equation to describe how the temperature of an object changes over time relative to the surrounding environment. The law is represented by the equation \(k(T(t)-T_a)=\frac{dT}{dt}\) where \(T(t)\) is the temperature of the object at time \(t\), \(T_a\) is the ambient or surrounding temperature, and \(k\) is the cooling or heating constant dependent on the properties of the object and environment.

Understanding how to manipulate and solve differential equations is crucial in many fields including physics, engineering, and even finance. The process involves calculating derivatives, which represent how a function is changing at any given point, and integrating, which allows us to find a function given its rate of change.
Temperature Function
The temperature function in physics problems, such as those involving Newton's Law of Cooling, is a mathematical expression that shows how temperature \(T\) varies with time \(t\). In the given exercise, we've determined the temperature function to be \(T(t)=20-15e^{kt}\).

This function is derived from solving the differential equation and incorporates environmental conditions and properties of the cooling or heating object. The temperature function is fundamental since it not only allows us to predict the temperature at any given time but also to conduct analysis on temperature-related phenomena. For example, it can determine at what time the temperature will reach a certain percentage of the ambient condition, like the 90% required in the exercise.
Separation of Variables
Separation of variables is a method for solving differential equations, where we 'separate' the variables (usually 'x' and 'y', or in our case 'T' and 't') onto two different sides of the equation. The goal is to rewrite the equation so that each side contains only one variable and its differentials. For the provided exercise, the step \(\frac{1}{T(t)-20} \frac{dT}{dt} = kdt\) involves separating the temperature dependent part from the time dependent part.

Once we have \(\frac{dT}{T(t)-20} = kdt\), we can integrate both sides independently, which is a powerful strategy for solving many types of differential equations. Understanding separation of variables can significantly simplify the process of finding solutions to otherwise complex differential equations.
Logarithmic Function
Logarithmic functions are the inverses of exponential functions and play a vital role in solving certain differential equations, including those we encounter in Newton's Law of Cooling. Following the separation of variables, integrating both sides leads to a natural logarithm function \(\ln|T(t)-20| = kt + C\).

In our milk temperature problem, logarithms allow us to solve for \(t\), the variable representing time, because the inverse operation of exponentiation (used in the temperature function \(T(t)=20+Ae^{kt}\)) is the logarithm. Thus, if we know the change in temperature, we can use the natural logarithm to find the time or rate at which that change occurred, which is exactly what was done to find at what time the milk reaches 90% of the ambient temperature.

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