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?

Step by step solution

01

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.

02

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: \(m = \frac{100}{160} = \frac{5}{8}\)

03

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: \(^{\circ}\text{A} = \frac{5}{8}\cdot(^{\circ}\text{C}) + \frac{225}{8}\)

04

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 \(^{\circ}\text{A}\):

05

Solve for Fahrenheit

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

06

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 \(75^{\circ}\text{C}\).

07

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 \(70.25^{\circ}\text{F}\).

08

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 \(84.375^{\circ}\text{A}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!