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Chapter 3: Problem 8

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 \(50^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
It will take the body 15 minutes to cool from \(100^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) in air maintained at \(50^{\circ} \mathrm{C}\).

Step by step solution

01

Understand Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of temperature of an object is proportional to the difference between its temperature and the ambient temperature. Mathematically, it's represented as: \[\frac{dT}{dt} = -k(T - T_A)\] where: - \(T\) is the temperature of the object at a given time, - \(T_A\) is the ambient temperature, - \(k\) is a positive proportionality constant, - \(t\) is the time elapsed, - \(dt\) is time differential - \(dT\) is the rate of change of temperature. Our goal is to find the time it takes for the object to cool from \(100^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) in air maintained at \(50^{\circ} \mathrm{C}\).
02

Separate variables and integrate

To find the formula for the temperature of the object at any given time, separate variables and integrate both sides. Divide both sides by \((T - T_A)\) and multiply by \(dt\): \[\frac{dT}{T - T_A} = -k \cdot dt\] Now, integrate both sides: \[\int_{T_1}^{T_2} \frac{dT}{T - T_A} = -k \int_{t_1}^{t_2} dt\] Where: - \(T_1\) and \(T_2\) are initial and final temperature values, respectively. - \(t_1\) and \(t_2\) are initial and final time moments, respectively.
03

Evaluate the integrals for the first cooling scenario

Given that the body cools from \(60^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in \(15\) minutes and the ambient temperature is maintained at \(30^{\circ} \mathrm{C}\), we obtain the following equation: \[\int_{60}^{50} \frac{dT}{T - 30} = -k \int_{0}^{15} dt\] Evaluate the left side: \[-\ln\Big|\frac{T - 30}{80-30}\Big| \Big|_{60}^{50} = -k \cdot (15 - 0)\] Hence, \[-\ln\Big(\frac{20}{50}\Big) = 15k\] Now, we can solve for the constant of proportionality \(k\): \[k = \frac{-\ln(0.4)}{15}\]
04

Evaluate the integrals for the second cooling scenario

Now, we are asked to find the time it takes for the body to cool from \(100^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) in air maintained at \(50^{\circ} \mathrm{C}\). Let this time be \(t'\). \[\int_{100}^{80} \frac{dT}{T - 50} = -k \int_{0}^{t'} dt\] Evaluate the left side: \[-\ln\Big|\frac{T - 50}{100-50}\Big| \Big|_{100}^{80} = -k \cdot (t' - 0)\] Hence, \[-\ln(0.4) = t' \cdot \frac{-\ln(0.4)}{15}\]
05

Solve for the unknown time \(t'\)

Now, we can solve for the unknown time \(t'\): \[t' = 15\] Therefore, it will take the body 15 minutes to cool from \(100^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) in air maintained at \(50^{\circ} \mathrm{C}\).

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that involve derivatives of a function.
They help us understand how quantities change over time or space.

In the context of Newton's Law of Cooling, we deal with a differential equation where the rate of change of temperature is related to the temperature difference between the object and its surroundings.
This relationship allows us to model how the temperature of an object decreases over time as it cools in an environment with a constant temperature.

The differential equation applied is: \[\frac{dT}{dt} = -k(T - T_A)\]- Here, \(\frac{dT}{dt}\) represents the rate at which the temperature changes.
- \(T\) is the temperature of the object at time \(t\),
- \(T_A\) is the ambient temperature.
- \(k\) is a constant that quantifies the cooling rate, specific to the object's material properties and conditions.
Integration
Integration is the mathematical process of finding the accumulated quantity over an interval.
In basic terms, it helps us "add up" small changes over a period or space to get a total change.
For Newton's Law of Cooling, integration is used to solve the differential equation.

We separate variables to re-arrange the equation into a form that can be integrated:\[\frac{dT}{T - T_A} = -k \, dt\]This equation is structured to integrate both sides:\[\int \frac{dT}{T - T_A} = -k \int dt\]- The left side, involving \(dT\), sums temperature changes.
- The right side, involving \(dt\), sums time intervals.

Because integration reverses differentiation, it helps us find the temperature as a function of time, revealing how long it takes to reach a certain temperature.
Rate of Change
The rate of change is a concept that's all about how quickly something varies over time or another parameter.
In this problem, it focuses on how the temperature of an object decreases in relation to the temperature of its surroundings.
The differential equation, \(\frac{dT}{dt} = -k(T - T_A)\), directly represents this rate.

Key points include:
  • The rate of change is proportional to the difference in temperature between the object and its environment.
  • If this difference is large, the rate of cooling is faster; if small, slower.
  • The negative sign suggests a decrease in temperature over time.

Knowing the rate of change lets you predict how fast or slow the body reaches the desired temperature.
Temperature Cooling
Temperature cooling is a common real-world phenomenon described by Newton's Law of Cooling.
All warm objects naturally tend to cool down to match the temperature of their surrounding air or environment.

The cooling process can be understood through:
  • Initial Temperature: The starting temperature of the body, \(T_1\).
  • Final Temperature: The temperature the body cools to, \(T_2\).
  • Ambient Temperature: The constant surrounding temperature, \(T_A\).
  • Time: Duration taken to cool from \(T_1\) to \(T_2\).

Applying these parameters,- You can clearly set up the cooling equation and solve for unknowns like time.- For example, solving the equation:\[\int_{T_1}^{T_2} \frac{dT}{T - T_A} = -k \int_{t_1}^{t_2} dt\]gives you the time needed for an object to reach a certain temperature, given the conditions.

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