Chapter 1, Problem 1.58QP

Interpretation Introduction

Interpretation:

The temperature at which the numerical reading on a Celsius thermometer equal that on a Fahrenheit thermometer has to be determined.

Concept Introduction:

Kelvin scale:

• The absolute temperature scale is called as Kelvin scale.
• The smallest temperature by theoretical is 0K, it is known as the absolute zero temperature.
• The relationship between degrees Celsius and Kelvin is given by the equation given below:

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

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

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

Answer to Problem 1.58QP

The temperature at which the numerical reading on a Celsius thermometer equal that on a Fahrenheit thermometer is $-40$.

Explanation of Solution

Different units are there to represent temperature scales.

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

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

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

$\begin{array}{c}{}^{\text{o}}\text{C}\text{=}\left({}^{\text{o}}\text{C}\times 9/5\right)+32\\ {}^{\text{o}}\text{C}-\left({}^{\text{o}}\text{C}\times 9/5\right)=32\\ -4/5{\times }^{\text{o}}\text{C}=32\\ {}^{\text{o}}\text{C}=-32\times \left(5/4\right)\\ {}^{\text{o}}\text{C}=-40\end{array}$

Thus the temperature at which the numerical reading on a Celsius thermometer equal that on a Fahrenheit thermometer is $-40$

Conclusion

The temperature at which the numerical reading on a Celsius thermometer equal that on a Fahrenheit thermometer was calculated.