Problem 49
Question
An ice cube tray filled with tap water is placed in the freezer, and the temperature of the water is changing at the rate of \(-12 e^{-0.2 t}\) degrees Fahrenheit per hour after \(t\) hours. The original temperature of the tap water was 70 degrees. a. Find a formula for the temperature of water that has been in the freezer for \(t\) hours. b. When will the ice be ready? (Water freezes at 32 degrees. \()\)
Step-by-Step Solution
Verified Answer
The formula is \(T(t) = 60 e^{-0.2 t} + 10\). Ice is ready in approximately 5 hours.
1Step 1: Set up the problem
We start by noting that we have a rate of change of temperature given by \(-12 e^{-0.2 t}\) degrees per hour. We need to find the temperature function itself, starting from the initial temperature of 70 degrees.
2Step 2: Integrate the rate of temperature change
To find the temperature function, integrate the rate of change \(-12 e^{-0.2 t}\) with respect to time \(t\). This will give us the general form of the temperature function. The integral is \(-12\int e^{-0.2 t} \, dt = 60 e^{-0.2 t} + C\), where \(C\) is the constant of integration.
3Step 3: Find the constant of integration
Use the initial condition that the temperature at \(t = 0\) is 70 degrees. Thus, \(T(0) = 60 e^{-0.2 \times 0} + C = 70\). This simplifies to \(60 + C = 70\), leading to \(C = 10\).
4Step 4: Write the temperature function
Substitute \(C = 10\) back into the temperature function. Hence, the formula for the temperature after \(t\) hours is \(T(t) = 60 e^{-0.2 t} + 10\).
5Step 5: Determine when the ice is ready
We are asked to find when the water reaches 32 degrees. Set \(T(t) = 32\) and solve for \(t\). This leads to the equation \(60 e^{-0.2 t} + 10 = 32\). Simplifying, we find \(e^{-0.2 t} = \frac{22}{60}\). Take the natural logarithm of both sides: \(-0.2 t = \ln\left(\frac{22}{60}\right)\). Solve for \(t\): \(t = -\frac{\ln\left(\frac{22}{60}\right)}{0.2}\).
6Step 6: Calculate the time
Calculate \(t\) using the expression from the previous step. \(t = -\frac{\ln(0.3667)}{0.2} \approx 5.02\) hours. So, it takes approximately 5 hours for the water to freeze.
Key Concepts
IntegrationRate of ChangeInitial Condition
Integration
Integration is a fundamental concept in calculus, especially when dealing with rates of change like in our ice cube example. When we have a known rate of change and want to find the original function, we utilize integration. It's like working backwards from finding the speed to determining how far we've traveled.
In our problem, the rate of temperature change is given by the function \(-12 e^{-0.2 t}\). To determine the actual temperature function, \(T(t)\), we integrate this rate of change with respect to time \(t\). The integral provides us not only with how fast the temperature changes but rebuilds the function that tells us the temperature at any given time.
By performing the integral, we find: \(-12\int e^{-0.2 t} \, dt = 60 e^{-0.2 t} + C\). This integral calculation gives us the basic temperature function, except we need to determine the value for \(C\), known as the constant of integration.
In our problem, the rate of temperature change is given by the function \(-12 e^{-0.2 t}\). To determine the actual temperature function, \(T(t)\), we integrate this rate of change with respect to time \(t\). The integral provides us not only with how fast the temperature changes but rebuilds the function that tells us the temperature at any given time.
By performing the integral, we find: \(-12\int e^{-0.2 t} \, dt = 60 e^{-0.2 t} + C\). This integral calculation gives us the basic temperature function, except we need to determine the value for \(C\), known as the constant of integration.
Rate of Change
The rate of change helps us understand how quickly a variable changes over time. In this problem, the rate of temperature change is described by the function \(-12 e^{-0.2 t}\). This function tells us that the temperature is decreasing and does so more slowly as time progresses.
The exponent \(-0.2 t\) shows that over time, as \(t\) increases, the negative effect grows slower. Thus, the initial drop in temperature is rapid, but this rate slows down as the liquid approaches freezing point. Understanding the rate of change is crucial as it affects the predicted times and behaviors resulting from the initial conditions set by the problem.
This concept is very practical in real-world applications, such as cooling rates in industrial processes or even predicting how long food items take to reach certain temperatures in a refrigerator or freezer.
The exponent \(-0.2 t\) shows that over time, as \(t\) increases, the negative effect grows slower. Thus, the initial drop in temperature is rapid, but this rate slows down as the liquid approaches freezing point. Understanding the rate of change is crucial as it affects the predicted times and behaviors resulting from the initial conditions set by the problem.
This concept is very practical in real-world applications, such as cooling rates in industrial processes or even predicting how long food items take to reach certain temperatures in a refrigerator or freezer.
Initial Condition
Initial conditions in calculus problems are used to solve for unknown constants that appear after integration. Without initial conditions, the solution would remain a family of curves rather than a precise function.
In our exercise, the initial condition is that the temperature of the water at time \(t = 0\) is 70 degrees Fahrenheit. Using this piece of information, we substitute into our previously found expression: \(T(0) = 60 e^{-0.2 \times 0} + C = 70\). By simplifying this, we find our constant \(C\) to be 10. With \(C\) known, the precise temperature function becomes \(T(t) = 60 e^{-0.2 t} + 10\).
This allows us to forecast the temperature at any future time \(t\). Initial conditions link the general integration solution back to the specific context or scenario you're interested in.
In our exercise, the initial condition is that the temperature of the water at time \(t = 0\) is 70 degrees Fahrenheit. Using this piece of information, we substitute into our previously found expression: \(T(0) = 60 e^{-0.2 \times 0} + C = 70\). By simplifying this, we find our constant \(C\) to be 10. With \(C\) known, the precise temperature function becomes \(T(t) = 60 e^{-0.2 t} + 10\).
This allows us to forecast the temperature at any future time \(t\). Initial conditions link the general integration solution back to the specific context or scenario you're interested in.
Other exercises in this chapter
Problem 48
Evaluate each definite integral. $$ \int_{2}^{4}\left(1+x^{-2}\right) d x $$
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Find each integral. [Hint: Try some algebra.] $$ \int(x+1)^{2} x^{3} d x $$
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Find the area bounded by the given curves. \(y=3 x^{2}-12 x\) and \(y=0\)
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Evaluate each definite integral. $$ \int_{1}^{2}\left(6 t^{2}-2 t^{-2}\right) d t $$
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