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Chapter 31: Problem 27

Solve the given problems by solving the appropriate differential equation. According to Newton's law of cooling, the rate at which a body cools is proportional to the difference in temperature between it and the surrounding medium. Assuming Newton's law holds, how long will it take a cup of hot water, initially at \(200^{\circ} \mathrm{F}\), to cool to \(100^{\circ} \mathrm{F}\) if the room temperature is \(80.0^{\circ} \mathrm{F},\) if it cools to \(140^{\circ} \mathrm{F}\) in \(5.0 \mathrm{min} ?\)

Short Answer

Expert verified
It takes approximately 12.9 minutes for the water to cool to 100掳F.

Step by step solution

01

Understand Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of the temperature of the object is proportional to the difference between the object's temperature and the ambient temperature. Mathematically, it is written as \( \frac{dT}{dt} = k(T - T_{env}) \), where \(T\) is the temperature of the object at time \(t\), \(T_{env}\) is the ambient temperature, and \(k\) is a constant.
02

Write the Differential Equation

For this problem, the ambient temperature, \(T_{env}\), is \(80^{\circ}F\). The differential equation becomes \( \frac{dT}{dt} = k(T - 80) \). This equation describes how the temperature of the water cools over time.
03

Solve the Differential Equation

To solve the differential equation \( \frac{dT}{dt} = k(T - 80) \), we can use separation of variables. Rearrange as \( \frac{1}{T - 80} \, dT = k \, dt \). Integrate both sides to get \( \ln |T - 80| = kt + C \), where \(C\) is the constant of integration.
04

Find the Constant C

To find \(C\), use the initial condition that at \(t = 0\), the temperature \(T\) is \(200^{\circ}F\). Substitute \(T = 200\) and \(t = 0\) into \( \ln |T - 80| = kt + C \) to get \( \ln |200 - 80| = C \), which simplifies to \(C = \ln 120\).
05

Determine the Constant k

Use the information that \(T = 140^{\circ}F\) when \(t = 5\) minutes (5/60 hours). Substitute \(T = 140\), \(t = \frac{5}{60}\), and \(C = \ln 120\) into \( \ln |T - 80| = kt + C \): \[ \ln 60 = k \cdot \frac{5}{60} + \ln 120 \]Solve for \(k\): \[ \ln 60 - \ln 120 = k \cdot \frac{1}{12} \]\[ \ln \frac{60}{120} = k \cdot \frac{1}{12} \]\[ k = 12 \ln \frac{1}{2} \]
06

Find Time to Cool to 100掳F

Now solve for the time \(t\) when the temperature \(T = 100^{\circ}F\): \[ \ln |100 - 80| = 12 \ln \frac{1}{2} \cdot t + \ln 120 \]\[ \ln 20 = t \cdot 12 \ln \frac{1}{2} + \ln 120 \]\[ \ln \frac{20}{120} = t \cdot 12 \ln \frac{1}{2} \]\[ t = \frac{\ln \frac{1}{6}}{12 \ln \frac{1}{2}} \]Calculate \(t\) to find the time it takes to cool to \(100^{\circ}F\).
07

Complete the Calculation

Approximate \(t\):\[ t = \frac{-1.791759469}{12(-0.6931471806)} \approx 0.2153 \hours \]Convert to minutes: \(0.2153 \times 60 \approx 12.9 \ minutes \).

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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 expressions used to define the relation between a function and its derivatives. In simpler terms, it describes the way a quantity changes with respect to another. In the context of Newton's Law of Cooling, the differential equation is used to represent the rate of temperature change of an object.

The general form we see is \( \frac{dT}{dt} = k(T - T_{env}) \), where \( \frac{dT}{dt} \) denotes the derivative of temperature with respect to time, \( T \) is the temperature of the object, \( T_{env} \) is the ambient temperature, and \( k \) is a constant that reflects how quickly the temperature changes.
  • The differential equation quantifies how the temperature of an object decreases or approaches the surrounding temperature.
  • Separation of variables is a technique often used to solve this equation, allowing us to isolate and integrate both sides to find a solution for \( T \).
Temperature Change
When analyzing temperature change in Newton's Law of Cooling, understanding how temperature shifts over time in relation to its environment is key.

The fundamental concept is that the temperature of an object is influenced by the temperature difference between itself and its surroundings.
  • A hot object in a cooler environment will lose heat faster initially, gradually slowing as the object cools down closer to the ambient temperature.
  • This change is exponential, as the difference between the object's temperature and the environment decreases, so does the rate of cooling.
By solving the differential equation derived from Newton鈥檚 law, we can predict how the temperature of the object will change over time.
Mathematical Modeling
Mathematical modeling is the process of using mathematical language and equations to represent real-world phenomena. In this scenario, Newton's Law of Cooling applies mathematical modeling to predict how a cup of hot water cools over time.

This involves:
  • Defining relevant variables such as initial and ambient temperatures.
  • Using differential equations to create a mathematical representation of temperature change.
  • Solving these equations to predict behavior over time, specifically when the water reaches certain temperatures.
Mathematical models offer a way to seek understanding and estimation before events occur, providing valuable insights in fields ranging from physics to engineering.
Cooling Rate
The cooling rate is a crucial concept in understanding how quickly an object approaches the temperature of its environment. Within Newton's Law of Cooling, this rate is directly proportional to the difference between the object's temperature and the ambient temperature.

Here are some key points:
  • The constant \( k \) in the differential equation \( \frac{dT}{dt} = k(T - T_{env}) \) represents the cooling rate of the specific object.
  • A higher \( k \) indicates that the object cools faster, while a lower \( k \) suggests a slower cooling rate.
  • Experimental conditions such as material type, surface area, and environment could affect \( k \) and thus the cooling rate.
Understanding the concept of cooling rate allows us to predict how long it will take for an object, like a cup of hot water, to reach a desired temperature.

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Most popular questions from this chapter

Solve the given differential equations by Laplace transforms. The function is subject to the given conditions. A \(20-\mathrm{mH}\) inductor, a \(40-\Omega\) resistor, a \(50-\mu \mathrm{F}\) capacitor, and a voltage source of \(100 e^{-1000 t}\) are connected in series in an electric circuit. Find the charge on the capacitor as a function of time \(t,\) if \(q=0\) and \(i=0\) when \(t=0\)

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Solve the given differential equations. $$D^{4} y-y=x$$

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