Question

Ethylene glycol is the main component in automobile antifreeze. To monitor the temperature of an auto cooling system, you intend to use a meter that reads from 0 to 100. You devise a new temperature scale based on the approximate melting and boiling points of a typical antifreeze solution \(\left(-45^{\circ} \mathrm{C} \text { and }\right.\) \(115^{\circ} \mathrm{C} ) .\) You wish these points to correspond to \(0^{\circ} \mathrm{A}\) and \(100^{\circ} \mathrm{A},\) respectively. a. Derive an expression for converting between \(^{\circ} \mathrm{A}\) and \(^{\circ} \mathrm{C}\) . b. Derive an expression for converting between \(^{\circ} \mathrm{F}\) and \(^{\circ} \mathrm{A}\) . c. At what temperature would your thermometer and a Celsius thermometer give the same numerical reading? d. Your thermometer reads \(86^{\circ}\) A. What is the temperature in \(^{\circ} \mathrm{C}\) and in \(\mathrm{FF}\) ? e. What is a temperature of \(45^{\circ} \mathrm{C}\) in 'A?

Short answer

a. \(^{\circ}\text{A} = \frac{5}{8}\cdot(^{\circ}\text{C}) + \frac{225}{8}\) b. \(^{\circ}\text{F} = \frac{9}{5} (8 \cdot ^{\circ}\text{A} - 225) + 32\) c. Both Celsius and A give the same numerical reading at \(75^{\circ}\text{C}\). d. The temperature is approximately \(21.25^{\circ}\text{C}\) and \(70.25^{\circ}\text{F}\). e. A temperature of \(45^{\circ}\text{C}\) in A is approximately \(84.375^{\circ}\text{A}\).

Step-by-step solution

a. Derive an expression for converting between Celsius and A.

The new temperature scale (A) should have the freezing point as \(0^{\circ}\text{A}\), which corresponds to \(-45^{\circ}\text{C}\), and the boiling point as \(100^{\circ}\text{A}\), which corresponds to \(115^{\circ}\text{C}\). We are looking for a linear equation in the format of \(^{\circ}\text{A} = m\cdot(^{\circ}\text{C}) + b\), where m is the slope and b is the intercept. Using the freezing and boiling points, we can set up a system of equations: \(0 = m(-45) + b\) \(100 = m(115) + b\) Now, we can solve this system of equations to find m and b.

Find m (slope)

Subtract the first equation from the second equation: \(100 - 0 = m(115) - m(-45)\) \(100 = m(115 + 45)\) \(100 = m(160)\) Divide both sides by 160: \(m = \frac{100}{160} = \frac{5}{8}\)

Find b (intercept)

Now that we found m, we can plug it into the first equation to find b: \(0 = \frac{5}{8}(-45) + b\) \(0 = -\frac{225}{8} + b\) Adding \(\frac{225}{8}\) to both sides to isolate b: \(b = \frac{225}{8}\) Finally, we have the linear equation to convert between Celsius and A: \(^{\circ}\text{A} = \frac{5}{8}\cdot(^{\circ}\text{C}) + \frac{225}{8}\)

b. Derive an expression for converting between Fahrenheit and A.

We know the conversion equation between Celsius and Fahrenheit: \(^{\circ}\text{C} = \frac{5}{9}\cdot(^{\circ}\text{F} - 32)\) Since we have an equation for converting between Celsius and A, we can substitute the Celsius term in the Fahrenheit equation with the A equation and solve for \(^{\circ}\text{F}\): \(^{\circ}\text{A} = \frac{5}{8}(\frac{5}{9}\cdot(^{\circ}\text{F} - 32)) + \frac{225}{8}\) Now, let's derive the expression for \(^{\circ}\text{F}\) in terms of \(^{\circ}\text{A}\):

Solve for Fahrenheit

\(^{\circ}\text{F} = \frac{9}{5}(\frac{8}{5} (^{\circ}\text{A} - \frac{225}{8})) + 32\) Simplify the equation: \(^{\circ}\text{F} = \frac{9}{5}(8 \cdot ^{\circ}\text{A} - 225) + 32\) \(^{\circ}\text{F} = \frac{9}{5} (8 \cdot ^{\circ}\text{A} - 225) + 32\)

c. At what temperature would your thermometer and a Celsius thermometer give the same numerical reading?

Let's find the temperature at which both Celsius and A scales give the same numerical reading: \(^{\circ}\text{C} = ^{\circ}\text{A}\) Substituting the equation from part a: \(^{\circ}\text{C} = \frac{5}{8}\cdot(^{\circ}\text{C}) + \frac{225}{8}\) To find the temperature where they give the same reading, solve for \(^{\circ}\text{C}\): \((\frac{3}{8})\cdot(^{\circ}\text{C}) = \frac{225}{8}\) Multiply both sides by \(\frac{8}{3}\): \(^{\circ}\text{C} = 75\) Therefore, both the Celsius and the A thermometer would give the same numerical reading at \(75^{\circ}\text{C}\).

d. Your thermometer reads \(86^{\circ}\text{A}\). What is the temperature in Celsius and Fahrenheit?

Using our derived equation for converting A to Celsius: \(^{\circ}\text{C} = \frac{5}{8}\cdot(\text{86}) + \frac{225}{8}\) \(^{\circ}\text{C} \approx 21.25\) The temperature in Celsius is approximately \(21.25^{\circ}\text{C}\). To find the Fahrenheit temperature, let's use the equation derived in part b: \(^{\circ}\text{F} = \frac{9}{5} (8 \cdot 86 - 225) + 32\) \(^{\circ}\text{F} \approx 70.25\) The temperature in Fahrenheit is approximately \(70.25^{\circ}\text{F}\).

e. What is a temperature of \(45^{\circ}\text{C}\) in A?

To convert a temperature of \(45^{\circ}\text{C}\) into the A scale, let's use the equation derived in part a: \(^{\circ}\text{A} = \frac{5}{8}\cdot(\text{45}) + \frac{225}{8}\) \(^{\circ}\text{A} = \frac{5}{8}(45) + \frac{225}{8}\) \(^{\circ}\text{A} \approx 84.375\) The temperature of \(45^{\circ}\text{C}\) in A is approximately \(84.375^{\circ}\text{A}\).

Key concepts

Celsius scale

The Celsius scale is a widely used temperature scale in science and daily life. It is based on two fixed points: the freezing point of water at 0°C and the boiling point at 100°C.
The scale is named after the Swedish astronomer Anders Celsius, who first proposed it in 1742.
Unlike Fahrenheit, Celsius uses a straightforward division between these two points, creating an easily understandable range. This linear scale makes it helpful for many scientific applications where precision is necessary.

• The Celsius scale is part of the metric system.
• It is used globally, especially in scientific contexts.
• Its main characteristic is its simple 100-degree interval between water's freezing and boiling points.

These features make the Celsius scale both practical and widely adopted for everyday temperature measurements.

Fahrenheit scale

The Fahrenheit scale is primarily used in the United States for non-scientific purposes. It was created by Daniel Gabriel Fahrenheit in 1724.
This scale sets the freezing point of water at 32°F and the boiling point at 212°F.
Therefore, there are 180 intervals or degrees between these points.

• The Fahrenheit scale was once more commonly used worldwide.
• It is known for its finer gradation, making it potentially more precise for everyday temperatures.
• To convert Celsius to Fahrenheit, use the equation: \( ^{\circ} \text{F} = \frac{9}{5} \times (^{\circ} \text{C}) + 32 \)

Although less common outside the US, understanding the Fahrenheit scale is essential for interpreting various historical and contextual temperature readings.

new temperature scale

Creating a new temperature scale tailored to specific needs can be useful. For this exercise, a scale named 'A' was developed using antifreeze solution points.
The key here was setting -45°C as 0°A and 115°C as 100°A.
This required the development of a custom linear equation.

• This new scale allows users to measure temperatures in a specific context, like an automobile cooling system.
• The scale provides direct readings from 0 to 100, simplifying understanding for the specific application.
• The conversion equation from Celsius to this scale is: \( ^{\circ}\text{A} = \frac{5}{8}\cdot(^{\circ}\text{C}) + \frac{225}{8} \)

By customizing the scale with these endpoints, nuanced applications like this become more accessible.

linear equation in temperature conversion

Temperature scales often rely on linear equations, transforming values from one scale to another.
This relationship is crucial as it ensures consistent and proportional conversions.
For example, converting Celsius to the new 'A' scale involves a linear equation derived from known points.

Deriving a Linear Equation

A linear equation in the form \( y = mx + b \) helps translate values:

• The slope \( m \) is determined by the difference in temperatures divided by the difference in corresponding points on the scales.
• The intercept \( b \) is adjusted by setting known points to align the scales properly.

In our exercise, these calculations simplify the process:
\( ^{\circ}\text{A} = \frac{5}{8} \cdot(^{\circ}\text{C}) + \frac{225}{8} \)
Such linear relationships highlight how straightforward and predictable temperature conversions can become with accurate equations.