Ethylene glycol is the main component in automobile...

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 $\left(-45^{\circ} \mathrm{C}\right.$ and $\left.115^{\circ} \mathrm{C}\right)$. 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 $86^{\circ} \mathrm{A} .$ What is the temperature in ${ }^{\circ} \mathrm{C}$ and in ${ }^{\circ} \mathrm{F}$ ? e. What is a temperature of $45^{\circ} \mathrm{C}$ in ${ }^{\circ} \mathrm{A}$ ?

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 $\left(-45^{\circ} \mathrm{C}\right.$ and $\left.115^{\circ} \mathrm{C}\right)$. 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 $86^{\circ} \mathrm{A} .$ What is the temperature in ${ }^{\circ} \mathrm{C}$ and in ${ }^{\circ} \mathrm{F}$ ?
e. What is a temperature of $45^{\circ} \mathrm{C}$ in ${ }^{\circ} \mathrm{A}$ ?

Key Concepts

Reference Points Calibration

Reference points calibration is the method of assigning specific numerical values to two known fixed points on a temperature scale (for example, a melting point and a boiling point) to determine the scale's linear relationship. These points enable the derivation of the slope and intercept needed to convert temperatures between scales.

Celsius–Fahrenheit Conversion

The Celsius–Fahrenheit conversion is a specific application of linear transformation where temperatures are converted using the formula F = (9/5) × C + 32. This formula accounts for both the different zero points and the unit size differences between the Celsius and Fahrenheit scales.

Temperature Scales

Temperature scales are systems of measurement based on fixed reference points, such as the freezing and boiling points of water or, in custom cases, any two reproducible points. They allow us to quantify thermal energy and compare temperatures consistently across different systems.

Linear Transformation

A linear transformation in temperature conversion involves creating an equation in the form of T_new = slope × T_old + intercept. This transformation preserves the order and spacing of temperature readings while accommodating differences in zero points and unit sizes between scales.

Transcript

00:01 All right, so we know that the scale that we're working with is going to go from negative 45 to 115.

00:13 Oh, that's 2 degrees celsius.

00:21 There we go.

00:27 Negative 45 is going to correspond to 0, and the 115 corresponds to 100.

00:36 So between negative 45 and 115, there's 160 degrees.

00:51 Let me get rid of the units because that's confusing.

00:53 So 160 degrees for every 100 degrees celsius.

01:11 So then we're going to use it's 1 .6.

01:23 And then our starting point is going to be the negative 45.

01:34 That's for part a.

01:37 For part b, we know that degrees fahrenheit is going to be 1 .8 times degrees celsius plus 32.

01:51 So if we wanted to get from a to f, we can kind of just manipulate this equation.

01:55 That this plus 45 divided by 1 .6 is equal to 1 degree celsius.

02:05 So degrees fahrenheit will just plug in 1 .8 times a plus 45 divided by 1 .6 plus 32, and then we can simplify it.

02:23 So degrees fahrenheit is equal to, let's see, 1 .8 divided by 1 .6 is 1 .125.

02:30 And then that times 45 is 50 .625 and then we can add 32 to that to get all the constants together so plus 82 .6 approximately or we could just leave it 82 .625.

02:53 All right so now we have those expressions.

02:58 Now at what temperature would the degrees in a and the celsius give the same numerical reading? so we want basically this.

03:19 So we could take our equation and we could just plug in degrees celsius equals 1 .6 degrees celsius minus 45.

03:29 All i did is replace the a with the c.

03:32 We'll offer c because it's the same as a...

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