Chapter 1, Problem 1.113QP

a)

Interpretation:

Interpretation Introduction

It should be determined the temperature at which the Celsius and Fahrenheit values are numerically equal

Concept introduction:

• Conversional formula from Celsius to Fahrenheit,

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

• Conversional formula from Fahrenheit to Celsius

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

• Conversional formula from Celsius to Kelvin,

$\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

To determine: The temperature at which the Celsius and Fahrenheit values are numerically equal

Explanation of Solution

Explanation

The conversional formula from Celsius to Fahrenheit is

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

It is needed to determine the temperature at which the Celsius and Fahrenheit values are numerically equal That is $°C=°F$.

To determine the temperature that numerically equal to Celsius and Fahrenheit values, it is needed to replace $\text{°F}$ with $\text{°C}$ in the above equation. That is,

$\begin{array}{l}\text{°C}\text{=}\left({}^{\text{o}}\text{C×}\frac{\text{9}}{\text{5}}\right)\text{+}\text{32°F}\\ \text{°C}\text{=}\frac{\text{9}}{\text{5}}{}^{\text{o}}\text{C}\text{+}\text{32}\\ \text{°C}\text{-}\frac{\text{9}}{\text{5}}{}^{\text{o}}\text{C=}\text{32}\\ \left(\text{1-}\frac{\text{9}}{\text{5}}\right)\text{°C}\text{=32}\\ \text{-}\frac{\text{4}}{\text{5}}\text{°C}\text{=}\text{32}\\ \text{°C}\text{=}\text{-}\frac{\text{32×5}}{\text{4}}\text{=}\text{-40°}\\ \end{array}$

So, it can be conclude that $\text{-40°C}\text{=}\text{-40°F}$

b)

Interpretation Introduction

Interpretation:

It should be determined the temperature at which the Kelvin and Fahrenheit values are numerically equal

Concept introduction:

• Conversional formula from Celsius to Fahrenheit,

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

• Conversional formula from Fahrenheit to Celsius

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

• Conversional formula from Celsius to Kelvin,

$\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

To determine: The temperature at which the Kelvin and Fahrenheit values are numerically equal

Explanation of Solution

Explanation

Conversional formula from Celsius to Fahrenheit is

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

Conversional formula from Celsius to Kelvin is

$\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

It is needed to determine the temperature at which Kelvin and Fahrenheit values are numerically equal That is $K=°F$.

To determine the numerically equal temperature it is needed to set the conversion factors equal to one another. Then solve for $\text{°C}$ which is the variable common to both equations. This value converted to both $\text{°F}$ and K

That is,

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

$\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

If $K=°F$

The above conversion equations becomes,

$\begin{array}{l}\left({}^{\text{o}}\text{C×}\frac{\text{9}}{\text{5}}\right)\text{+}\text{32}\text{=}\text{°C+}\text{273}\text{.15}\\ \text{1}\text{.8°C+32}\text{=°C+273}\text{.15}\\ \text{0}\text{.8°C}\text{=}\text{241}\text{.15}\\ \text{°C}\text{=}\text{301}\text{.44}\end{array}$

This value should be used to solve $\text{K}\text{or}\text{°F}$

$\begin{array}{l}\text{K=}\text{301}\text{.44°C+}\text{273}\text{.15}\text{=}\text{574}\text{.59K}\\ {\text{F}}^{\text{o}}\text{=}\left(\text{301}{\text{.44}}^{\text{o}}\text{C×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32F}}^{\text{o}}\text{=}\text{574}{\text{.59}}^{\text{o}}\text{F}\end{array}$

That is, $574.59K=574.59°F$

C)

Interpretation Introduction

Interpretation:

It should be checked that, is there any temperature at which the Celsius and Kelvin values are numerically equal

Concept introduction:

• Conversional formula from Celsius to Fahrenheit,

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

• Conversional formula from Fahrenheit to Celsius

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

• Conversional formula from Celsius to Kelvin,

$\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

To check: Is there any temperature at which the Celsius and Kelvin values are numerically equal

Explanation of Solution

Conversional formula from Celsius to Kelvin is

$\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

The values of Kelvin are always equal to a value that is 273.15, which is greater than the Celsius value. So there is no temperature at which the values are numerically equal