Problem 72 - Chemistry: The Central Science

Question

Suppose you decide to define your own temperature scale with units of \(\mathrm{O}\), using the freezing point \(\left(13{ }^{\circ} \mathrm{C}\right)\) and boiling point \(\left(360^{\circ} \mathrm{C}\right)\) of oleic acid, the main component of olive oil. If you set the freezing point of oleic acid as \(0^{\circ} \mathrm{O}\) and the boiling point as \(100^{\circ} \mathrm{O}\), what is the freezing point of water on this new scale?

Step-by-Step Solution

Answer

The freezing point of water on the Oleic scale is approximately \(-3.76^{\circ} \mathrm{O}\).

Step 1: Identify known values and conversion points

We have been given the following conversion points: - Freezing point of oleic acid: \(13^{\circ} \mathrm{C} = 0^{\circ} \mathrm{O}\) - Boiling point of oleic acid: \(360^{\circ} \mathrm{C} = 100^{\circ} \mathrm{O}\) - Freezing point of water: \(0^{\circ} \mathrm{C} = ?^{\circ} \mathrm{O}\) Our objective is to calculate the freezing point of water in degree Oleic.

Step 2: Establish relationship between Celsius and Oleic

Let's represent the Celsius temperatures as \(C\) and the Oleic temperatures as \(O\). To establish a relationship between the two scales, we will also need to introduce a proportional constant, \(k\), and an intercept, \(b\). We can write the conversion equation as follows: \[O = kC + b\] Using the given conversion points, we can create two equations: For the freezing point of oleic acid: \[0 = k(13) + b \Rightarrow b = -13k \hspace{20px} (1)\] For the boiling point of oleic acid: \[100 = k(360) + b \hspace{20px} (2)\]

Step 3: Find the values of k and b

Next, we need to determine the values of the proportional constant, \(k\), and the intercept, \(b\). We can easily find them by solving the two equations (1) and (2) simultaneously. Let's substitute the value of \(b\) from equation (1) into equation (2): \[100 = 360k - 13k\] \[100 = 347k\] Now solve for \(k\): \[k = \frac{100}{347}\] Now, substitute the value of \(k\) back into equation (1) to find the value of \(b\): \[b = -13 \cdot \frac{100}{347}\]

Step 4: Find the freezing point of water in degree Oleic

To find the freezing point of water in degree Oleic, we will use the conversion equation and the values of \(k\) and \(b\) that we have just found. Plug in the Celsius freezing point of water into the equation: \[O = \left(\frac{100}{347}\right)(0) - 13\left(\frac{100}{347}\right)\] Simplify the equation and solve for \(O\): \[O = -\frac{1300}{347}\] So, the freezing point of water on the Oleic scale is approximately \(-3.76^{\circ} \mathrm{O}\).

Key Concepts

Temperature Scales

Temperature scales are used to measure how hot or cold something is, using specific reference points to define the scale. Different scales can be created depending on the reference points chosen and their relationship.
Common scales include Celsius, Fahrenheit, and Kelvin, but in this example, a fictional scale called Oleic is used.

The Oleic scale defines:

• Freezing point of oleic acid as 0°O
• Boiling point of oleic acid as 100°O

In creating a new scale, it's crucial to establish the corresponding values on another known scale, like Celsius, to understand conversions.

Oleic Acid

Oleic acid is an unsaturated fatty acid that is a significant component of olive oil. Its physical properties, such as freezing ( (<13°C) ) and boiling points ( (<360°C) ), can vary based on external conditions.

In this problem, oleic acid provides the specific reference points for a new temperature scale. By setting its freezing and boiling points to an arbitrary 0°O and 100°O, respectively, it draws a parallel to how the Celsius scale uses the freezing and boiling points of water.

Calculation of Freezing Points

The calculation of freezing points often involves understanding the relationship between various temperature scales. In this exercise, we calculate the freezing point of water on the Oleic scale using known conversion points.

To find this, a linear equation is formulated linking Celsius ((
\[ O = kC + b \]
Specific steps include:

• Identifying known points and creating equations
• Solving for the proportional constant \(k\) and intercept \(b\)
• Finding the Oleic scale freezing point of water using these values

Resultantly, the transformation yields a water freezing point of approximately -3.76°O.

Celsius

Celsius is one of the most widely used temperature scales internationally. It sets the freezing point of water at 0°C and the boiling point at 100°C under standard conditions. From this baseline, other scales can be related to Celsius.

While working with Celsius in the context of creating a new scale:

• It offers a clear framework to measure temperature in a directly relatable manner.
• The two reference points are crucial for deriving conversion factors for any custom scale, like Oleic.

This inter-scale relationship helps physical scientists and students understand not just how temperature is quantified, but how it varies across different systems.

Proportional Constants

Proportional constants are coefficients that define the relationship between two variables, in this case, two temperature scales.

When establishing a conversion between Celsius and Oleic scales, the proportionality constant ((
Steps to calculate \(k\):

• Use known points on both scales to build linear equations
• Solve simultaneously to find the specific value

The proportional constant calculated was \(\frac{100}{347}\) which underpins the conversion equation. It allows for transforming any temperature from the Celsius system to the Oleic system, demonstrating proportionality within defined boundaries.