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 \({ }^{\circ} \mathrm{A}\) ?

Step by step solution

01

Set up the equation

We will use a linear function to represent the relationship between A-scale (\({}^{\circ}\mathrm{A}\)) and Celsius (\({}^{\circ}\mathrm{C}\)) temperatures. We can set up the equation in the form: \[ {}^{\circ} \mathrm{A} = k({}^{\circ}\mathrm{C}) + b \]

02

Use known temperature points to find k and b

We know that \(-45^{\circ} \mathrm{C}\) corresponds to \(0^{\circ}\mathrm{A}\), and \(115^{\circ}\mathrm{C}\) corresponds to \(100^{\circ}\mathrm{A}\). Use these points to create two linear equations: \[ 0 = k(-45) + b \] \[ 100 = k(115) + b \]

03

Solve the system of equations

Solving the system of equations above, we get: \[ k = 4/3 \] \[ b = 60 \]

04

Write the conversion equation

Using the values of k and b found, the conversion equation from Celsius to A-scale is: \[ {}^{\circ} \mathrm{A} = \frac{4}{3} ({}^{\circ}\mathrm{C}) + 60 \] Now, we will derive an expression for converting \({ }^{\circ} \mathrm{F}\) and \({ }^{\circ} \mathrm{A}\). #b. Derive an expression for converting between \({ }^{\circ} \mathrm{F}\) and \({ }^{\circ} \mathrm{A}\).#

05

Write the equation to convert Fahrenheit to Celsius

We know the equation to convert between Fahrenheit and Celsius: \[ {}^{\circ}\mathrm{C}=\frac{5}{9} ({}^{\circ}\mathrm{F}-32) \]

06

Substitute Celsius conversion into A-scale conversion

Replace the Celsius term (\({ }^{\circ}\mathrm{C}\)) in the A-scale conversion equation (derived in part a) with the Fahrenheit conversion: \[ {}^{\circ}\mathrm{A} = \frac{4}{3}\left(\frac{5}{9} ({}^{\circ}\mathrm{F}-32)\right) + 60 \]

07

Simplify the equation

Simplify the equation to get the expression for converting Fahrenheit to A-scale: \[ {}^{\circ}\mathrm{A} = \frac{40}{27} ({}^{\circ}\mathrm{F}) - \frac{1180}{27} \] #c. At what temperature would your thermometer and a Celsius thermometer give the same numerical reading?#

08

Set A-scale and Celsius scales equal

Set the A-scale and Celsius conversion equations equal to each other to find the temperature where the numerical reading is the same: \[ {}^{\circ}\mathrm{C} = \frac{4}{3} ({}^{\circ}\mathrm{C}) + 60 \]

09

Solve for \({ }^{\circ}\mathrm{C}\)

Solve the equation above for \({ }^{\circ}\mathrm{C}\): \[ {}^{\circ}\mathrm{C} = -36 \] So, the two thermometers would give the same numerical reading at \(-36^{\circ}\mathrm{C}\). #d. Your thermometer reads \(86^{\circ} \mathrm{A} .\) What is the temperature in \({ }^{\circ} \mathrm{C}\) and in \({ }^{\circ} \mathrm{F}\)?#

10

Convert A-scale to Celsius

Use the A-scale to Celsius conversion equation to find the Celsius temperature: \[ {}^{\circ} \mathrm{C} = \frac{3}{4} (86{}^{\circ}\mathrm{A} - 60) \] \[ {}^{\circ} \mathrm{C} = 19.5 \]

11

Convert Celsius to Fahrenheit

Use the Celsius to Fahrenheit conversion equation to find the Fahrenheit temperature: \[ {}^{\circ}\mathrm{F} = \frac{9}{5} (19.5{}^{\circ}\mathrm{C}) + 32 \] \[ {}^{\circ}\mathrm{F} = 67.1 \] So, when your thermometer reads \(86^{\circ}\mathrm{A}\), the temperature in Celsius is \(19.5^{\circ}\mathrm{C}\) and in Fahrenheit is \(67.1^{\circ}\mathrm{F}\). #e. What is a temperature of \(45^{\circ} \mathrm{C}\) in \({ }^{\circ} \mathrm{A}\)?#

12

Convert Celsius to A-scale

Use the Celsius to A-scale conversion equation to find the A-scale temperature: \[ {}^{\circ}\mathrm{A} = \frac{4}{3} (45{}^{\circ}\mathrm{C}) + 60 \] \[ {}^{\circ}\mathrm{A} = 140 \] So, a temperature of \(45^{\circ}\mathrm{C}\) corresponds to \(140^{\circ}\mathrm{A}\).

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