Problem 34 - Precalculus with Calculus Previews

Question

The rate at which a body cools also depends on its exposed surface area \(S\). If \(S\) is a constant, then a modification of (4) is $$ T(t)=T_{m}+\left(T_{0}-T_{m}\right) e^{k S t}, \quad k<0 $$ Suppose two cups \(A\) and \(B\) are filled with coffee at the same time. Initially the temperature of the coffee is \(150^{\circ} \mathrm{F}\). The exposed surface area of the coffee in \(\operatorname{cup} B\) is twice the surface area of the coffee in cup \(A\). After 30 min, the temperature of the coffee in \(\operatorname{cup} A\) is \(100^{\circ} \mathrm{F}\). If \(T_{m}=70^{\circ} \mathrm{F}\), what is the temperature of the coffee in \(\operatorname{cup} B\) after 30 \(\min ?\)

Step-by-Step Solution

Answer

The temperature of cup B after 30 minutes is 81.25°F.

Step 1: Write the cooling formula for cup A

The cooling formula is given by \( T(t) = T_{m} + (T_{0} - T_{m}) e^{k S t} \). For cup A, with \( T_m = 70^{\circ} \text{F} \), \( T_0 = 150^{\circ} \text{F} \), and \( T(t=30) = 100^{\circ} \text{F} \), we substitute these values to find \( kS \) as follows:\[ 100 = 70 + (150 - 70) e^{kS \cdot 30} \]This simplifies to:\[ 30 = 80 e^{30kS} \]\[ e^{30kS} = \frac{3}{8} \]

Step 2: Solve for \(kS\) for cup A

Taking the natural logarithm of both sides, we get:\[ 30kS = \ln\left(\frac{3}{8}\right) \]\[ kS = \frac{1}{30} \ln\left(\frac{3}{8}\right) \]

Step 3: Apply the surface area ratio for cup B

Given that the surface area of cup B is twice that of cup A, the expression for \( k \cdot S_B \) becomes:\[ k \cdot S_B = 2kS_A \]

Step 4: Write the cooling formula for cup B

For cup B, the formula becomes:\[ T_B(t) = 70 + (150 - 70) e^{2kS_A \cdot 30} \]Substitute \( kS_A \) from Step 2:\[ T_B(30) = 70 + 80 \left(\frac{3}{8}\right)^2 \]

Step 5: Calculate the temperature of cup B

Simplify the expression from Step 4:\[ T_B(30) = 70 + 80 \cdot \left(\frac{9}{64}\right) \]\[ T_B(30) = 70 + 11.25 \]\[ T_B(30) = 81.25 \] F

Key Concepts

Exponential Decay

In the context of thermodynamics and temperature change, exponential decay describes how the temperature of an object decreases over time. It follows a predictable pattern where the rate of cooling slows down as the object approaches the temperature of its surroundings. This pattern is captured in the formula:

• \( T(t) = T_{m} + (T_{0} - T_{m}) e^{kSt} \)

Here, \( T(t) \) is the temperature at time \( t \), \( T_{m} \) is the ambient temperature, \( T_{0} \) is the initial temperature, and \( k \) is the cooling constant which is typically negative. The term \( e^{kSt} \) represents the exponential decay factor, which ensures that the temperature decreases as time progresses.
To put it simply, this equation models how much and how quickly an object's temperature will drop, decaying towards the ambient temperature at an exponential rate. The larger the absolute value of \( k \), the faster the object cools. It's useful for predicting how liquids like coffee lose heat over time.

Surface Area Effect

Surface area plays a crucial role in how quickly an object cools. According to Newton's Law of Cooling, a greater surface area allows more heat to escape, speeding up the cooling process. This principle is particularly relevant when comparing different scenarios with varying surfaces exposed to the air.
In the provided coffee cooling problem, two cups of coffee have different surface areas exposed to the air. Cup B, having twice the surface area of cup A, cools faster due to the larger surface area. This is mathematically represented by the surface area term \( S \) in the cooling formula \( T(t) = T_{m} + (T_{0} - T_{m}) e^{kSt} \). Essentially, increasing \( S \) increases the rate at which heat is lost.
The practical consequence of the surface area effect is noticeable in everyday life. For example, spilling a cup of hot coffee on a wide plate will cool it faster than if left in a tall mug, simply because more of the coffee's surface is exposed to air.

Coffee Cooling Problem

The coffee cooling problem gives us a real-world application of Newton's Law of Cooling, showcasing how changes in conditions impact results. In this scenario, two coffee cups cool down over the same time period, with different surface areas affecting their cooling rates.
To find the final temperature of the coffee in cup B, we begin by determining how cup A cools over time. By using the given formula \( T(t) = T_{m} + (T_{0} - T_{m}) e^{kSt} \) and the known temperatures after 30 minutes, we can calculate \( k \) for cup A. Knowing that cup B's surface area is double, we adjust the equation to reflect the doubled effect on cooling.
This setup not only highlights the key concepts of cooling but also teaches how adjustments in physical properties can impact the outcome. It reflects the nature of exponential decay and the importance of surface area in real-life physics applications, beyond just academic exercises. This makes it a solid example for understanding the practical implications of theoretical physics laws.