Problem 26 - Precalculus

Question

Use the following formula for Newton’s Law of Cooling: If you take a hot dinner out of the oven and place it on the kitchen countertop, the dinner cools until it reaches the temperature of the kitchen. Likewise, a glass of ice set on a table in a room eventually melts into a glass of water at that room temperature. The rate at which the hot dinner cools or the ice in the glass melts at any given time is proportional to the difference between its temperature and the temperature of its surroundings (in this case, the room). This is called Newton's law of cooling (or warming) and is modeled by $$T=T_{S}+\left(T_{0}-T_{S}\right) e^{-k t}$$ where \(T\) is the temperature of an object at time \(t, T_{s}\) is the temperature of the surrounding medium, \(T_{0}\) is the temperature of the object at time \(t=0, t\) is the time, and \(k\) is a constant. A cold drink is taken out of an ice chest with a temperature of \(38^{\circ} \mathrm{F}\) and placed on a picnic table with a surrounding temperature of \(75^{\circ} \mathrm{F}\). After 5 minutes, the temperature of the drink is \(45^{\circ} \mathrm{F}\). What will the temperature of the drink be 20 minutes after it is taken out of the chest?

Step-by-Step Solution

Answer

The temperature of the drink will be approximately 59.8°F after 20 minutes.

Step 1: Understand the Formula

The problem involves Newton's Law of Cooling, where the formula is given as \( T = T_{S} + (T_{0} - T_{S}) e^{-kt} \). Here, \( T \) is the temperature at time \( t \), \( T_{S} \) is the surrounding temperature, \( T_{0} \) is the initial temperature, and \( k \) is a constant related to the cooling rate.

Step 2: Identify Given Values

Given values from the problem: \( T_0 = 38^{\circ} \mathrm{F} \), \( T_S = 75^{\circ} \mathrm{F} \), at \( t = 5 \) minutes, \( T = 45^{\circ} \mathrm{F} \). The goal is to find \( T \) at \( t = 20 \) minutes.

Step 3: Find the Cooling Constant \( k \)

Use the formula \( 45 = 75 + (38 - 75) e^{-5k} \). Simplify to get \( -30 = -37e^{-5k} \), therefore \( e^{-5k} = \frac{30}{37} \). Take the natural logarithm: \( -5k = \ln \left(\frac{30}{37}\right) \). Thus, \( k = -\frac{1}{5} \ln \left(\frac{30}{37}\right) \).

Step 4: Substitute to Find \( T \) at 20 minutes

Substitute \( k \) and \( t = 20 \) into the formula: \( T = 75 + (38 - 75) e^{-k \times 20} \). Use the previously calculated \( k \) to simplify \( e^{-20k} = \left(\frac{30}{37}\right)^{4} \). Calculate \( T \) to find \( T = 75 - 37 \left(\frac{30}{37}\right)^4 \).

Step 5: Calculate Final Temperature

Plug the values into the expression: \( T \approx 75 - 37 \cdot \left(\frac{30}{37}\right)^4 \approx 59.77^{\circ} \mathrm{F} \).

Step 6: Conclusion

The temperature of the drink 20 minutes after it is taken out of the chest will be approximately \( 59.8^{\circ} \mathrm{F} \).

Key Concepts

Temperature Modeling

The concept of temperature modeling using Newton's Law of Cooling allows us to predict how the temperature of an object changes over time. This is particularly useful in everyday situations where objects are removed from environments with controlled temperatures, such as an oven or refrigerator, and placed in different surroundings.
When modeling temperature changes, we use the formula\[ T = T_{S} + (T_{0} - T_{S}) e^{-kt} \]In this equation:

• \(T\) is the temperature of the object at time \(t\).
• \(T_{S}\) is the constant temperature of the surrounding environment.
• \(T_{0}\) represents the initial temperature of the object.
• \(k\) is a constant that determines how quickly the object exchanges heat with its surroundings.

Temperature modeling helps us better understand and predict how conditions change, leading to applications in cooking, preserving food, scientific experiments, and even calibrating equipment.

Exponential Decay

Exponential decay is a fundamental mathematical concept that describes processes where quantities decrease at a rate proportional to their current value. In the context of Newton's Law of Cooling, the decay term \(e^{-kt}\) captures how the temperature difference between an object and its surroundings reduces over time.
This process is termed 'decay' because:

• The rate at which the temperature decreases is not constant but diminishes as the temperature of the object approaches that of its surroundings.
• The exponential function \(e^{-kt}\) is always positive but becomes smaller as \(t\) increases, indicating the slowing rate of temperature change.

Understanding exponential decay helps in appreciating why hot objects cool faster initially and then slow down as they approach ambient temperature. It's an elegant demonstration of how natural processes tend to stabilize over time, moving toward a state of equilibrium.

Rate of Change

The rate of change in Newton's Law of Cooling quantifies how quickly the temperature of an object changes over time. This rate is crucial for predicting the behavior of the cooling or warming process.
Key points about the rate of change include:

• It is directly linked to the parameter \(k\), the cooling constant, which dictates how fast the temperature approaches the ambient level.
• A larger \(k\) value means the object cools or warms more rapidly.
• The rate of change is not constant; it decreases as the temperature difference between the object and its surroundings decreases.

In practical terms, the rate of change helps us decide when food might reach an ideal temperature for serving, how quickly an ice pack might become less effective, or when equipment might reach operational temperature. By analyzing the rate of change, we gain insight into the dynamics of thermal systems and can make more informed predictions and decisions based on those insights.