Problem 78 - 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 new temperature scale (O) is approximately -3.75 °O.

Step 1: Identify given values and temperature scales

We are given a problem that is defined in two different temperature scales: Celsius (C) and the new scale called O. Given values: - Freezing point of oleic acid in Celsius: \(13^{\circ}\mathrm{C}\) - Boiling point of oleic acid in Celsius: \(360^{\circ}\mathrm{C}\) - Freezing point of oleic acid in new scale O: \(0^{\circ}\mathrm{O}\) - Boiling point of oleic acid in new scale O: \(100^{\circ}\mathrm{O}\) - Freezing point of water in Celsius: \(0^{\circ}\mathrm{C}\) (from general knowledge) We need to find the freezing point of water in the new scale O.

Step 2: Find the ratio between two scales

The first step in finding the relationship between two temperature scales is to find the ratio between their units. To do this, we will use the difference between the values of the two points in both scales. In Celsius: - Difference between freezing and boiling points of oleic acid: \(360 - 13 = 347^{\circ}\mathrm{C}\) In new scale O: - Difference between freezing and boiling points of oleic acid: \(100 - 0 = 100^{\circ}\mathrm{O}\) Therefore, the ratio between the two scales is: Ratio = \(\frac{\Delta O}{\Delta C} = \frac{100^{\circ}\mathrm{O}}{347^{\circ}\mathrm{C}}\)

Step 3: Find the relationship between the two scales

Now, we need to create a linear relationship between the Celsius and O temperature scales. Let's denote the temperature in Celsius by T(C) and in O by T(O). Since we know the ratio between the two scales, we can set up an equation: \(T(O) - T_O(0) = \frac{100^{\circ}\mathrm{O}}{347^{\circ}\mathrm{C}}\cdot(T(C) - T_C(0))\) where \(T_O(0)\) is the freezing point of oleic acid in O (0 °O), and \(T_C(0)\) is the freezing point of oleic acid in Celsius (13 °C).

Step 4: Find the freezing point of water in the new scale

Now, we can use the relationship we found in Step 3 to find the freezing point of water in the new O scale. We'll input the known freezing point of water in Celsius (0 °C) and solve for T(O). \(T(O) - 0^{\circ}\mathrm{O} = \frac{100^{\circ}\mathrm{O}}{347^{\circ}\mathrm{C}}\cdot(0^{\circ}\mathrm{C} - 13^{\circ}\mathrm{C})\) \(T(O) = \frac{100^{\circ}\mathrm{O}}{347^{\circ}\mathrm{C}}\cdot(-13^{\circ}\mathrm{C})\) \(T(O) = -\frac{1300^{\circ}\mathrm{O}}{347}\) \(T(O) \approx -3.75^{\circ}\mathrm{O}\) Therefore, the freezing point of water on this new temperature scale is approximately -3.75 °O.

Key Concepts

Celsius scale

The Celsius scale, also known as the centigrade scale, is a widely used temperature scale in which the freezing point of water is set at 0 °C and the boiling point at 100 °C under standard atmospheric conditions. It is named after the Swedish astronomer Anders Celsius, who developed this scale in 1742. The Celsius scale is part of the metric system, which makes it extremely useful for scientific measurements worldwide.
This scale is linear, meaning each degree represents an equal temperature increment. This characteristic helps in comparing temperatures and in conversions between other scales, such as Fahrenheit or Kelvin. It's important to understand that Celsius is based on the physical properties of water, a substance that is abundant and has a solid-liquid-gas transition at convenient temperatures, making it reliable for calibration.
When converting Celsius to another scale, a linear relationship is often used, where the difference in degrees across two temperature points can be directly translated to another scale using a ratio. This concept is fundamental when creating a new scale, like the O temperature scale derived in your exercise with oleic acid.

Temperature ratio

A temperature ratio is a comparative relationship that allows you to convert temperatures from one scale to another. In the given exercise, you defined a new scale, O, using known points of oleic acid: its freezing point at 13 °C and boiling point at 360 °C. In the O scale, these are 0 °O and 100 °O respectively.
To find the ratio between these scales, calculate the temperature range difference in both scale systems. For Celsius, the range from oleic acid's freezing to boiling point is 347 °C. For the O scale, it is 100 °O. Thus the ratio is:\[\text{Ratio} = \frac{100^{\circ} O}{347^{\circ} C}\]
This ratio tells us how much one degree change in the Celsius scale equates to a change in the O scale. Such a ratio is vital because it provides a tool to translate temperatures between the two scales directly, allowing for accurate conversions and a deeper understanding of the relationship between these temperature scales.

O temperature scale

Creating the O temperature scale involves choosing key points, such as the freezing and boiling points of a substance, much as Celsius does with water. In the exercise described, oleic acid's freezing and boiling points become the defining references.
Here's how you can derive the freezing point of water on the O scale:- Use the derived temperature ratio \( \frac{100^{\circ} O}{347^{\circ} C} \).
- Set up the linear relationship based on this ratio:\[T(O) = \frac{100^{\circ} O}{347^{\circ} C} \cdot (T(C) - 13^{\circ} C)\]
To find the freezing point of water (0 °C) in the O scale, substitute into the relationship:\[T(O) = \frac{100^{\circ} O}{347^{\circ} C} \cdot (0^{\circ} C - 13^{\circ} C)\]Calculating this, you find that the freezing point of water in the O scale is approximately -3.75 °O. Understanding new temperature scales helps illustrate how temperature is a relative measure that depends on the system chosen for measurement.