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Chapter 8: Problem 60

A glass of lemonade with a temperature of \(40^{\circ} \mathrm{F}\) is placed in a room with a constant temperature of \(70^{\circ} \mathrm{F}\), and 1 hour later its temperature is \(52^{\circ} \mathrm{F}\). Show that \(t\) hours after the lemonade is placed in the room its temperature is approximated by \(T=70-30 e^{-0.5 t}.\)

Short Answer

Expert verified
The temperature is approximately \( T = 70 - 30e^{-0.5t} \).

Step by step solution

01

Identify the Differential Equation

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. We can represent this by the differential equation \( \frac{dT}{dt} = -k(T - T_r) \), where \( T \) is the temperature of the lemonade, \( T_r = 70\degree F \) is the room temperature, and \( k \) is a positive constant.
02

Introduce Initial Conditions

We are given that at \( t = 0 \), the temperature of the lemonade \( T = 40\degree F \). After 1 hour (\( t = 1 \)), the temperature \( T = 52\degree F \). These provide our initial conditions for solving the differential equation.
03

Solve the Differential Equation

By separation of variables, the differential equation becomes \( \frac{dT}{T - 70} = -k \, dt \). Integrating both sides, we get \( \ln|T - 70| = -kt + C \), where \( C \) is the constant of integration. Exponentiating both sides produces \( |T - 70| = e^{-kt + C} = Ce^{-kt} \), so \( T - 70 = Ce^{-kt} \).
04

Determine the Constant of Integration C

Using the initial condition \( T(0) = 40 \), we substitute into the equation: \( 40 - 70 = C \times e^0 \rightarrow C = -30 \). Thus the equation simplifies to \( T - 70 = -30e^{-kt} \).
05

Determine the Constant k

Using the temperature after 1 hour \( T(1) = 52 \), substitute into the equation: \( 52 - 70 = -30e^{-k(1)} \). This becomes \( -18 = -30e^{-k} \), leading to \( e^{-k} = \frac{18}{30} = 0.6 \). Taking the natural log: \( -k = \ln(0.6) \rightarrow k \approx 0.5108 \).
06

Finalize the Temperature Expression

Substitute \( k \) back into the solution, we have \( T(t) = 70 - 30e^{-0.5108t} \). Given that we need an approximation, the formula \( T = 70 - 30e^{-0.5t} \) approximates the temperature well for this scenario.

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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 fundamental tools used in mathematics to describe how a particular quantity changes. In the context of Newton's Law of Cooling, they express how the temperature changes over time. The differential equation derived from Newton's Law is written as \( \frac{dT}{dt} = -k(T - T_r) \). This equation states that the rate of change of the temperature \( T \) of the lemonade is proportional to the difference between \( T \) and the room temperature \( T_r \). Here, \( k \) is a constant that will later be determined.

Solving differential equations often gives us functions that describe a situation over time, predicting future outcomes based on initial conditions. By understanding the nature of differential equations, you gain the power to model a range of natural phenomena.
  • They describe how something changes over time or space.
  • They offer insights into how we can expect systems to behave under different conditions.
  • They require initial conditions to provide unique solutions.
Initial Conditions
Initial conditions are crucial when solving differential equations. They allow us to find a specific solution tailored to our problem. For the lemonade problem, we start with an initial temperature of the lemonade, \( T(0) = 40^{\circ} F \). This is given when \( t = 0 \).

Additionally, after 1 hour, the temperature is \( T(1) = 52^{\circ} F \). These conditions help us determine constants within our solution, such as \( C \) in the equation \( |T - 70| = Ce^{-kt} \). By providing these conditions, we can solve the equation uniquely for our scenario and predict the lemonade's temperature over time.
  • Ensure the solution fits the specific scenario.
  • Help calculate unknown constants in equations.
  • Verify the predicted outcomes are accurate over time.
Separation of Variables
Separation of variables is a powerful technique to solve differential equations. It involves rearranging the equation so that each variable appears on different sides. In the case of our lemonade problem, we have \( \frac{dT}{T - 70} = -k \, dt \).

This means we've separated the variables: temperatures, \( T \,\) on one side, and time, \( t \,\) on the other. We can then integrate both sides independently, leading us to \( \ln|T - 70| = -kt + C \). This integration step is crucial as it provides the core part of our solution function for \( T(t) \).
  • Facilitates solving complex differential equations.
  • Allows integration of each side separately.
  • Makes it possible to find a function that fits the changes over time.
By mastering this technique, you can approach not only cooling problems but a vast array of scenarios involving differential changes.

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Most popular questions from this chapter

Use Euler's Method with the given step size \(\Delta x\) or \(\Delta t\) to approximate the solution of the initial-value problem over the stated interval. Present your answer as a table and as a graph. $$d y / d x=x-y^{2}, y(0)=1,0 \leq x \leq 2, \Delta x=0.25$$

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\). $$y^{\prime}-(1+x)\left(1+y^{2}\right)=0$$

Determine whether the statement is true or false. Explain your answer. If a population is growing exponentially, then the time it takes the population to quadruple is independent of the size of the population.

Solve the differential equation. If you have a CAS with implicit plotting capability, use the CAS to generate five integral curves for the equation. $$y^{\prime}=\frac{y}{1+y^{2}}$$

Determine whether the statement is true or false. Explain your answer. The equation $$ \left(\frac{d y}{d x}\right)^{2}=\frac{d y}{d x}+2 y $$ is an example of a second-order differential equation.
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