/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 The rate at which a body cools a... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 20

The rate at which a body cools also depends on its exposed surface area \(S\). If \(S\) is a constant, then a modification of (2) is $$\frac{d T}{d t}=k S\left(T-T_{m}\right)$$ where \(k<0\) and \(T_{m}\) is a constant. Suppose that two cups \(A\) and \(B\) are filled with coffee at the same time. Initially, the temperature of the coffee is \(150^{\circ} \mathrm{F}\). The exposed surface area of the coffee in cup \(B\) is twice the surface area of the coffee in cup \(A\). After 30 min the temperature of the coffee in cup \(A\) is \(100^{\circ} \mathrm{F}\). If \(T_{m}=70^{\circ} \mathrm{F}\), then what is the temperature of the coffee in cup \(B\) after 30 min?

Short Answer

Expert verified
The temperature of the coffee in cup B after 30 minutes is approximately 81.25掳F.

Step by step solution

01

Set up the differential equation for each cup

The modified Newton's Law of Cooling is given by \( \frac{dT}{dt} = kS(T - T_m) \). For cup A, let the exposed surface area be \( S_A \), so the equation becomes \( \frac{dT_A}{dt} = kS_A(T_A - 70) \). For cup B, the surface area is twice that of cup A, \( S_B = 2S_A \), hence \( \frac{dT_B}{dt} = k(2S_A)(T_B - 70) \).
02

Integrate the differential equation for cup A

Rearrange and integrate the differential equation for cup A: \[ \frac{dT_A}{T_A - 70} = kS_A \, dt \] Integrate both sides to obtain: \[ \ln|T_A - 70| = kS_A t + C_1 \] Solve for \( T_A \): \[ T_A = 70 + Ce^{kS_A t} \] where \( C \) is a constant determined by initial conditions.
03

Apply initial and boundary conditions for cup A

Initially, \( T_A(0) = 150 \), so \( 150 = 70 + Ce^{kS_A \cdot 0} \), leading to \( C = 80 \). After 30 minutes, \( T_A(30) = 100 \), giving: \[ 100 = 70 + 80e^{kS_A \times 30} \] Solve this for \( kS_A \): \[ 30 = 80e^{30kS_A} \] \[ e^{30kS_A} = \frac{3}{8} \] \[ 30kS_A = \ln\left(\frac{3}{8}\right) \] \[ kS_A = \frac{1}{30} \ln\left(\frac{3}{8}\right) \]
04

Integrate the differential equation for cup B

For cup B, the equation is \( \frac{dT_B}{dt} = k(2S_A)(T_B - 70) \). Integrate:\[ \frac{dT_B}{T_B - 70} = 2kS_A \, dt \]Integration gives:\[ \ln|T_B - 70| = 2kS_A t + C_2 \]Solve for \( T_B \):\[ T_B = 70 + C'e^{2kS_A t} \] where \( C' \) is a constant determined by initial conditions.
05

Apply initial conditions for cup B and calculate final temperature

Initially, \( T_B(0) = 150 \), so \( C' = 80 \). Use \( kS_A = \frac{1}{30} \ln\left(\frac{3}{8}\right) \) to find the temperature after 30 minutes:\[ T_B(30) = 70 + 80e^{30(2kS_A)} \]\[ T_B(30) = 70 + 80 \left( \frac{3}{8} \right)^2 \]\[ T_B(30) = 70 + 80 \times \frac{9}{64} \]\[ T_B(30) \approx 81.25 \]Thus, the temperature of the coffee in cup B after 30 minutes is approximately \( 81.25^{\circ} \mathrm{F} \).

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Key Concepts

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

Differential Equation
A differential equation is a mathematical equation that relates a function to its derivatives. In the context of Newton's Law of Cooling, these equations help describe how the temperature of an object changes over time.
The specific differential equation modified for cooling is given by:
  • \( \frac{d T}{d t} = k S (T - T_m) \)
Here, \( \frac{d T}{d t} \) represents the rate of change of temperature with time, \( k \) is a constant indicating cooling efficiency, \( S \) is the surface area, \( T \) is the current temperature, and \( T_m \) is the ambient temperature or the surrounding temperature.
Solving this differential equation gives us an exponential function that allows us to predict how the temperature of the object decreases over time.
Cooling Rate
The cooling rate refers to how quickly the temperature of a body decreases over time. It is directly influenced by several factors, including the surface area and the temperature difference between the body and its environment.
In the equation \( \frac{d T}{d t} = k S (T - T_m) \):
  • The term \( k S (T - T_m) \) defines the rate at which cooling occurs.
  • The constant \( k \) is negative, ensuring that the temperature of the object moves towards the ambient temperature \( T_m \).
  • This rate depends on two things: the exposed surface area \( S \), and the difference \( T - T_m \).
The greater the difference between \( T \) and \( T_m \), the faster the cooling rate. Similarly, a larger surface area \( S \) increases the cooling rate since more of the body's surface is in contact with the surrounding environment.
Surface Area
Surface area, noted as \( S \) in the differential equation, plays a critical role in Newton's Law of Cooling.
When a larger surface area is exposed, more heat can be transferred between the object and its environment. This accelerates the cooling process.
  • For instance, in the exercise scenario, cup B has a surface area twice that of cup A.
  • This change in \( S \) reflects in our differential equation as it is multiplied by a factor of two for cup B.
This increased surface contact facilitates faster heat dissipation, which explains why cup B's temperature drops more rapidly compared to cup A over the same time period. Thus, knowing the surface area is crucial in calculating how quickly temperature changes occur.
Temperature Change
Temperature change represents the shift in temperature of an object as it cools towards the ambient temperature. Measuring this change involves analyzing initial temperature conditions and how they evolve over time based on Newton's Law of Cooling.
As given in the exercise, initially, both cups had a temperature of \( 150^{\circ} \mathrm{F} \).
  • For cup A, after 30 minutes, the temperature drops to \( 100^{\circ} \mathrm{F} \).
  • Cup B, which has more surface area, cools even more, reaching approximately \( 81.25^{\circ} \mathrm{F} \).
This change is calculated through the integrated form of the differential equation, \( T = T_m + Ce^{kSt} \), and it depends on factors like surface area and the initial temperature difference from the ambient temperature. By looking at exponential decay based on this relation, we can predict future temperatures over different periods.

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Most popular questions from this chapter

A 30 -volt electromotive force is applied to an \(L R\) -series circuit in which the inductance is 0.1 henry and the resistance is 50 ohms. Find the current \(i(t)\) if \(i(0)=0 .\) Determine the current as \(t \rightarrow \infty\)

When interest is compounded continuously, the amount of money increases at a rate proportional to the amount \(S\) present at time \(t\) that is, \(d S / d t=r S,\) where \(r\) is the annual rate of interest. (a) Find the amount of money accrued at the end of 5 years when \(\$ 5000\) is deposited in a savings account drawing \(5 \frac{3}{4} \%\) annual interest compounded continuously. (b) In how many years will the initial sum deposited have doubled? (c) Use a calculator to compare the amount obtained in part (a) with the amount \(S=5000\left(1+\frac{1}{4}(0.0575)\right)^{5(4)}\) that is accrued when interest is compounded quarterly.

The differential equation \(d P / d t=(k \cos t) P,\) where \(k\) is a positive constant, is a mathematical model for a population \(P(t)\) that undergoes yearly seasonal fluctuations. Solve the equation subject to \(P(0)=P_{0}\) Use a graphing utility to graph the solution for different choices of \(P_{0}\)

When a vertical beam of light passes through a transparent medium, the rate at which its intensity \(I\) decreases is proportional to \(I(t),\) where \(t\) represents the thickness of the medium (in feet). In clear seawater, the intensity 3 feet below the surface is \(25 \%\) of the initial intensity \(I_{0}\) of the incident beam. What is the intensity of the beam 15 feet below the surface?

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