Problem 22 - Applied 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 \(10: 00 \mathrm{pm}\), 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 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 indoor temperature at 7:00 am is approximately 48.1°F, so pipes are not freezing. The initial assumption of constant outdoor temperature may not hold, so if it got colder outside, the indoor temperature would be lower.

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. It can be expressed as the differential equation: \[ \frac{dT}{dt} = -k(T - T_{a}) \] where, - \( T \) is the temperature of the object at time \( t \), - \( T_a \) is the ambient temperature (assumed constant here), - \( k \) is a positive constant that depends on the characteristics of the object and the surroundings.

Step 2: Determine the Variables and Constants

In this problem:- \( T_a = 10^{\circ} \text{F} \) since this is the outside temperature,- \( T(0) = 68^{\circ} \text{F} \) is the initial indoor temperature at 1 pm,- At 10 pm, \( T(9) = 57^{\circ} \text{F} \), and - We need to find the indoor temperature at 7 am the next day, which is 18 hours after 1 pm.The differential equation becomes: \[ \frac{dT}{dt} = -k(T - 10) \]

Step 3: Solve the Differential Equation

To solve the differential equation, we separate variables and integrate:\[ \frac{dT}{dt} = -k(T - 10) \] \[ \int \frac{1}{T - 10} \, dT = -\int k \, dt \] This gives us: \[ \ln|T - 10| = -kt + C \] Exponentiating both sides: \[ T - 10 = Ce^{-kt} \] Where \( C \) is the constant of integration.

Step 4: Determine Constants Using Initial Conditions

Given \( T(0) = 68 \), we plug into our solution:\[ 68 - 10 = Ce^0 \] \[ C = 58 \] Our equation becomes:\[ T(t) = 10 + 58e^{-kt} \] Using the information that \( T(9) = 57 \), solve for \( k \):\[ 57 = 10 + 58e^{-9k} \] \[ 47 = 58e^{-9k} \] \[ e^{-9k} = \frac{47}{58} \] \[ -9k = \ln \left( \frac{47}{58} \right) \] \[ k = -\frac{1}{9} \ln \left( \frac{47}{58} \right) \]

Step 5: Estimate Temperature at 7:00 AM

Now that we have \( k \), we can estimate the temperature at 7 am:\[ T(18) = 10 + 58e^{-18k} \] Substitute \( k \):\[ T(18) = 10 + 58 \left( \frac{47}{58} \right)^2 \] Calculate:\[ T(18) \approx 10 + 58 \times 0.656 \approx 48.1^{\circ} \text{F} \]

Step 6: Assess Risk to Water Pipes

Considering water freezes at \( 32^{\circ} \text{F} \), the house at \( 48.1^{\circ} \text{F} \) is above this threshold. There is no immediate risk of pipes freezing unless further temperature drop happens.

Step 7: Evaluate Assumptions and Effects

We assumed the outside temperature \( T_a \) was constant at \( 10^{\circ} \text{F} \). If the outside temperature actually dropped further after 10 pm, the indoor temperature would decrease more and our estimate would be lower. Revising the estimate would depend on actual outdoor temperature data.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. These are vital tools for modeling how quantities change over time. In the context of Newton's Law of Cooling, the differential equation takes the form:

• \( \frac{dT}{dt} = -k(T - T_{a}) \)

This equation tells us that the rate of change of temperature \( T \) is proportional to the difference between \( T \) and the ambient temperature \( T_a \). Here:

• \( \frac{dT}{dt} \) is the rate of temperature change over time.
• \( k \) is a constant that reflects how quickly the temperature changes in response to the differential, depending on specific conditions.

Understanding this concept allows you to predict how quickly a particular environment will return to equilibrium with its surroundings after a change in temperature.

Ambient Temperature

Ambient temperature refers to the temperature of the surrounding environment of a particular system. In discussions involving Newton's Law of Cooling, this is an integral concept, as the rate of temperature change depends on the difference between the object's temperature and the ambient temperature.

In our exercise, the ambient temperature is held constant at \( 10^{\circ} \text{F} \), which is the temperature outside the house. This is a key assumption, as any changes to the ambient temperature could drastically influence the cooling process.

• A consistent ambient temperature implies that the indoor temperature change depends solely on the initial conditions and the cooling constant \( k \).
• If the ambient temperature were to change, the original calculations would need to be revised to reflect the new conditions.

Temperature Change

Temperature change over time is governed by several factors, especially in the framework of Newton's Law of Cooling. Here, understanding the temperature change helps to assess how the warmth gradually dissipates in a house over the period of power loss.

The initial condition set our starting point at \( 68^{\circ} \text{F} \). Between 1 pm and 10 pm, the house's temperature dropped to \( 57^{\circ} \text{F} \), indicating the rate dictated by our differential equation.

• The differential equation incorporates these initial values to predict further temperature changes, such as to \( 48.1^{\circ} \text{F} \) at 7 am.
• Understanding these changes is crucial for making practical decisions, like evaluating the potential risk to water pipes.

Freezing Point

The freezing point is a critical threshold for assessing potential environmental risks within temperature-dependent scenarios. Water, for instance, freezes at \( 32^{\circ} \text{F} \) under normal atmospheric pressure.

In our exercise, determining whether your house's water pipes are at risk depends on the forecasted indoor temperature. Given the estimate of \( 48.1^{\circ} \text{F} \), the temperature remains safely above freezing.

• To ensure safety, it's important to regularly monitor if forecasted temperature drop could lead pipes closer to the freezing point.
• Factors like insulation, wind chill, and duration of exposure also play a part in real-life risk assessments.