Problem 22 - Applied Calculus

Question

At \(1: 00 \mathrm{pm}\) 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 \(10: 00 \mathrm{pm}\), 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 \(7: 00 \mathrm{am}\) 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 estimated temperature at 7:00 am is \( 52.5^{\circ} \mathrm{F} \); no immediate worry for pipes. If outside temp drops more, adjust estimate downwards.

Step 1: Understand 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 its own temperature and the ambient temperature. Mathematically, this is described as: \[ \frac{dT}{dt} = -k(T - T_{out}) \]where \( T \) is the temperature in the house, \( T_{out} \) is the outdoor temperature, and \( k \) is a constant of proportionality.

Step 2: Write the Differential Equation

Based on Newton's Law of Cooling, the differential equation modeling the problem is: \[ \frac{dT}{dt} = -k(T - 10) \]because the outside temperature \( T_{out} = 10^{\circ} \mathrm{F} \).

Step 3: Solve the Differential Equation

The differential equation \( \frac{dT}{dt} = -k(T - 10) \) is separable. We can separate variables and integrate:\[ \int \frac{1}{T-10} \, dT = -k \int \, dt \]Integrating gives:\[ \ln|T-10| = -kt + C \]Thus, solving for \( T \), we have:\[ T(t) = 10 + Ce^{-kt} \]

Step 4: Use Initial Conditions to Find Constants

We use the fact that at \( t=0 \) (1:00 pm), \( T(0) = 68 \):\[ 68 = 10 + Ce^{0} \Rightarrow C = 58 \]Now, at \( t=9 \) hours (10:00 pm), \( T(9) = 57 \):\[ 57 = 10 + 58e^{-9k} \]\[ 47 = 58e^{-9k} \]\[ e^{-9k} = \frac{47}{58} \]

Step 5: Solve for k

Take the natural log to solve for \( k \):\[ -9k = \ln \left( \frac{47}{58} \right) \]\[ k = -\frac{1}{9} \ln \left( \frac{47}{58} \right) \]

Step 6: Estimate Temperature at 7:00 am

At 7:00 am, \( t = 18 \) hours:\[ T(18) = 10 + 58e^{-18k} \]Substitute \( k \approx -\frac{1}{9} \ln \left( \frac{47}{58} \right) \) and calculate \( T(18) \). After computing, you will find that \( T(18) \approx 52.5^{\circ} \text{F} \). Since this temperature is above freezing, the pipes should be fine.

Step 7: Consider Assumptions About Outer Temperature

Assume constant external temperature of \( 10^{\circ} \text{F} \), which may not be true. If outside gets colder, the estimate \( 52.5^{\circ} \mathrm{F} \) is too high and vice versa if it gets warmer.

Key Concepts

Differential Equations

Differential equations are mathematical tools used to describe how a quantity changes with respect to another. They are equations that involve a function and its derivatives. In our case, the function represents the temperature in the house, and its derivative represents the rate of change of this temperature over time.

For Newton's Law of Cooling, the differential equation is written as \( \frac{dT}{dt} = -k(T - T_{out}) \). This equation shows that the rate at which the temperature (\( T \)) changes is proportional to the difference between the current temperature and the ambient temperature outside (\( T_{out} \)). This is a first-order linear differential equation and is fundamental for modeling temperature changes in systems exposed to external temperature influences.

In the problem at hand, solving this differential equation helps us predict how the temperature in the house will change over time given the constant external temperature.

Rate of Change

The 'rate of change' in the context of this problem is a key element of Newton's Law of Cooling. It defines how quickly or slowly the temperature in the house changes over time. The differential equation \( \frac{dT}{dt} = -k(T - T_{out}) \) explicitly models this rate.

Here, \( \frac{dT}{dt} \) is the rate of change of the temperature in the house. It is dependent on two main factors:

• The difference \((T - T_{out})\), which is how much warmer the inside is compared to the outside.
• The constant \( k \), which determines how quickly the temperature change happens. This constant is influenced by factors like building insulation and airflow.

The equation tells us: the larger the temperature difference or the greater the value of \( k \), the faster the rate of cooling (or warming) of the house. This relationship allows us to predict future temperatures if we know the current conditions.

Ambient Temperature

Ambient temperature is the temperature of the surrounding environment and plays a crucial role in Newton's Law of Cooling. It directly influences the rate of temperature change in a system such as a house.

In the given problem, the ambient temperature is the outdoor temperature, which is assumed to be constantemente at \( 10^{\circ} \mathrm{F} \). This assumption simplifies our calculations by treating the surrounding air temperature as stable, though in reality, it might fluctuate.

This stable ambient temperature acts as a baseline from which the house's temperature diverges. The greater the difference between the ambient temperature and the interior temperature, the faster the house cools down. Without the ambient temperature factor in our equation, there would be no natural tendency for the house's temperature to decrease, further highlighting its importance in temperature modeling.

Temperature Modeling

Temperature modeling is the process of predicting the future temperature of a space based on current conditions and established mathematical relationships.

Using Newton's Law of Cooling, we create a model with the differential equation \( \frac{dT}{dt} = -k(T - T_{out}) \) to estimate how the temperature in the house changes.

The model is established by:

• Setting the initial condition, such as the temperature in the house at the start.
• Calculating the constant \( k \) using known temperature data at different times.
• Predicting the temperature at future times with the equation \( T(t) = T_{out} + Ce^{-kt} \), where \( C \) is a constant derived from initial conditions.

In practical terms, this modeling helps determine if conditions inside the house reach a critical point, such as the freezing point, potentially affecting aspects like comfort or the risk of damage to plumbing. Effective temperature modeling allows homeowners to anticipate and mitigate potential problems caused by temperature fluctuations.