Question

On a new Jekyll temperature scale, water freezes at \(17^{\circ} \mathrm{J}\) and boils at \(97^{\circ} \mathrm{J} .\) On another new temperature scale, the Hyde scale, water freezes at \(0^{\circ} \mathrm{H}\) and boils at \(120^{\circ} \mathrm{H}\). If methyl alcohol boils at 84 \({ }^{\circ} \mathrm{H},\) what is its boiling point on the Jekyll scale?

Short answer

The boiling point of methyl alcohol on the Jekyll scale.

Step-by-step solution

Establish the Linear Relationship

First, we recognize that the relationship between the Jekyll and Hyde temperature scales is linear. We can use two points to define this relationship: the freezing and boiling points of water. On the Jekyll scale, these are 17 and 97 degrees respectively, and on the Hyde scale, they are 0 and 120 degrees respectively.

Define the Transformation Equation

Let's denote the Jekyll temperature as 'J' and the Hyde temperature as 'H'. Now we can set up a linear function that transforms H into J. Using the two given points, we create the equation \( J = a * H + b \), where 'a' is the scale factor and 'b' is the y-intercept. Since we have two pairs of corresponding temperatures, we can form two equations based on these points: \[ J = a * H + b \rightarrow \begin{cases} 97 = a * 120 + b \ 17 = a * 0 + b \end{cases} \]

Calculate the Scale Factor and Y-Intercept

Solve the system of equations to find 'a' and 'b'. From the second equation, we immediately get \( b = 17 \). Substituting 'b' into the first equation, we get \( 97 = 120a + 17 \). To find 'a', we solve this equation: \[ a = \frac{97 - 17}{120} \]

Find the Methyl Alcohol Boiling Point on the Jekyll Scale

With the known value of 'a' and 'b', now substitute the Hyde temperature for methyl alcohol boiling, which is 84 degrees, into the transformation equation \( J = a * H + b \), and solve for 'J' to find the boiling point of methyl alcohol on the Jekyll scale.

Key concepts

Linear Relationship in Chemistry

Understanding the concept of a linear relationship is essential in various areas of chemistry, especially when dealing with relationships between different scales of measurement. A linear relationship refers to a direct proportionality between two variables, where one variable is a constant multiple of the other, plus a constant. This concept is vital in chemistry for converting units, understanding concentration dilutions, and, as demonstrated in our exercise, converting between different temperature scales.

When two temperature scales are linearly related, any temperature on one scale can be converted to the corresponding temperature on the other scale using the equation of a straight line, which is of the form \( y = mx + b \). In this equation, 'y' represents the temperature on one scale, 'x' represents the temperature on the other scale, 'm' is the slope or scale factor, and 'b' is the y-intercept, representing the value of 'y' when 'x' is zero. The concept of the linear relationship is crucial for chemists when calibrating instruments or interpreting data across different measurement systems.

Thermodynamic Temperature Scales

Thermodynamic temperature scales are used to measure the average kinetic energy of particles in a substance. The most common thermodynamic temperature scale is the Celsius scale, which is based on the properties of water, with 0°C representing the freezing point and 100°C the boiling point at one atmosphere of pressure. However, scientists also work with other scales like Kelvin, Fahrenheit, and, in our exercise example, the hypothetical Jekyll and Hyde scales.

The Kelvin scale is particularly important in thermodynamics because it is an absolute scale with its zero point at absolute zero, the temperature at which molecular motion ceases. When working with different scales, chemists must be familiar with the fixed points (like the freezing and boiling points of water) and the linear transformation needed to convert between scales. By understanding how these scales are constructed and related to each other, scientists can accurately measure and communicate temperatures in different contexts.

Solving System of Equations

To solve a system of equations means finding the values for the variables that make all the equations true at the same time. In the context of our exercise, we used a system of linear equations which is a set of two or more linear equations involving the same set of variables. In chemistry, systems of equations can be used to balance chemical reactions, determine reaction yields, and, as highlighted in the exercise, to find a conversion factor between different temperature scales.

In our example, the system was composed of two equations derived from known points on the Jekyll and Hyde temperature scales. By solving the system, we found the scale factor 'a' and y-intercept 'b', critical in deriving the conversion formula. There are several methods for solving systems, including substitution, elimination, and using matrix operations. In our case, substitution was used to efficiently find the values of 'a' and 'b' which are then applied to convert any given temperature from one scale to another.