Problem 29 - General Chemistry

Question

Say you live in a climate where the temperature ranges from \(-100^{\circ} \mathrm{F}\) to \(20^{\circ} \mathrm{F}\) and you want to define a new temperature scale, YS (YS is the "Your Scale" temperature scale), which defines this range as \(0.0^{\circ} \mathrm{YS}\) to \(100.0^{\circ} \mathrm{YS}\). a) Come up with an equation that would allow you to convert between \({ }^{\circ} \mathrm{F}\) and \({ }^{\circ} \mathrm{YS}\). b) Using your equation, what would be the temperature in \({ }^{\circ} \mathrm{F}\) if it were \(66^{\circ} \mathrm{YS} ?\)

Step-by-Step Solution

Answer

The temperature is -20.8°F at 66°YS.

Step 1: Identify the Linear Relationship

The problem involves converting Fahrenheit (°F) to a new scale (°YS). Since we want to map a range from °F to °YS linearly, we use the linear equation format: \[Y = mX + c\] where \(X\) is the temperature in °F, \(Y\) is the temperature in °YS, \(m\) is the slope, and \(c\) is the intercept. We need to find \(m\) and \(c\).

Step 2: Determine the Slope (m)

To find the slope, calculate the change in the YS scale over the change in the Fahrenheit scale. Given the range from -100°F to 20°F maps to 0°YS to 100°YS, we compute:\[m = \frac{100 - 0}{20 + 100} = \frac{100}{120} = \frac{5}{6}\].

Step 3: Calculate the Intercept (c)

Using one of the known points, substitute into the linear equation to solve for \(c\). We use -100°F and 0°YS:\[0 = \frac{5}{6}(-100) + c\]Solving for \(c\), we get:\[0 = -\frac{500}{6} + c\]\[c = \frac{500}{6} = \frac{250}{3}\].

Step 4: Formulate the Equation

Using \(m\) and \(c\) in \(Y = mX + c\), the conversion equation is:\[Y = \frac{5}{6}X + \frac{250}{3}\].

Step 5: Convert 66°YS to Fahrenheit

We now convert 66°YS to °F using the rearranged equation. Set \(Y = 66\), and solve for \(X\):\[66 = \frac{5}{6}X + \frac{250}{3}\]Subtract \(\frac{250}{3}\) from both sides:\[66 - \frac{250}{3} = \frac{5}{6}X\]Convert 66 to a fraction with the same denominator:\[\frac{198}{3} - \frac{250}{3} = \frac{5}{6}X\]\[\frac{-52}{3} = \frac{5}{6}X\]Solve for X:\[X = \frac{-52}{3} \times \frac{6}{5} = \frac{-104}{5} = -20.8\].

Key Concepts

Fahrenheit to YS conversion

Converting temperatures from Fahrenheit (°F) to the newly created YS scale involves a linear transformation. In this exercise, we aim to understand how a specific range in Fahrenheit relates to a newly defined range in an imaginary temperature scale called YS.
Imagine you have two endpoints:

• -100°F translates to 0°YS
• 20°F translates to 100°YS

By identifying these points, we can map these temperatures directly. The goal is to develop a linear equation that represents this mapping, enabling seamless conversion between the two scales.

linear equation in temperature scales

The core of converting °F to °YS revolves around a linear equation because the relationship between the two temperature scales is linear.
The general formula for a linear equation is:\[ Y = mX + c \]Here,

• \(Y\) is the temperature in YS scale
• \(X\) is the temperature in Fahrenheit
• \(m\) is the slope
• \(c\) is the y-intercept

The slope \(m\) is calculated as the change in YS divided by the change in Fahrenheit. This gives us the rate at which °YS changes for each degree Fahrenheit.
The y-intercept \(c\) is found by plugging one of the Fahrenheit endpoints into the equation after solving for \(m\). Here, -100°F supplying a 0°YS allows easy calculation of \(c\), the offset needed in our equation.

temperature scale transformation

Using the linear equation derived from known endpoints, you can transform °F to °YS through a specific mathematical process. We start by framing the equation:
\[ Y = \frac{5}{6}X + \frac{250}{3} \]This formula is derived by first calculating the slope \(m = \frac{5}{6}\) and the intercept \(c = \frac{250}{3}\).
The slope represents how much the YS temperature changes relative to changes in Fahrenheit. Calculating \(c\) involves using a endpoint, such as -100°F coinciding with 0°YS. This gives us a method to convert any Fahrenheit temperature into YS using simple substitution in this equation.

creating custom temperature scales

Designing a custom temperature scale like YS requires creativity and mathematical understanding. This can be useful in specific contexts or domains where unique measurement standards are needed.
To create your own scale, follow these general steps:

• Determine a range of interest in an existing scale (like Fahrenheit)
• Define desired endpoints that make sense in your custom scale
• Calculate the slope \(m\) using the changes in your defined points
• Identify the intercept \(c\) from your chosen equations

Once these parameters are set, you have a robust equation that allows you to easily shift between established and custom scales, effectively broadening your potential for measurement comparisons.