Problem 25 - 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. An apple pie is taken out of the oven with an internal temperature of \(325^{\circ} \mathrm{F}\). It is placed on a rack in a room with a temperature of \(72^{\circ} \mathrm{F}\). After 10 minutes, the temperature of the pie is \(200^{\circ} \mathrm{F}\). What will the temperature of the pie be 30 minutes after coming out of the oven?

Step-by-Step Solution

Answer

The pie will be approximately \(100.5^{\circ}F\) after 30 minutes.

Step 1: Identify the Given Values

We need to identify the given values to use in the formula.- The oven pizza temperature is initially, \(T_0 = 325^{\circ}F\).- The room temperature (surrounding temperature), \(T_S = 72^{\circ}F\).- The temperature of the pie after 10 minutes is \(T = 200^{\circ}F\), when \(t = 10\) minutes.

Step 2: Apply Initial Values to the Formula

We apply the known values to solve for the cooling constant \(k\) using the formula:\[ T = T_{S} + (T_{0} - T_{S}) e^{-k t} \]Substitute the values after 10 minutes:\[ 200 = 72 + (325 - 72) e^{-10k} \]

Step 3: Simplify and Solve for k

Subtract 72 from both sides:\[ 128 = 253 e^{-10k} \]Now, divide by 253:\[ \frac{128}{253} = e^{-10k} \]Take the natural logarithm of both sides to solve for \(-10k\):\[ \ln\left( \frac{128}{253} \right) = -10k \]Solve for \(k\):\[ k = -\frac{1}{10} \ln\left( \frac{128}{253} \right) \approx 0.0729 \]

Step 4: Calculate Temperature After 30 Minutes

With \(k\) known, calculate the temperature after 30 minutes:\[ T = 72 + (325 - 72) e^{-0.0729 \times 30} \]Calculate:\[ T = 72 + 253 e^{-2.187} \]Evaluating \(e^{-2.187}\):\[ T = 72 + 253 \times 0.1125 \]So, \[ T \approx 72 + 28.5 \approx 100.5^{\circ}F \]

Key Concepts

Exponential Decay

Newton's Law of Cooling is an excellent example of exponential decay in action. Exponential decay describes how a quantity decreases rapidly at first, then levels off over time. In this scenario, the temperature difference between an object and its surroundings decreases exponentially. This means that a hot object doesn't cool at a constant rate; instead, the rate of cooling reduces as it approaches the ambient temperature.
Newton's formula for cooling is expressed as: \[ T(t) = T_S + (T_0 - T_S) e^{-kt} \]

• \(T(t)\) is the temperature at time \(t\)
• \(T_S\) is the surrounding temperature
• \(T_0\) is the initial temperature at time zero
• \(k\) is a constant that determines the rate of cooling

As \(t\) increases, the exponential term \(e^{-kt}\) decreases towards zero, making the temperature \(T(t)\) tend towards \(T_S\). This phenemenon shows how strong the pull is towards balance or equilibrium in nature. Understanding exponential decay aids in predicting cooling patterns in biology, physics, and engineering.

Temperature Differentiation

Temperature differentiation refers to the rate at which an object's temperature differs from its surroundings. In Newton’s Law of Cooling, this differentiation is vital because it influences the speed at which an object cools.
Initially, when there is a significant temperature difference between the object and its environment, the cooling process is rapid. As the temperatures begin to converge, the rate of cooling slows down.
How this works in practice:

• A larger difference means a faster initial cooling rate. For example, a pie coming out of a hot oven will cool very quickly at first when placed in a much cooler room.
• A smaller temperature difference results in a slower cooling rate. As the pie's temperature approaches room temperature, the rate at which it loses heat slows down significantly.

This temperature differentiation is why a pie might start hot at 325°F and reduce quickly to 200°F, but take much longer to cool from 200°F to near room temperature. Monitoring this change helps identify cooling behaviors and can guide in efficient energy use control and prediction of time before the object becomes cool enough to handle.

Mathematical Modeling

Mathematical modeling in physics often involves creating equations or formulas that closely represent a real-world process. Newton's Law of Cooling is a prime example. This equation models how the temperature of an object changes over time within a different temperature environment.
Through this modeling:

• It helps in predicting future states of the system, allowing strategic planning. For instance, chefs can determine how long it will take for a hot dish to become cool enough to preserve or serve.
• Enables simulation of alterations - if the environmental temperature or object dimensions change, the model anticipates the impact on cooling stages.
• Supports testing hypotheses about thermal conductivity properties in materials for physics and engineering.

Ensures accuracy in predicting outcomes helps in understanding concepts deeply, allowing the refinement of theories. Mathematical models are simplified versions of reality and they depend on certain assumptions, such as a consistent surrounding temperature. Discrepancies can occur if those assumptions vary. Yet, mathematical modeling stands as a cornerstone in solving practical problems through theoretical reasoning.