Problem 41 - Thomas Calculus

Question

Cooling soup Suppose that a cup of soup cooled from \(90^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) after 10 \(\mathrm{min}\) in a room whose temperature was \(20^{\circ} \mathrm{C}\) . Use Newton's Law of Cooling to answer the following questions. \begin{equation} \begin{array}{l}{\text { a. How much longer would it take the soup to cool to } 35^{\circ} \mathrm{C} \text { ? }} \\ {\text { b. Instead of being left to stand in the room, the cup of } 90^{\circ} \mathrm{C} \text { . }} \\ {\text { soup is put in a freezer whose temperature is }-15^{\circ} \mathrm{C} \text { . How }} \\ {\text { long will it take the soup to cool from } 90^{\circ} \mathrm{C} \text { to } 35^{\circ} \mathrm{C} \text { ? }}\end{array} \end{equation}

Step-by-Step Solution

Answer

a) Approximately 28 more minutes. b) About 19 minutes in total to cool.

Step 1: Understanding Newton's Law of Cooling

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature. The formula is given by \( T(t) = T_{ ext{ambient}} + (T_0 - T_{ ext{ambient}}) e^{-kt} \), where \( T(t) \) is the temperature at time \( t \), \( T_0 \) is the initial temperature, \( T_{ ext{ambient}} \) is the ambient temperature, and \( k \) is the cooling constant.

Step 2: Determine the cooling constant (k) for the room

We know the initial temperature \( T_0 = 90^{\circ}C \), the ambient temperature \( T_{ ext{ambient}} = 20^{\circ}C \), and the temperature at 10 minutes \( T(10) = 60^{\circ}C \). Plug these into the formula: \( 60 = 20 + (90 - 20)e^{-10k} \). Solving this equation for \( k \), we have:\( 40 = 70e^{-10k} \) \( e^{-10k} = \frac{4}{7} \) \( -10k = \ln(\frac{4}{7}) \) \( k = -\frac{1}{10}\ln(\frac{4}{7}) \).

Step 3: Calculate time to cool to 35°C in the room

We will again use the formula \( T(t) = T_{ ext{ambient}} + (T_0 - T_{ ext{ambient}}) e^{-kt} \). Substitute: \( 35 = 20 + 70e^{-kt} \). This changes to \( 15 = 70e^{-kt} \), so \( e^{-kt} = \frac{3}{14} \). Solving this equation for \( t \), \( -kt = \ln(\frac{3}{14}) \) Substitute \( k \) from Step 2: \( \frac{1}{10}\ln(\frac{4}{7}) \cdot t = \ln(\frac{3}{14}) \) Solving for \( t \), \( t = 10 \times \frac{\ln(\frac{3}{14})}{\ln(\frac{4}{7})} \). Compute to find \( t \).

Step 4: Find the time in the freezer with ambient temperature -15°C

Now we have \( T_{ ext{ambient}} = -15^{\circ}C \). Use the cooling constant \( k \) calculated for the room with \( T_0 = 90^{\circ}C \) and solve: \( 35 = -15 + (90 + 15)e^{-kt} \). It simplifies to: \( 50 = 105e^{-kt} \) \( e^{-kt} = \frac{10}{21} \). Substitute \( k \) from Step 2: \( -kt = \ln(\frac{10}{21}) \) \( t = \frac{\ln(\frac{10}{21})}{k} \). Compute \( t \) using the known value of \( k \).

Key Concepts

Cooling Constant

The cooling constant, often denoted as "k," is pivotal in understanding Newton's Law of Cooling. It quantifies how quickly the temperature of an object approaches the ambient temperature. If you've ever noticed that hot soup left on a kitchen counter cools more rapidly at first and then slows, you're witnessing the cooling constant in action.
Newton's Law of Cooling predicts that the rate at which an object loses heat is proportional to the difference between its temperature and the ambient temperature. Mathematically, we express it as:

• The temperature at time \( t \) is given by \( T(t) = T_{\text{ambient}} + (T_0 - T_{\text{ambient}}) e^{-kt} \).

In this formula:

• \( T(t) \) is the temperature at time \( t \).
• \( T_0 \) is the initial temperature.
• \( T_{\text{ambient}} \) is the ambient temperature.
• \( e^{-kt} \) is the exponential decay representing the cooling process.

Solving for \( k \) typically requires knowledge of the initial conditions and a temperature measurement taken at a known time. Once determined, \( k \) allows us to predict how long it will take for an object to reach various temperatures under stable conditions.

Ambient Temperature

In the context of Newton's Law of Cooling, the ambient temperature \( T_{\text{ambient}} \) is the constant temperature of the surrounding environment where the cooling process takes place. It serves as a baseline or reference point for the cooling equation.
Ambient temperature plays a crucial role because:

• It dictates the asymptotic limit that the object's temperature will approach over time.
• When the ambient temperature varies (for instance, moving a cup of soup from a room to a freezer), the time taken for the object to cool to a desired temperature changes significantly.

For example, in the exercise, the soup starts in a room at \( 20^{\circ} \text{C} \), and we later consider it in a freezer at \( -15^{\circ} \text{C} \). The colder ambient temperature of the freezer would naturally speed up the cooling process compared to the warmer room. Therefore, understanding and quantifying \( T_{\text{ambient}} \) is key to predicting cooling behavior accurately.

Temperature Rate of Change

The temperature rate of change is a fundamental aspect of Newton's Law of Cooling. It refers to how swiftly or slowly an object's temperature changes as it equilibrates with its surroundings. This is mathematically represented by the equation:

• \( \frac{dT}{dt} = -k(T - T_{\text{ambient}}) \)

Here, \( \frac{dT}{dt} \) represents the rate of change of temperature with respect to time. The negative sign indicates that the temperature of the object decreases over time as it approaches the ambient temperature.
Key factors affecting the temperature rate of change include:

• The difference between the object's temperature and the ambient temperature: The greater this difference, the faster the object cools initially.
• The cooling constant \( k \): A larger \( k \) means a quicker temperature change, while a smaller \( k \) indicates a slower change.

Understanding this concept allows us to predict when an object will reach a specific temperature and to tailor conditions if we wish to accelerate or decelerate the cooling process.