91Ó°ÊÓ

Ncwton's Law of Coling. An object has an initial temperature of \(T_{0}\) and is placed into a surrounding medium with a lower temperature \(C\). The temperature \(T\) of the cooling object drops at a rate that is proportional to the difference \(T-C\). That is, $$ \frac{d T}{d t}=-k(T-C) $$ where \(k\) is a positive constant and \(t\) is time. a) Use the techniques of this section and the condition \(T(0)=T_{0}\) to solve for \(T(t)\) b) A metal object that has been heated to \(143^{\circ} \mathrm{F}\) is placed into a room that is kept at a constant \(70^{\circ} \mathrm{F} .\) After \(30 \mathrm{~min}\), it is observed to cool to \(117^{\circ} \mathrm{F}\). How long will it take the object to cool to \(90^{\circ} \mathrm{F}\) ?

Step by step solution

01

Set Up the Differential Equation

Use the given differential equation \ \( \frac{d T}{d t} = -k(T - C) \) \ along with the initial condition \( T(0) = T_0 \).

02

Separate Variables and Integrate

Rewrite the equation as \ \( \frac{d T}{T - C} = -k \, dt \). \ Then integrate both sides: \ \( \int \frac{1}{T - C} \, dT = - \int k \, dt \).

03

Solve the Integrals

The integral on the left side gives \ \( \ln |T - C| \) and the integral on the right side gives \ \( -kt \) plus a constant of integration: \ \( \ln |T - C| = -kt + C_1 \).

04

Simplify the Equation

Exponentiate both sides to remove the natural logarithm: \ \( |T - C| = e^{-kt + C_1} = e^{C_1} \, e^{-kt} \). \ Let \( A = e^{C_1} \), resulting in \ \( T - C = A \, e^{-kt} \).

05

Apply the Initial Condition

Use the initial condition \( T(0) = T_0 \) to determine the value of \( A \): \ \( T_0 - C = A \cdot e^{0} \) \ which simplifies to \ \( T_0 - C = A \). \ Therefore, \( A = T_0 - C \).

06

Write the Final Solution

Substitute \( A \) back into the equation: \ \( T(t) - C = (T_0 - C) \, e^{-kt} \). \ Thus, \ \( T(t) = C + (T_0 - C) \, e^{-kt} \).

07

Apply Given Values

For part b, use \( T_0 = 143^{\circ} \mathrm{F} \), \( C = 70^{\circ} \mathrm{F} \), and the observed temperature after 30 minutes (\(T(30) = 117^{\circ} \mathrm{F}\)): \ \( 117 = 70 + (143 - 70) \, e^{-30k} \).

08

Solve for k

Simplify and solve for \( k \): \ \( 47 = 73 \, e^{-30k} \) \ \( \frac{47}{73} = e^{-30k} \) \ Take the natural logarithm of both sides: \ \( \ln \left( \frac{47}{73} \right) = -30k \) \ \( k = - \frac{1}{30} \ln \left( \frac{47}{73} \right) \).

09

Determine Time for Cooling to 90°F

Use the determined \( k \) value and set up the equation for \( T(t) = 90^{\circ} \mathrm{F} \): \ \( 90 = 70 + (143 - 70) \, e^{-kt} \). \ \( 20 = 73 \, e^{-kt} \). \ \( \frac{20}{73} = e^{-kt} \). \ Take the natural logarithm of both sides: \ \( \ln \left( \frac{20}{73} \right) = -kt \), and solve for \( t \): \ \( t = - \frac{1}{k} \ln \left( \frac{20}{73} \right) \). \ Substitute the value of \( k \).

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Key Concepts

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

Differential Equations

A differential equation involves functions and their derivatives. In this exercise, Newton's Law of Cooling uses the differential equation \( \frac{d T}{d t} = -k(T - C) \). This equation expresses the rate of temperature change over time.
To solve such equations, it is crucial to rearrange terms and integrate both sides. This approach helps transform the differential equation into an algebraic equation, making it easier to find the function that describes how temperature varies with time.

Exponential Decay

Exponential decay describes processes where the quantity decreases at a rate proportional to its current value. Here, the temperature difference \( T - C \) decays exponentially.
We express this relationship as \( T(t) = C + (T_0 - C) e^{-kt} \), where
- \( T_0 \) is the initial temperature,
- \( C \) is the ambient temperature,
- \( k \) is a constant determining the decay rate.
Exponential decay curves are negatively sloped and asymptotically approach the ambient temperature.

Initial Value Problem

An initial value problem specifies the value of the unknown function at a given point. This information allows for solving differential equations accurately.
For Newton's Law of Cooling, the initial condition is \( T(0) = T_0 \). Given this starting temperature, we can integrate the differential equation and determine the specific solution. Applying the initial value, we solve for constants like \( A = T_0 - C \).
Thus, initial conditions are essential for tailoring general solutions to specific real-world scenarios.

Natural Logarithms

Natural logarithms, denoted as \( \ln \), are logarithms with the base \( e \) (approximately 2.718). They are vital in solving exponential growth and decay problems, including Newton's Law of Cooling.
For instance, to solve for time \( t \), we use the natural log:
\( \frac{T - C}{A} = e^{-kt} \). Taking the natural log of both sides yields \( \ln \( \frac{T - C}{A} \) = -kt \).
Consequently, the natural logarithm helps convert exponential equations into linear ones, simplifying the solution process.
In the provided solution, natural logs help determine the constant \( k \), leading to the precise time necessary for temperature changes.