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Solve the differential equation for Newton's Law of Cooling to find the temperature function in the following cases. Then answer any additional questions. An iron rod is removed from a blacksmith's forge at a temperature of \(900^{\circ} \mathrm{C}\). Assume \(k=0.02\) and the rod cools in a room with a temperature of \(30^{\circ} \mathrm{C}\). When does the temperature of the rod reach \(100^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The iron rod reaches a temperature of \(100^{\circ} \mathrm{C}\) after approximately \(134.86\) minutes.

Step by step solution

01

Rewrite the differential equation with the given information

Plug in the given values of \(k\) and \(T_a\) into the Newton's Law of Cooling differential equation: $$\frac{dT(t)}{dt} = -0.02(T(t) - 30)$$
02

Separate the variables

To solve this differential equation, we need to separate the variables. To do this, we rearrange the equation: $$\frac{dT(t)}{T(t) - 30} = -0.02 \, dt$$
03

Integrate both sides

Now, we will integrate both sides of the equation with respect to their respective variables: $$\int \frac{dT(t)}{T(t) - 30} = -0.02 \int dt$$ On the left side, we have a simple logarithmic integral: $$\ln|T(t) - 30| = -0.02t + C$$
04

Solve for temperature function T(t)

Now, we will find the temperature function \(T(t)\). First, we will exponentiate both sides of the equation to get: $$T(t) - 30 = e^{-0.02t + C}$$ Notice that we can write \(e^{-0.02t + C} = e^{-0.02t} \cdot e^C\). Since \(e^C\) is another constant, we can define \(A = e^C\) and rewrite the previous expression as: $$T(t) - 30 = Ae^{-0.02t}$$ Then, we solve for \(T(t)\): $$T(t) = 30 + Ae^{-0.02t}$$
05

Solve for A using the initial temperature condition

We are given that \(T(0) = 900^{\circ} \mathrm{C}\). Plugging this into our expression for \(T(t)\): $$900 = 30 + Ae^{-0.02(0)}$$ Solve for A: $$A = 870$$ Thus, the temperature function is given by: $$T(t) = 30 + 870e^{-0.02t}$$
06

Find the time when the temperature reaches \(100^{\circ} \mathrm{C}\)

Now, we want to find the time \(t\) when the temperature \(T(t) = 100^{\circ} \mathrm{C}\). Plugging this value into our temperature function: $$100 = 30 + 870e^{-0.02t}$$ Rearrange and solve for \(t\): $$e^{-0.02t} = \frac{70}{870}$$ $$-0.02t = \ln{\frac{70}{870}}$$ $$t = \frac{\ln{\frac{70}{870}}}{-0.02} $$ Now, use a calculator to find the numerical value of \(t\): $$t \approx 134.86 \, \mathrm{minutes}$$ So, the temperature of the rod reaches \(100^{\circ} \mathrm{C}\) after approximately \(134.86\) minutes.

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Key Concepts

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

Differential Equations
Differential equations play a crucial role in modeling various real-world scenarios, such as heat transfer, population growth, and motion. They are equations that involve an unknown function and its derivatives. In the context of Newton's Law of Cooling, the differential equation models how a substance's temperature changes with time.

For Newton's Law of Cooling, the differential equation is given by:
  • \(\frac{dT(t)}{dt} = -k(T(t) - T_a)\)
where \(T(t)\) is the temperature of the object at time \(t\), \(T_a\) is the ambient temperature, and \(k\) is a positive constant that depends on the properties of the material. This equation suggests that the rate of change of temperature is proportional to the difference in temperature between the object and its surroundings.

The solution to this type of differential equation helps us find the temperature function \(T(t)\) over time, which is essential to answering problems like when an iron rod will cool to a certain temperature.
Initial Value Problems
Solving initial value problems is key in applying differential equations to practical scenarios. An initial value problem specifies the value of the unknown function at a particular point, which is crucial for determining the constant involved in the solution.

When dealing with Newton's Law of Cooling, an initial condition is often given, such as the initial temperature of the object. In our exercise, the condition is:
  • \(T(0) = 900^{\circ} \mathrm{C}\)
This tells us the temperature at time \(t = 0\).
By using the initial condition in the solution of the differential equation, we find the specific value for the constant. In this case, the presence of an initial value allows us to determine \(A\), making our temperature function specific to the problem."
Separation of Variables
Separation of variables is a straightforward method for solving differential equations, particularly when the variables of the equation can be isolated on opposite sides. This technique is pivotal in solving Newton's Law of Cooling problems.

The process begins by rearranging the differential equation so that each variable and its differential are on different sides of the equation:
  • \(\frac{dT(t)}{T(t) - 30} = -0.02 \, dt\)
Once separated, each side can be integrated independently to progress towards the solution. The beauty of separation of variables is in its simplicity, turning a potentially complex problem into integrals that are easier to handle.
After integrating, we obtain an equation with a logarithmic expression on one side and a linear expression involving time on the other. This step is crucial in developing the function that describes the system's behavior over time.
Exponential Functions
Exponential functions appear frequently in solutions to differential equations like those encountered with Newton's Law of Cooling. These functions exhibit the characteristic constant ratio of change over equal increments of the variable.

As a solution to the differential equation, the temperature function takes the form:
  • \(T(t) = 30 + Ae^{-0.02t}\)
Here, the exponential part \(Ae^{-0.02t}\) represents the rate at which the temperature changes. Such expressions are essential in modeling decay processes, as they reflect how rapidly or slowly an object cools.

Exponentiating after integration helps us isolate the temperature function itself, as logarithmic integration initially provides a natural log expression. Understanding the role of exponential functions in this context helps clarify why temperature diminishes exponentially over time.

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