Problem 27 - Calculus

Question

At 1: 00 pm one winter afternoon, there is a power failure at your house in Wisconsin, and your heat does not work without electricity. When the power goes out, it is \(68^{\circ} \mathrm{F}\) in your house. At 10: 00 pm, it is \(57^{\circ} \mathrm{F}\) in the house, and you notice that it is \(10^{\circ} \mathrm{F}\) outside. (a) Assuming that the temperature, \(T,\) in your home obeys Newton's Law of Cooling, write the differential equation satisfied by \(T\) (b) Solve the differential equation to estimate the temperature in the house when you get up at 7: 00 am the next morning. Should you worry about your water pipes freezing? (c) What assumption did you make in part (a) about the temperature outside? Given this (probably incorrect) assumption, would you revise your estimate up or down? Why?

Step-by-Step Solution

Answer

The house temperature at 7:00 AM is approximately \(50.4^{\circ} F\). You should worry about pipes, given lower actual outdoor temperatures.

Step 1: Understanding Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the ambient temperature. Mathematically, this can be expressed as \( \frac{dT}{dt} = -k (T - T_a) \), where \( T \) is the temperature of the object, \( T_a \) is the ambient (or surrounding) temperature, and \( k \) is a positive constant.

Step 2: Setting Up the Differential Equation

Given the information, at 1:00 pm, \( T(0) = 68^{\circ} F \) and at 10:00 pm, \( T(9) = 57^{\circ} F \). The ambient temperature \( T_a = 10^{\circ} F \). Using Newton's Law of Cooling, the differential equation becomes \( \frac{dT}{dt} = -k (T - 10) \).

Step 3: Solving the Differential Equation

The differential equation \( \frac{dT}{dt} = -k (T - 10) \) is separable. Rearrange and integrate: \( \int \frac{1}{T - 10} \, dT = -k \int \, dt \). This gives \( \ln |T - 10| = -kt + C \). Solving for \( T \), we get \( T = 10 + Ce^{-kt} \). Using initial condition \( T(0) = 68 \), we find \( C = 58 \), so \( T(t) = 10 + 58e^{-kt} \).

Step 4: Determine the Rate Constant "k"

Using \( T(9) = 57 \), set \( 57 = 10 + 58e^{-9k} \). So \( 47 = 58e^{-9k} \), leading to \( e^{-9k} = \frac{47}{58} \). Solving for \( k \), we find \( k = - \frac{1}{9}\ln\left(\frac{47}{58}\right) \approx 0.0228 \).

Step 5: Estimate the Temperature at 7:00 AM

Calculate \( T(18) \) for 7:00 AM (18 hours after 1:00 PM): \( T(18) = 10 + 58e^{-18 imes 0.0228}\). This results in \( T(18) \approx 50.4^{\circ} F \).

Step 6: Answering Part (c) and Evaluating Assumptions

The assumption was that the outside temperature remained constant at \( 10^{\circ} F \). In reality, temperatures may vary, usually dropping overnight, which would increase the cooling rate, thus your estimate would likely be revised downward, suggesting potentially colder conditions.

Key Concepts

Differential Equation

At the heart of Newton's Law of Cooling is a differential equation, a powerful mathematical tool used to describe how a quantity changes over time. In this context, the quantity being considered is the temperature inside the house. A differential equation relating to temperature change needs to take into account the difference between the current temperature and the ambient temperature.

The differential equation provided by the law is: \[ \frac{dT}{dt} = -k (T - T_a) \]Here,

• \(\frac{dT}{dt}\) represents the rate of change of temperature over time.
• \(T\) is the temperature of the house, which changes as time passes.
• \(T_a\) is the constant ambient temperature outside, which is significantly lower, at \(10^{\circ} F\).
• \(k\) is a positive constant that represents how quickly the temperature approaches the ambient temperature, unique to each situation.

The equation highlights how temperature inside the house is inevitably creeping closer to the surrounding chill.

Rate of Change of Temperature

Behind every interesting physical phenomenon is the subtle dance of change over time, captured mathematically through the rate of change. In the context of cooling, it provides insight into how swiftly the temperature in your house is dropping.

From Newton's Law of Cooling, the rate of change of temperature is given by:\[ \frac{dT}{dt} = -k (T - T_a) \]This encapsulates:

• **Rate of change (\(\frac{dT}{dt} \))**: It quantifies how temperature shifts moment by moment.
• **Difference factor (\(T - T_a \))**: It is the driving force, calculated by the current temperature minus ambient temperature.
• *Dissipation constant (\(k \))*: It adjusts the speed based on environmental and material properties.

Understanding this rate allows homeowners to predict future conditions, determining the coldness risk inside as the night drags on.

Ambient Temperature

The concept of ambient temperature plays a crucial role in determining how quickly a body cools down. It represents the temperature of the environment surrounding the object—in this case, the temperature outside the house in Wisconsin.

Imagine your house as a sealed container; the warmth inside is trying desperately to blend with the cold of its surroundings. This outside temperature, termed ambient temperature, was noted as \(10^{\circ}F\).

Its constancy is an underlying assumption in our model, providing a stable reference for the calculation. However, in real world scenarios, ambient temperature can fluctuate wildly, especially overnight. This deviation means our estimates might need adjusting, as colder ambient temperatures could result in quicker cooling inside.

Separable Differential Equation

A separable differential equation is a particular type of differential equation where the variables can be separated on different sides of the equation. This property simplifies solving the equation greatly. Your answer becomes more approachable, just like piecing together an easy puzzle.

In Newton's Law of Cooling, we use this characteristic to decouple the temperature variable, \(T\), and time, \(t\). Rearranging gives:\[\int \frac{1}{T - 10} \, dT = -k \int \, dt\]This rearranged form allows us to integrate each side separately, resulting in: \[ \ln |T - 10| = -kt + C \]The solution lets us express temperature explicitly, making it possible to use the initial conditions given in the exercise to solve for specific constants, like \(C\), ensuring we model the cooling process accurately for your home.

Such techniques turn complex differential puzzles into manageable models of understanding temperature dynamics within our living spaces.