Problem 42 - College Algebra

Question

Use this scenario: A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 69° F room. After fifteen minutes, the internal temperature of the soup was 95° F. To the nearest degree, what will the temperature be after 2 and a half hours?

Step-by-Step Solution

Answer

The temperature will be approximately 75°F after 2.5 hours.

Step 1: Understand the Situation

We have a pot of soup cooling down from an initial temperature of 100°F in a room with a constant temperature of 69°F. Using Newton's Law of Cooling, we can predict the temperature over time. This states that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature.

Step 2: Set Up Newton's Law of Cooling

Newton's Law of Cooling is given by the formula \( T(t) = T_a + (T_0 - T_a) e^{-kt} \), where \( T(t) \) is the temperature at time \( t \), \( T_a \) is the ambient temperature, \( T_0 \) is the initial temperature, and \( k \) is a cooling constant. We need to determine \( k \) using the information given after 15 minutes.

Step 3: Solve for the Cooling Constant 'k'

Plug in the values at 15 minutes into the formula: \( 95 = 69 + (100 - 69) e^{-15k} \). This simplifies to \( 26 = 31 e^{-15k} \). Solve this equation by dividing both sides by 31 and taking the natural logarithm to isolate \( k \):\[ e^{-15k} = \frac{26}{31} \]\[ -15k = \ln\left(\frac{26}{31}\right) \]\[ k = -\frac{1}{15} \ln\left(\frac{26}{31}\right) \]

Step 4: Calculate 'k'

Compute \( k \) using the natural logarithm:\[ k = -\frac{1}{15} \ln\left(\frac{26}{31}\right) \approx 0.011 \]

Step 5: Calculate Temperature after 2.5 hours

Use the formula again with \( t = 150 \) minutes (or 2.5 hours):\[ T(150) = 69 + (100 - 69) e^{-150 \times 0.011} \]Calculate the exponential:\[ T(150) = 69 + 31 e^{-1.65} \]\[ T(150) = 69 + 31 \times 0.192 \approx 69 + 5.95 \approx 75 \]

Step 6: Interpret the Result

Round the result to the nearest degree. The temperature of the soup after two and a half hours will be approximately 75°F.

Key Concepts

Understanding Exponential Decay

Exponential decay is a fundamental concept in modeling how quantities diminish over time. Imagine you have a hot cup of soup left in a cooler room. Over time, its temperature gradually decreases. This phenomenon is characterized by exponential decay, where the rate of cooling slows down as the temperature difference between the soup and room lessens.

In the context of Newton's Law of Cooling, this decay is not random but follows a predictable pattern described by an exponential function. This function utilizes a time factor in its exponent, which is why time plays a crucial role in how fast or slow the soup cools to the ambient temperature. The key takeaway is that as more time passes, the change in temperature becomes smaller, highlighting the essence of exponential decay.

Decoding the Cooling Constant

The cooling constant, represented by the symbol \( k \), is a crucial element in the formula used in Newton's Law of Cooling. This constant helps determine how fast or slow an object will cool down to match the surrounding temperature.

The process to find the cooling constant involves using the initial conditions provided, such as initial and measured temperatures at a specific time. By substituting these into the formula \( T(t) = T_a + (T_0 - T_a) e^{-kt} \), we solve for \( k \).

• If \( k \) is large, the object cools quickly.
• If \( k \) is small, the cooling process is slower.

For example, in the problem, we calculated \( k \) to be approximately 0.011, indicating a moderate rate of cooling.

Predicting Temperature Change

Temperature change prediction is all about using mathematical models to foresee how the temperature of an object will evolve over time. It's like having a crystal ball for thermodynamics.

By using Newton's Law of Cooling, we predict how the temperature of the soup will change. The formula \( T(t) = T_a + (T_0 - T_a) e^{-kt} \) allows us to input the amount of time that has passed and compute the future temperature.

For instance, when the task was to predict the soup's temperature after 2.5 hours (or 150 minutes), we substituted \( t = 150 \) into the formula, using our previously calculated \( k \). This precise prediction method shows the temperature would be approximately 75°F.

Such predictions are not just limited to soup; they are vital in fields like meteorology, cooking, and even industrial processes where temperature control is crucial.