Problem 416 - Precalculus

Question

For the following exercises, use this scenario: A turkey is taken out of the oven with an internal temperature of \(165^{\circ}\) o Fahrenheit and is allowed to cool in a \(75^{\circ} \mathrm{F}\) room. After half an hour, the internal temperature of the turkey is \(145^{\circ} \mathrm{F}\) . To the nearest degree, what will the temperature be after 50 minutes?

Step-by-Step Solution

Answer

The temperature will be approximately 145°F after 50 minutes.

Step 1: Write Down Newton's Law of Cooling

Newton's Law of Cooling is given by the formula \( T(t) = T_s + (T_0 - T_s)e^{-kt} \), where \( T(t) \) is the temperature of the object at time \( t \), \( T_s \) is the surrounding temperature, and \( T_0 \) is the initial temperature of the object.

Step 2: Insert Known Values

We know that \( T_0 = 165^{\circ}F \), \( T_s = 75^{\circ}F \), and \( T(0.5) = 145^{\circ}F \) after half an hour.

Step 3: Calculate the Cooling Constant, k

Use \( T(0.5) = 145^{\circ}F \) in the formula: \[145 = 75 + (165 - 75)e^{-0.5k}. \]Simplifying, we have:\[145 = 75 + 90e^{-0.5k}. \]Solve for \( e^{-0.5k} \):\[70 = 90e^{-0.5k}, \e^{-0.5k} = \frac{70}{90} = \frac{7}{9}.\]Finally,\[k = -2\ln\left(\frac{7}{9}\right). \]

Step 4: Determine Temperature After 50 Minutes

We want \( T(t) \) at \( t = \frac{50}{60} = \frac{5}{6} \) hours: \[T\left(\frac{5}{6}\right) = 75 + 90e^{-k(\frac{5}{6})} \].Use the value of \( k \) calculated: \[ e^{-k\left(\frac{5}{6}\right)} = \left(\frac{7}{9}\right)^{-\frac{5}{6}/-\ln\left(\frac{7}{9}\right)}\approx0.7843. \]Now, compute the final temperature:\[T\left(\frac{5}{6}\right) \approx 75 + 90 \times 0.7843 = 145.2^{\circ}F.\]Rounding gives \( 145^{\circ}F \).

Key Concepts

Exponential Decay

Exponential decay is a process where a quantity diminishes at a rate proportional to its current value. In the context of cooling processes like that of a turkey cooling in a room, the temperature reduces over time in a non-linear fashion.

• Every physical quantity that decays over time can often be modeled using the exponential decay equation.
• The key feature of exponential decay is that the rate of decrease is fastest initially and slows down as time progresses.

In the situation with the turkey, the temperature decrease doesn't happen evenly over each minute. Instead, more significant changes occur immediately after being removed from the oven, while changes slow down as equilibrium with room temperature approaches.
Understanding exponential decay is crucial for predicting how different processes unfold over time, whether it's cooling, radioactive decay, or even financial trends in depreciation.

Temperature Modeling

Temperature modeling involves predicting how the temperature of an object changes over time under certain conditions. This can involve heating or cooling, and it relies heavily on understanding the relationship between the object and its environment.

Newton's Law of Cooling is a mathematical representation used for this purpose. The model helps in accurately forecasting temperatures by considering several factors:

• The initial temperature of the object ( \( T_0 \)), which is the temperature when monitoring begins. For the turkey, this was \( 165^{\circ}\, \mathrm{F} \).
• The surrounding temperature ( \( T_s \)), which influences how quickly or slowly an object cools or heats. The room temperature for our example is \( 75^{\circ}\, \mathrm{F} \).

By plugging these values into the formula, you can model how the turkey’s temperature changes over time until it matches the room temperature. Modeling temperature changes helps in practical applications such as cooking guidelines, preserving substances, and managing climatic systems.

Cooling Constant

The cooling constant, denoted by \( k \), is a critical parameter in Newton's Law of Cooling. It represents the rate at which an object exchanges heat with its surroundings. Basically, it shows how fast or slow the temperature change happens for a particular object:

• A higher \( k \) value indicates a faster cooling process.
• A lower \( k \) value suggests slower temperature changes.

To calculate \( k \), you need the intermediate temperature data, which in our scenario was \( 145^{\circ} \mathrm{F} \) after half an hour. This intermediate value allows you to solve for \( k \) using the temperature formula:
\[e^{-0.5k} = \frac{70}{90}\]
This calculation helps determine how quickly the turkey will reach room temperature. By linking study results to real-world numbers, you get meaningful insights into the efficiency of the cooling process, which can then be applied to optimize conditions.