Chapter 1, Problem 68E

Ethylene glycol is the main component in automobile anti-freeze. 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 (−45°C and 115°C). You wish these points to correspond to 0°A and 100°A, respectively.

1. a. Derive an expression for converting between °A and °C.
2. b. Derive an expression for converting between °F and °A.
3. c. At what temperature would your thermometer and a Celsius thermometer give the same numerical reading?
4. d. Your thermometer reads 86°A. What is the temperature in °C and in °F?
5. e. What is a temperature of 45°C in °A?

Interpretation Introduction

Interpretation:

By using the given data, the answers for the given set of questions have to be given.

Concept introduction:

Conversion of temperature units in between degree Celsius, Kelvin and Fahrenheit is given by

$\begin{array}{l}{\text{Conversion of }}^{\text{o}}\text{F}\text{into}{}^{\text{o}}\text{C:}\\ {}^{\text{o}}\text{C}\text{=}\left({}^{\text{o}}\text{F}\text{-}{\text{32}}^{\text{o}}\text{F}\right)\text{×}\frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}\\ {\text{Conversion of }}^{\text{o}}\text{C}\text{into}{}^{\text{o}}\text{F:}\\ {}^{\text{o}}\text{F}\text{=}\left({}^{\text{o}}\text{C}\text{×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32}}^{\text{o}}\text{F}\\ {\text{Conversion of }}^{\text{o}}\text{C}\text{into}\text{K:}\\ \text{K = °C + 273}\text{.15}\end{array}$

Answer to Problem 68E

The relationship between ${}^{\text{o}}\text{C}\text{and}{}^{\text{o}}\text{A}$ is ${}^{\text{o}}\text{C}\text{=}\text{1}\text{.6}\text{×}{}^{\text{o}}\text{A}\text{-}\text{45}$

Explanation of Solution

To give: the relationship between ${}^{\text{o}}\text{C}\text{and}{}^{\text{o}}\text{A}$

The difference in degrees Celsius unit is $\text{160}{}^{\text{o}}\text{C,}$ but in Degrees ${}^{\text{o}}\text{A}$ unit is $\text{100}{}^{\text{o}}\text{A}\text{.}$

The relationship between ${}^{\text{o}}\text{C}\text{and}{}^{\text{o}}\text{A}$ is given below

$\begin{array}{l}\text{temperature}\text{in}{}^{\text{o}}\text{C}\text{=}{}^{\text{o}}\text{A}\text{×}\left(\frac{\text{8}}{\text{5}}\right)\text{-}\text{45}\\ \text{=}{}^{\text{o}}\text{A}\text{×}1.6\text{-}\text{45}\end{array}$

The relationship between ${}^{\text{o}}\text{C}\text{and}{}^{\text{o}}\text{A}$ is ${}^{\text{o}}\text{C}\text{=}\text{1}\text{.6}\text{×}{}^{\text{o}}\text{A}\text{-}\text{45}$

Interpretation Introduction

Interpretation:

By using the given data, the answers for the given set of questions have to be given.

Concept introduction:

Conversion of temperature units in between degree Celsius, Kelvin and Fahrenheit is given by

$\begin{array}{l}{\text{Conversion of }}^{\text{o}}\text{F}\text{into}{}^{\text{o}}\text{C:}\\ {}^{\text{o}}\text{C}\text{=}\left({}^{\text{o}}\text{F}\text{-}{\text{32}}^{\text{o}}\text{F}\right)\text{×}\frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}\\ {\text{Conversion of }}^{\text{o}}\text{C}\text{into}{}^{\text{o}}\text{F:}\\ {}^{\text{o}}\text{F}\text{=}\left({}^{\text{o}}\text{C}\text{×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32}}^{\text{o}}\text{F}\\ {\text{Conversion of }}^{\text{o}}\text{C}\text{into}\text{K:}\\ \text{K = °C + 273}\text{.15}\end{array}$

Answer to Problem 68E

The correct expression between ${}^{\text{o}}\text{F}\text{and}{}^{\text{o}}\text{A}$ units is ${}^{\text{o}}\text{F}\text{=}\text{2}\text{.88}{}^{\text{o}}\text{A}\text{-}\text{49}\text{.}$

Explanation of Solution

To give: the relationship between ${}^{\text{o}}\text{F}\text{and}{}^{\text{o}}\text{A}$

The relationship between ${}^{\text{o}}\text{F}\text{and}{}^{\text{o}}\text{C}$ is given by

${}^{\text{o}}\text{C}\text{=}\left({}^{\text{o}}\text{F}\text{-}{\text{32}}^{\text{o}}\text{F}\right)\text{×}\frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}$

Compare the two relationships

$\begin{array}{l}\left({}^{\text{o}}\text{F}\text{-}\text{32}\right)\text{×}\frac{\text{5}}{\text{9}}\text{=}\frac{\text{8}}{\text{5}}\text{×}\text{A}\text{-}\text{45}\\ \text{F}\text{-}\text{32 = }\left(\text{9/5}\right)\text{×}\left(\text{8/5}\right)\text{×}\text{A - }\left(\text{9/5}\right)\text{×}\left(\text{9x5}\right)\\ \text{F-32 = 72/25 A - 81}\text{=}\text{72/25 A - 81}\text{+ 32}\\ \text{F = 2}\text{.88 A - 49 }\end{array}$

The relationship between ${}^{\text{o}}\text{F}\text{and}{}^{\text{o}}\text{A}$ units is ${}^{\text{o}}\text{F}\text{=}\text{2}\text{.88}{}^{\text{o}}\text{A}\text{-}\text{49}\text{.}$

Interpretation Introduction

Interpretation:

By using the given data, the answers for the given set of questions have to be given.

Concept introduction:

Conversion of temperature units in between degree Celsius, Kelvin and Fahrenheit is given by

$\begin{array}{l}{\text{Conversion of }}^{\text{o}}\text{F}\text{into}{}^{\text{o}}\text{C:}\\ {}^{\text{o}}\text{C}\text{=}\left({}^{\text{o}}\text{F}\text{-}{\text{32}}^{\text{o}}\text{F}\right)\text{×}\frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}\\ {\text{Conversion of }}^{\text{o}}\text{C}\text{into}{}^{\text{o}}\text{F:}\\ {}^{\text{o}}\text{F}\text{=}\left({}^{\text{o}}\text{C}\text{×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32}}^{\text{o}}\text{F}\\ {\text{Conversion of }}^{\text{o}}\text{C}\text{into}\text{K:}\\ \text{K = °C + 273}\text{.15}\end{array}$

Answer to Problem 68E

The thermometer will give same reading at ${}^{\text{o}}\text{A}\text{=}\text{72}\text{=}{}^{\text{o}}\text{C}\text{.}$

Explanation of Solution

To determine: the temperature at which the thermometer same reading in ${}^{\text{o}}\text{A}\text{and}{}^{\text{o}}\text{C}$ units

Apply the relationship between ${}^{\text{o}}\text{C}\text{and}{}^{\text{o}}\text{A}$

$\begin{array}{l}{}^{\text{o}}\text{C}\text{=}\text{1}\text{.6}\text{×}{}^{\text{o}}\text{A}\text{-}\text{45}\\ \text{When}\text{A}\text{=}\text{C}\\ \text{A}\text{=}\text{1}\text{.6}\text{×}\text{A}\text{-}\text{45}\\ \text{45}\text{=}\left(\text{1/1}\text{.6}\right)\text{A}\text{=}\text{0}\text{.625A}\\ \text{A}\text{=}\text{72}\end{array}$

The thermometer will give same reading at ${}^{\text{o}}\text{A}\text{=}\text{72}\text{=}{}^{\text{o}}\text{C}\text{.}$

Interpretation Introduction

Interpretation:

By using the given data, the answers for the given set of questions have to be given.

Concept introduction:

Conversion of temperature units in between degree Celsius, Kelvin and Fahrenheit is given by

$\begin{array}{l}{\text{Conversion of }}^{\text{o}}\text{F}\text{into}{}^{\text{o}}\text{C:}\\ {}^{\text{o}}\text{C}\text{=}\left({}^{\text{o}}\text{F}\text{-}{\text{32}}^{\text{o}}\text{F}\right)\text{×}\frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}\\ {\text{Conversion of }}^{\text{o}}\text{C}\text{into}{}^{\text{o}}\text{F:}\\ {}^{\text{o}}\text{F}\text{=}\left({}^{\text{o}}\text{C}\text{×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32}}^{\text{o}}\text{F}\\ {\text{Conversion of }}^{\text{o}}\text{C}\text{into}\text{K:}\\ \text{K = °C + 273}\text{.15}\end{array}$

Answer to Problem 68E

The given temperature in degrees Celsius unit is $\text{92}\text{.6}{}^{\text{o}}\text{C,}$ in degrees Fahrenheit unit is $\text{198}\text{.7}{}^{\text{o}}\text{F}\text{.}$

Explanation of Solution

To determine: the given temperature in ${}^{\text{o}}\text{C}\text{and}{}^{\text{o}}\text{F}$ units

Given information

$\text{86}{}^{\text{o}}\text{A}$

Apply the relationship between ${}^{\text{o}}\text{C}\text{and}{}^{\text{o}}\text{A}$

$\begin{array}{l}{}^{\text{o}}\text{C}\text{=}\text{1}\text{.6}\text{×}{}^{\text{o}}\text{A}\text{-}\text{45}\\ \text{=}\text{1}\text{.6}\text{×}\text{86}\text{-}\text{45}\\ \text{=}\text{92}\text{.6}{}^{\text{o}}\text{C}\end{array}$

Apply the relationship between ${}^{\text{o}}\text{C}\text{and}{}^{\text{o}}\text{F}$

$\begin{array}{l}{}^{\text{o}}\text{F}\text{=}\left({}^{\text{o}}\text{C}\text{×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32}}^{\text{o}}\text{F}\\ \text{=}\left(\text{92}\text{.6}{}^{\text{o}}\text{C}\text{×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32}}^{\text{o}}\text{F}\\ \text{=}\text{198}\text{.7}{}^{\text{o}}\text{F}\end{array}$

The given temperature in degrees Celsius unit is $\text{92}\text{.6}{}^{\text{o}}\text{C,}$ in degrees Fahrenheit unit is $\text{198}\text{.7}{}^{\text{o}}\text{F}\text{.}$

Interpretation Introduction

Interpretation:

By using the given data, the answers for the given set of questions have to be given.

Concept introduction:

Conversion of temperature units in between degree Celsius, Kelvin and Fahrenheit is given by

$\begin{array}{l}{\text{Conversion of }}^{\text{o}}\text{F}\text{into}{}^{\text{o}}\text{C:}\\ {}^{\text{o}}\text{C}\text{=}\left({}^{\text{o}}\text{F}\text{-}{\text{32}}^{\text{o}}\text{F}\right)\text{×}\frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}\\ {\text{Conversion of }}^{\text{o}}\text{C}\text{into}{}^{\text{o}}\text{F:}\\ {}^{\text{o}}\text{F}\text{=}\left({}^{\text{o}}\text{C}\text{×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32}}^{\text{o}}\text{F}\\ {\text{Conversion of }}^{\text{o}}\text{C}\text{into}\text{K:}\\ \text{K = °C + 273}\text{.15}\end{array}$

Answer to Problem 68E

The given temperature in ${}^{\text{o}}\text{A}$ unit is $\text{56}\text{.25}{}^{\text{o}}\text{A}\text{.}$

Explanation of Solution

To determine: the given temperature in ${}^{\text{o}}\text{A}$ unit

Given information

$\text{45}{}^{\text{o}}\text{C}$

Apply the relationship between ${}^{\text{o}}\text{C}\text{and}{}^{\text{o}}\text{A}$

$\begin{array}{l}{}^{\text{o}}\text{C}\text{=}\text{1}\text{.6}\text{×}{}^{\text{o}}\text{A}\text{-}\text{45}\\ \text{45}\text{+}\text{45}\text{=}\text{1}\text{.6}{}^{\text{o}}\text{A}\\ \text{A}\text{=}\text{90/1}\text{.6}\\ \text{=}\text{56}{\text{.25 }}^{\text{o}}\text{A}\end{array}$

The given temperature in ${}^{\text{o}}\text{A}$ unit is $\text{56}\text{.25}{}^{\text{o}}\text{A}\text{.}$