Problem 100 - Elementary and Intermediate Algebra

Question

After the contents of a package of JELL-O are combined with boiling water, the mixture is placed in a refrigerator whose temperature remains a constant \(42^{\circ} \mathrm{F}\). Estimate the number of hours \(t\) that it will take for the JELL-O to cool to \(50^{\circ} \mathrm{F}\) using the formula \(t=-\frac{1}{0.9} \ln \frac{50-T_{r}}{200-T_{r}}\) where \(T_{r}\) is the temperature of the refrigerator.

Step-by-Step Solution

Answer

It will take approximately 3.31 hours for the JELL-O to cool to 50°F.

Step 1: Understanding the Problem

We need to find the time it takes for the JELL-O mixture to cool from its initial temperature of 200°F to 50°F when placed in a refrigerator at a constant 42°F. We will use the given formula for time, which is dependent on the refrigerator temperature, to calculate this.

Step 2: Identify Given Information

The problem provides the refrigerator temperature, \( T_r = 42 \)°F. The initial JELL-O temperature is 200°F, and it needs to cool down to 50°F.

Step 3: Substitute Values in the Formula

We substitute \( T_r = 42 \) into the given formula: \[ t = -\frac{1}{0.9} \ln \frac{50 - 42}{200 - 42} \].

Step 4: Calculate the Temperature Differences

Calculate \( 50 - 42 = 8 \) and \( 200 - 42 = 158 \). So, the expression inside the logarithm becomes \( \frac{8}{158} \).

Step 5: Evaluate the Logarithm

First, calculate \( \frac{8}{158} \approx 0.05063 \). Then find the natural logarithm: \( \ln(0.05063) \approx -2.982 \).

Step 6: Calculate the Time \( t \)

Use \( t = -\frac{1}{0.9} \times (-2.982) \) to find the time. This becomes \( t \approx \frac{2.982}{0.9} \approx 3.313 \).

Step 7: Round the Result

Round 3.313 to a reasonable number of decimal places for practical use. Therefore, \( t \approx 3.31 \) hours.

Key Concepts

Exponential Decay Formula

When you hear the term "decay," it usually suggests reduction or decline. In mathematics, the exponential decay formula describes how quantities diminish over time. It's pivotal in cooling time calculations like in our JELL-O problem.

**Why Use Exponential Decay?**

• It's a great way to model processes where quantities decrease steadily but more rapidly at first.
• Certain physical processes like cooling, radioactive decay, or draining of a tank work under similar principles.

For the given JELL-O problem, the exponential decay equation is laid out as:\[t = -\frac{1}{k} \ln \left(\frac{T - T_r}{T_0 - T_r}\right)\]where:

• \(t\) = time for cooling
• \(T\) = final temperature
• \(T_0\) = initial temperature
• \(T_r\) = constant refrigerator temperature

This formula accounts for how fast or slow the temperature of an object approaches its surrounding temperature. Parameter \(k\) shows the decay rate characteristic to a specific cooling situation. As time progresses, the JELL-O cools, exponentially decreasing the difference between the JELL-O's temperature and that of the refrigerator.

Natural Logarithm in Algebra

The natural logarithm, denoted as \(\ln\), is a cornerstone of algebra, especially in problems involving exponential growth or decay. But what exactly is a natural logarithm? And why might it be critical in calculations like cooling time?

**What is a Natural Logarithm?**

• The natural logarithm \(\ln(x)\) answers the question: To what power must the base \(e\) (approximately 2.718) be raised to get the value \(x\)?
• It's central in calculus and algebra due to its properties with respect to exponential equations.

For cooling or decay problems, \(\ln\) is utilized to "undo" the exponential behavior. It helps simplify expressions that describe changing quantities. In the JELL-O scenario, calculating \(\ln(\frac{8}{158})\) allowed us to derive the cooling time.

By applying the natural logarithm, identifying timing in temperature descent scenarios becomes easier and practically solvable. It's your reliable tool whenever an exponential term is involved in algebraic manipulation.

Temperature Conversion

Temperature is measured in various units, but understanding how to convert between them is a fundamental skill. In cooling problems, knowing your way around Fahrenheit or Celsius systems may come in handy.

**Understanding the Fahrenheit Scale**

• In the United States, temperature measurements often rely on the Fahrenheit scale.
• On this scale, water freezes at 32°F and boils at 212°F, setting the context for this cooling problem.

The specific JELL-O exercise stays within Fahrenheit, so no conversion is needed in this scenario. Nonetheless, if you ever need to switch between Celsius and Fahrenheit, remember:

• Fahrenheit to Celsius: \(C = \frac{5}{9}(F - 32)\)
• Celsius to Fahrenheit: \(F = \frac{9}{5}C + 32\)

Temperature conversion is helpful when problems present data in different metric systems. Mastery over converting temperatures improves analytical flexibility across diverse scientific contexts, ensuring accuracy in interpreting cooling time results.