Question

An object is taken from an oven at \(300^{\circ} \mathrm{F}\) and left to cool in a room at \(75^{\circ} \mathrm{F}\). If the temperature fell to \(200^{\circ} \mathrm{F}\) in \(\frac{1}{2}\) hour, what will it be after 3 hours?

Short answer

The temperature after 3 hours will be approximately \(82.96^{\circ}\text{F}\).

Step-by-step solution

Understand Newton's Law of Cooling

Newton's Law of Cooling states that the rate at which an object cools is proportional to the difference in temperature between the object and its surrounding environment. The formula is given by: \[ T(t) = T_s + (T_0 - T_s) e^{-kt} \] where \( T(t) \) is the temperature at time \( t \), \( T_s \) is the surrounding temperature, \( T_0 \) is the initial temperature, and \( k \) is the cooling constant.

Identify Given Values

We are given: - Initial temperature, \( T_0 = 300^{\circ} \text{F} \)- Surrounding temperature, \( T_s = 75^{\circ} \text{F} \)- Temperature at \( t = \frac{1}{2} \text{ hour} \), \( T(0.5) = 200^{\circ} \text{F} \)- We need to find \( T(3) \).

Solve for Cooling Constant \( k \)

Using the known data point, \( T(0.5) = 200^{\circ} \text{F} \), substitute the given temperatures into Newton's Law of Cooling.\[ 200 = 75 + (300 - 75)e^{-k(0.5)} \] Solving for \( k \) helps us understand how quickly the temperature is changing.

Simplification and Calculation

Start solving for \( k \):\[ 200 - 75 = 225 e^{-0.5k} \]\[ 125 = 225 e^{-0.5k} \]\[ \frac{125}{225} = e^{-0.5k} \]\[ \ln\left(\frac{125}{225}\right) = -0.5k \]\[ k = -2 \ln\left(\frac{125}{225}\right) \]

Calculate Future Temperature

Now use the value of \( k \) to find the temperature after 3 hours:\[ T(3) = 75 + 225 e^{-3k} \] Substitute \( k \) into the equation and solve for \( T(3) \).

Final Calculation

After substituting the value of \( k \) and calculating using a calculator, compute the temperature:\[ T(3) = 75 + 225 \times \left(\frac{125}{225}\right)^{6} \]\[ T(3) \approx 82.96^{\circ} \text{F} \]

Key concepts

Cooling Constant

The cooling constant, often represented as \( k \) in Newton's Law of Cooling, is crucial in determining the rate at which an object cools down. It quantifies how quickly the temperature of an object changes towards the surrounding temperature.
For instance, when an object is removed from a hot oven and placed in a cooler room, it doesn't just lose heat immediately. Instead, it cools down gradually, following an exponential pattern. This is where the cooling constant comes into play.
In our example, we found \( k \) using the equation \[ \ln\left(\frac{125}{225}\right) = -0.5k \]. By solving, we determine \( k = -2 \ln\left(\frac{125}{225}\right) \).
Understanding the cooling constant is essential for predicting how long it will take for an object to reach a particular temperature. It provides a numerical representation of the cooling efficiency of a particular situation.

• Decrease in \( k \): Slower cooling.
• Increase in \( k \): Faster cooling.

Temperature Change

Temperature change, in this context, refers to how the temperature of the object declines over time. One of the key considerations when applying Newton's Law of Cooling is the importance of initial conditions and surroundings.
In the exercise, the initial temperature of the object was at \( 300^{\circ}\text{F} \), and it cooled to \( 200^{\circ}\text{F} \) over half an hour in a room with a steady temperature of \( 75^{\circ}\text{F} \).
To determine the temperature at any given time, the temperature change is calculated using the formula:

• \[ T(t) = T_s + (T_0 - T_s) e^{-kt} \]

This indicates how the object's temperature approaches the surrounding temperature, slowing down as the difference in temperatures decreases.
Knowing how fast a temperature changes over time allows us to practically apply this knowledge in everyday situations such as cooking, engineering, and environmental science.

Exponential Decay

Exponential decay is a fundamental concept in Newton's Law of Cooling. This concept describes how the temperature of an object decreases at a rate proportional to its temperature difference from the surroundings.
The equation \[ T(t) = T_s + (T_0 - T_s) e^{-kt} \] showcases exponential decay clearly. Here, the term \( e^{-kt} \) represents the exponential decay factor, dynamically changing over time.
As time progresses, the value of \( e^{-kt} \) shrinks, leading to a steady decline in the temperature's difference from its surroundings. This is what "decay" refers to — a diminishing process rather than a linear drop. Let's break it down further:

• Early stages: The object cools quickly because there's a large temperature difference initially.
• Later stages: The cooling rate slows down as the object's temperature nears the surrounding temperature.

Grasping exponential decay is pivotal for predicting thermal behavior in objects transitioning from hotter to cooler environments effectively.