Problem 41 - Thomas Calculus in SI Units

Question

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. a. How much longer would it take the soup to cool to \(35^{\circ} \mathrm{C} ?\) b. Instead of being left to stand in the room, the cup of \(90^{\circ} \mathrm{C}\) soup is put in a freezer whose temperature is \(-15^{\circ} \mathrm{C}\). How long will it take the soup to cool from \(90^{\circ} \mathrm{C}\) to \(35^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

a) It takes an additional 17.08 minutes. b) It takes 9.58 minutes in the freezer.

Step 1: Write the Newton's Law of Cooling Formula

Newton's Law of Cooling can be expressed as: \[ T(t) = T_s + (T_0 - T_s) e^{-kt} \]Where:- \( T(t) \) is the temperature of the object at time \( t \).- \( T_s \) is the surrounding temperature.- \( T_0 \) is the initial temperature of the object.- \( k \) is the cooling constant.- \( t \) is the time.

Step 2: Determine the Cooling Constant (k)

We can use the given data to determine \( k \). When \( t = 10 \) minutes, \( T(10) = 60^{\circ} \), \( T_0 = 90^{\circ} \), and \( T_s = 20^{\circ} \). Plug these into the formula:\[ 60 = 20 + (90 - 20) e^{-10k} \]\[ 40 = 70 e^{-10k} \]\[ \frac{4}{7} = e^{-10k} \]Taking the natural logarithm of both sides gives: \[ -10k = \ln\left(\frac{4}{7}\right) \]Thus, \[ k = -\frac{1}{10} \ln\left(\frac{4}{7}\right) \]

Step 3: Calculate Time to Cool to 35°C in Room

To find the time taken to cool to \( 35^{\circ} \) in the room:\[ 35 = 20 + (90 - 20) e^{-kt_{room}} \]\[ 15 = 70 e^{-kt_{room}} \]\[ \frac{15}{70} = e^{-kt_{room}} \]Taking the log,\[ t_{room} = -\frac{1}{k} \ln\left(\frac{15}{70}\right) \]Substituting \( k \): \[ t_{room} = \frac{10}{\ln(4/7)} \cdot \ln(15/70) \]Calculate this with the value of \( k \) from Step 2.

Step 4: Compute Additional Time Beyond First 10 Minutes

We know it takes 10 minutes to reach \( 60^{\circ} \). Calculate the additional time \( t_{add} \) required to go from \( 60^{\circ} \) to \( 35^{\circ} \):\[ t_{add} = t_{room} - 10 \]

Step 5: Calculate Time in Freezer to 35°C

Now consider the soup placed in a freezer with \( T_s = -15^{\circ} \) initially at \( 90^{\circ} \):\[ 35 = -15 + (90 + 15) e^{-k_{freezer} t} \]\[ 50 = 105 e^{-kt} \]\[ \frac{50}{105} = e^{-kt} \]So,\[ t = -\frac{1}{k} \ln \left( \frac{50}{105} \right) \]Use the value of \( k \) found from Step 2.

Key Concepts

Cooling Constant

A crucial concept in Newton's Law of Cooling is the cooling constant, often represented by the symbol 'k'. This constant characterizes how quickly an object cools down in a given environment. The cooling constant is unique to each scenario and depends on the properties of the object being cooled and the surrounding environment. To find the cooling constant, we use the formula:

• When you monitor an object's temperature change over a known time, you can solve for 'k'.
• In our soup example, we calculated that 'k' by knowing the soup's temperature dropped from \(90^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) in 10 minutes with a room temperature of \(20^{\circ} \mathrm{C}\). This particular setup gave us the value for 'k' using mathematical manipulation involving logarithms.
• The formula becomes: \(-10k = \ln\left(\frac{4}{7}\right)\). Solving this gives you 'k', which can then be used in further calculations to predict future cooling times.

Understanding 'k' helps in modeling the cooling process accurately for any further predictions about the object's temperature behavior.

Temperature Change

Temperature change is the core outcome we observe when applying Newton's Law of Cooling. In the context of our soup example, we look at how the soup cools from its initial temperature to a lower one over time. The formula we use for temperature change is:

• \(T(t) = T_s + (T_0 - T_s) e^{-kt}\)
• Here, \(T(t)\) is the temperature at time \(t\), \(T_s\) is the surrounding temperature, and \(T_0\) is the initial temperature of the object.
• Temperature change is strongly influenced by both the cooling constant and the surrounding environment's temperature. For instance, the soup's temperature decrease from \(90^{\circ}\) to \(60^{\circ}\) within a specific timeframe offers a basis to calculate 'k' and predict future temperature changes.

This concept of change helps in understanding the dynamics of temperature over time and allows us to make predictions about future states based on current observations.

Rate of Cooling

The rate of cooling refers to how quickly the temperature of an object decreases over time. This rate is crucial because it helps us understand how long it takes for an object to cool down to a desired temperature.

• The rate of cooling is determined by both the cooling constant 'k' and the temperature difference between the object and the surrounding environment.
• For the soup, the rate at which it cooled from \(90^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C}\) gives us a practical example of how fast temperature can fall in a specific environment.
• In scenarios like placing the soup in a freezer with a different surrounding temperature, \(T_s = -15^{\circ}\), we expect the cooling rate to change, as the environmental temperature has a substantial impact on the rate.

Understanding this rate is vital for applications where temperature control is crucial, such as food safety or chemical reactions.

Mathematical Modeling

Mathematical modeling is a powerful tool that uses mathematical expressions to represent real-world phenomena. Newton's Law of Cooling is a classic example of such a model, as it provides a structured way to understand and predict temperature changes over time.

• By employing the formula \( T(t) = T_s + (T_0 - T_s) e^{-kt} \), we create a model that represents the cooling process mathematically.
• This modeling approach allows us to calculate unknowns, such as how much longer it will take for our soup to cool further or how fast it will cool in a different environment like a freezer.
• In educational settings, mathematical modeling helps students grasp complex physical processes through simplified equations and can be applied across diverse fields beyond cooling, including finance and population dynamics.

Such models become a bridge between theoretical learning and practical application in real-world scenarios.