Chapter 5, Problem 68A

(a)

Interpretation Introduction

Interpretation:

The relation between the degree Celsius and degree X is to be derived on the basis of given figures.

Concept Introduction :

There are various units in which temperature is measured but kelvin is the SI unit of temperature. The temperate is converted from one unit to other units like as, from degree Celsius to kelvin, from degree Fahrenheit to degree Celsius or vice versa.

Conversion of the temperature from $\text{°F}\text{to}\text{°C}$ is done as follows:

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

Conversion of the temperature from $\text{°C}\text{to}\text{°F}$ is done as follows:

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

Conversion from degree Celsius to Kelvin is done as follows:

  $\text{K=°C+273}\text{.15K}$

Answer to Problem 68A

The relationship between $\text{°C}\text{and}\text{°F}$ is $\text{°X=}\left(\text{°C×}\frac{\text{5}}{\text{14}}\right)$ .

Explanation of Solution

The initial temperature in X-scale is $0°\text{X}$ and the final temperature is $50°\text{X}$ . The temperature difference is,

  $\text{50°X-0°X=50°X}$

Similarly, the initial temperature in Celsius-scale is $-10°\mathrm{C}$ and the final temperature is $13\text{0°C}$ . The temperature difference is,

  $\text{130°C-}\left(\text{-10°C}\right)\text{=140°C}$

This result in a ratio for degree Celsius to degree Fahrenheit of $\frac{140}{50}=\frac{14}{5}$

Then $0°\text{X}$ equals to the $-10°\text{C}$

The formula for $\text{°C}\text{to}\text{°F}$ is,

  $\text{°X=}\left(\text{°C×}\frac{\text{5}}{\text{14}}\right)\text{+0}$

(b)

Interpretation Introduction

Interpretation:

The temperature in units of $°\text{X}$ is to be calculated.

Concept Introduction :

There are various units in which temperature is measured but kelvin is the SI unit of temperature. The temperate is converted from one unit to other units like as, from degree Celsius to kelvin, from degree Fahrenheit to degree Celsius or vice versa.

Conversion of the temperature from $\text{°F}\text{to}\text{°C}$ is done as follows:

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

Conversion of the temperature from $\text{°C}\text{to}\text{°F}$ is done as follows:

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

Conversion from degree Celsius to Kelvin is done as follows:

  $\text{K=°C+273}\text{.15K}$

Answer to Problem 68A

The unit of the length in the metric system is most appropriate in size for measuring the size of the room is meter.

Explanation of Solution

Given information: outside temperature is, $T_{\text{outside}}=22°\text{C}$

Calculation:

  $\text{°X=}\left(\text{°C×}\frac{\text{5}}{\text{14}}\right)\text{+0}$

The temperature in terms of $°X$ is,

  $\text{°X=}\left(\text{22×}\frac{\text{5}}{\text{14}}\right)\text{=}\frac{\text{110}}{\text{14}}\text{=7}\text{.8°X}$

(c)

Interpretation Introduction

Interpretation:

The given temperature in degree X is to be converted into the degree Celsius, degree Fahrenheit and kelvin scale.

Concept Introduction :

There are various units in which temperature is measured but kelvin is the SI unit of temperature. The temperate is converted from one unit to other units like as, from degree Celsius to kelvin, from degree Fahrenheit to degree Celsius or vice versa.

Conversion of the temperature from $\text{°F}\text{to}\text{°C}$ is done as follows:

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

Conversion of the temperature from $\text{°C}\text{to}\text{°F}$ is done as follows:

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

Conversion from degree Celsius to Kelvin is done as follows:

  $\text{K=°C+273}\text{.15K}$

Answer to Problem 68A

The unit of the length in the metric system is most appropriate in size for measuring the distance from New York to London is kilometer.

Explanation of Solution

Firstly, convert the $58°Xinto°C$ by using the above formula, such that,

  $\begin{array}{l}\text{°X=}\left(\text{°C×}\frac{\text{5}}{\text{14}}\right)\\ \text{°C=}\frac{\text{14}}{\text{5}}\text{×°X}\\ \frac{\text{14}}{\text{5}}\text{×58=162}\text{.4°C}\end{array}$

Now, convert degree Celsius into kelvin, such that

  $\text{K=162}\text{.4°C+273}\text{.15K=435}\text{.5K}$

Now, convert the degree Celsius into degree Fahrenheit, such that

  $\text{°F=}\left(\text{162}\text{.4×}\frac{\text{9}}{\text{5}}\right)\text{+32=324}\text{.32°F}$