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Chapter 3: Problem 7

Assume Newton's Law of Cooling to solve the following problem: A body of temperature \(100^{\circ} \mathrm{F}\) is placed at time \(t=0\) in a medium the temperature of which is maintained at \(40^{\circ} \mathrm{F}\). At the end of \(10 \mathrm{~min}\), the body has cooled to a temperature of \(90^{\circ} \mathrm{F}\). (a) What is the temperature of the body at the end of \(30 \mathrm{~min}\) ? (b) When will the temperature of the body be \(50^{\circ} \mathrm{F}\) ?

Short Answer

Expert verified
The temperature of the body at the end of 30 minutes is approximately \(72.32^{\circ}F\), and the body will reach a temperature of \(50^{\circ}F\) at approximately \(69.31\) minutes.

Step by step solution

01

Write the differential equation for Newton's Law of Cooling

According to Newton's Law of Cooling, we have the following differential equation \[\frac{dT}{dt} = -k(T-T_m)\] where \(T\) is the temperature of the body at any time \(t\), \(T_m\) is the temperature of the medium (constant), and \(k\) is the constant of proportionality.
02

Solve the differential equation for T(t)

To find the function \(T(t)\), we need to solve the following separable differential equation: \[\frac{dT}{dt} = -k(T-T_m)\] We can rewrite it as: \[\frac{dT}{T - T_m} = - k dt\] Now, integrate both sides: \[\int{\frac{dT}{T - T_m}} = - k \int dt\] We get: \[\ln |T - T_m| = -kt + C\] where \(C\) is the constant of integration. Taking the exponent of both sides to remove the natural logarithm, we obtain: \[T(t) = T_m + (T_0 - T_m)e^{-kt}\] Here, \(T_0\) is the initial temperature of the body, and \(t\) is the time elapsed.
03

Substitute the given values to find the constant k

At \(t=0\), the initial temperature \(T_0\) is 100掳F, and at \(t=10\) min, the temperature \(T\) is 90掳F. The temperature of the medium, \(T_m\), is maintained at 40掳F. Substitute these values in the equation: \[90 = 40 + (100 - 40)e^{-10k}\] Simplifying the equation, we get: \[50 = 60e^{-10k}\] Divide both sides by 60: \[\frac{5}{6} = e^{-10k}\] Take the natural logarithm of both sides: \[\ln{\frac{5}{6}} = -10k\] Finally, solve for k: \[k = -\frac{1}{10} \ln{\frac{5}{6}}\]
04

Find the temperature of the body at the end of 30 min (part a)

Using the temperature function derived in Step 2 and the constant k found in Step 3, we can find the temperature of the body at \(t = 30\) min: \[T(30) = 40 + (100 - 40)e^{-30 \cdot (-\frac{1}{10} \ln\frac{5}{6})}\] Solve for \(T(30)\): \[T(30) \approx 72.32^\circ F\] So the temperature of the body at the end of 30 min is approximately 72.32掳F.
05

Find the time when the temperature of the body is 50掳F (part b)

Given the temperature function and the constant k, we need to solve the equation \(T(t) = 50\) for \(t\): \[50 = 40 + (100 - 40)e^{-t \cdot (-\frac{1}{10} \ln\frac{5}{6})}\] Simplify the equation: \[10 = 60e^{-t \cdot (-\frac{1}{10} \ln\frac{5}{6})}\] Divide both sides by 60: \[\frac{1}{6} = e^{-t \cdot (-\frac{1}{10} \ln\frac{5}{6})}\] Take the natural logarithm of both sides: \[\ln{\frac{1}{6}} = -t \cdot (-\frac{1}{10} \ln\frac{5}{6})\] Finally, solve for t: \[t = 40 \ln(6)\] So the temperature of the body will be 50掳F when \(t \approx 69.31\) min.

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Key Concepts

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

Ordinary Differential Equations
In the realm of calculus and mathematical modeling, ordinary differential equations (ODEs) play a pivotal role. These equations involve functions of a single variable and their derivatives. They express the rate of change of a physical quantity, such as temperature in the case of Newton's Law of Cooling, and often underlie the dynamic processes observed in nature and engineering.

An ordinary differential equation may look complicated, but it actually represents something quite intuitive: the relationship between an ever-changing quantity and the rate at which it changes. For instance, when examining the cooling of a body, we are concerned with how fast its temperature approaches that of its surrounding environment - this rate of change is precisely what an ODE can describe.

To solve an ODE means to find a function that satisfies the equation for all values within a certain range. In educational terms, if the ODE is the question asking 'how does this quantity change over time?', then the function we find is the answer telling us 'here's exactly how it changes' at any given moment.
Separable Differential Equations
A subclass of ordinary differential equations is the separable differential equations, which can be separated into two parts - one involving only the independent variable (like time) and the other involving only the dependent variable (like temperature).

This makes them relatively easier to solve, as you can treat each side of the equation independently. The separation process involves algebraic manipulation to put all the terms involving the dependent variable on one side of the equation and all the terms involving the independent variable on the other. Then, you integrate both sides, which brings you one step closer to solving for the function you're after.

In the context of Newton's Law of Cooling, we deal with separable equations. By isolating temperature-dependent terms on one side and time-dependent terms on the other, we can integrate and slowly chip away at the problem, making it manageable and bringing us closer to understanding the cooling process.
Exponential Decay
One of the patterns that frequently emerge from solving differential equations in natural sciences is exponential decay. This term describes a process where a quantity decreases at a rate proportional to its current value. What's fascinating about exponential decay is that it's a natural pattern found in a myriad of scenarios, from the cooling of coffee to the discharge of a capacitor in electronics.

Visualizing Exponential Decay

In the case of our cooling problem, the temperature difference between the object and the environment decreases exponentially over time - meaning it changes quickly at first, but this change gradually slows down as it nears the surrounding temperature.

Mathematically, this is described by a function involving the constant 'e', which is the base of natural logarithms. The presence of 'e' in our solution isn't just a mathematical quirk; it's indicative of this universal decay pattern that we can witness in many natural phenomena.
Integration in Calculus
The process of integration in calculus is a fundamental tool to solve differential equations and understand the accumulation of quantities. In essence, integration is the inverse operation of differentiation. Where differentiation gives us the rate at which something changes, integration tells us the total amount of change over an interval.

In problems like Newton's Law of Cooling, integration allows us to move from knowing the rate at which temperature changes to figuring out the temperature at any given time. By integrating both sides of our separable differential equation, we get a function that relates temperature to time.

Importance in Physics and Engineering

By mastering integration, students can unlock answers to countless problems in physics and engineering, not just in theory but in real-world applications as well. Understanding integration isn't just about solving a problem on a page; it's about interpreting what happens in the practical world, one integral at a time.

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Most popular questions from this chapter

At 10 A.M. a woman took a cup of hot instant coffee from her microwave oven and placed it on a nearby kitchen counter to cool. At this instant the temperature of the coffee was \(180^{\circ} \mathrm{F}\), and 10 minutes later it was \(160^{\circ} \mathrm{F}\). Assume the constant temperature of the kitchen was \(70^{\circ} \mathrm{F}\). (a) What was the temperature of the coffee at \(10: 15\) A.m.? (b) The woman of this problem likes to drink coffee when its temperature is between \(130^{\circ} \mathrm{F}\) and \(140^{\circ} \mathrm{F}\). Between what times should she have drunk the coffee of this problem?

Assume Newton's Law of Cooling to solve the following probiem: A body cools from \(60^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in \(15 \mathrm{~min}\) in air which is maintained at \(30^{\circ} \mathrm{C}\). How long will it take this body to cool from \(100^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) in air that is maintained at \(50^{\circ} \mathrm{C}\) ?

A shell weighing \(1 \mathrm{lb}\) is fired vertically upward from the earth's surface with a muzzle velocity of \(1000 \mathrm{ft} / \mathrm{sec}\). The air resistance (in pounds) is numerically equal to \(10^{-4} v^{2}\), where \(v\) is the velocity (in feet per second). (a) Find the velocity of the rising shell as a function of the time. (b) How long will the shell rise?

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Suppose a certain amount of money is invested and draws interest compounded continuously. (a) If the original amount doubles in two years, then what is the annual interest rate? (b) If the original amount increases \(50 \%\) in six months, then how long will it take the original amount to double?
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