Problem 20 - Prealgebra

Question

We have mentioned the two temperature scales, Fahrenheit and Celsius. Table 1 is intended to give you a more intuitive idea of the relationship between the two temperatures scales. (TABLE CANNOT COPY) Table 2 gives the formulas, in both symbols and words, that are used to convert between the two scales. (TABLE CANNOT COPY) Let \(C=100\) in the formula \(F=\frac{9}{5} C+32,\) and solve for \(F\) Does the value of \(F\) agree with the information in Table \(1 ?\)

Step-by-Step Solution

Answer

When \( C = 100 \), \( F = 212 \), which agrees with Table 1.

Step 1: Understand the formula

The formula given is \( F = \frac{9}{5}C + 32 \), where \( F \) is the temperature in Fahrenheit and \( C \) is the temperature in Celsius. Our task is to find \( F \) when \( C = 100 \).

Step 2: Substitute the value of C into the formula

Substitute \( C = 100 \) into the formula. This gives: \[ F = \frac{9}{5} \times 100 + 32 \]

Step 3: Calculate the product

Calculate \( \frac{9}{5} \times 100 \). First, calculate \( \frac{9}{5} = 1.8 \). Then multiply \( 1.8 \times 100 \) to get 180.

Step 4: Add to the constant in the formula

Add 32 to the result obtained in Step 3. Thus, \( F = 180 + 32 = 212 \).

Step 5: Verify with Table 1

Assuming Table 1 provides the equivalent Fahrenheit temperature for 100 Celsius as 212, the calculated value matches perfectly with Table 1. Therefore, the value of \( F = 212 \) agrees with the information in the table.

Key Concepts

Fahrenheit to Celsius

Converting temperatures from Fahrenheit to Celsius is a common task in science and daily life. The formula for converting Fahrenheit (F) to Celsius (C) is simple and relies on basic arithmetic. The formula is: \[ C = \frac{5}{9} (F - 32) \] Here's how to use it:

• First, subtract 32 from the Fahrenheit temperature.
• Then multiply the result by \(\frac{5}{9}\).

For instance, if we have 212°F and we want to convert it to Celsius:

• First, subtract 32: \(212 - 32 = 180\).
• Next, multiply by \(\frac{5}{9}\): \(180 \times \frac{5}{9} = 100\).

The answer is 100°C. This conversion is vital for people traveling between countries using different temperature metrics, chefs using recipes from abroad, and scientists conducting experiments under international collaborations.

Celsius to Fahrenheit

On the flip side, converting Celsius to Fahrenheit is equally straightforward. The formula is designed to adjust from the metric system to the imperial system. You would use the following formula to make the conversion: \[ F = \frac{9}{5} C + 32 \] To convert from Celsius to Fahrenheit:

• Multiply the Celsius temperature by \(\frac{9}{5}\).
• Add 32 to the result.

Let's work through an example. If the temperature is 100°C and you want to find out what it is in Fahrenheit:

• First, calculate \(\frac{9}{5} \times 100 = 180\).
• Then, add 32 to get the final result: \(180 + 32 = 212\).

So, 100°C equals 212°F. This conversion helps facilitate various tasks, such as reading temperature measurements in different countries or understanding global weather data.

Mathematical Formulas

Mathematics plays a crucial role in converting temperature scales. The formulas used in these conversions are derived from the relationship between the two units. Understanding these mathematical formulas not only helps in solving conversion problems but also enhances comprehension of heat and temperature concepts. Key aspects of the formulas to notice include:

• The conversion factor of \(\frac{9}{5}\) or \(\frac{5}{9}\).
• The addition or subtraction of 32 to account for the difference in zero points between the scales.

These formulas demonstrate how different systems can measure the same phenomenon, scale it linearly, and thus provide universal understanding. Such knowledge is instrumental in fields like physics, chemistry, and even daily life scenarios where temperature readings might need to be interpreted consistently across various systems. Understanding these formulas encourages logical thinking and problem-solving skills in dealing with real-world mathematical problems.