Chapter 1, Problem 9PP

(a) The boiling point of acetylene is $-28.1°\text{C}.$ Below what temperature. in kelvins and degrees Fahrenheit, is acetylene a liquid?

(b) The bailing point of helium is 4 K. Below what temperature, in degrees Celsius, is helium a liquid?

(c) Human body temperature is normally $98.6°\text{F}.$ What is this temperature on the Celsius and Kelvin scales?

Interpretation Introduction

Interpretation:

The temperature in terms of Kelvin and degree Celcius at which acetylene exists as a liquid state.

Explanation of Solution

The conversion of temperature from degree Celsius to degree Fahrenheit can be done using the following relation:

$T_{°\text{F}}=\frac{9}{5}\left(T_{°\text{C}}\right)+32$ $\left(1\right)$

Here, $T_{°\text{F}}$ is the temperature in degree Fahrenheit and $T_{°\text{C}}$ is the temperature in degree Celsius.

The conversion of temperature from degree Celsius to Kelvin can be done using the following relation:

$T_{\text{K}}=T_{°\text{C}}+273.15$ $\left(2\right)$

Here, $T_{\text{K}}$ is the temperature in Kelvin and $T_{°\text{C}}$ is the temperature in degree Celsius.

The conversion of temperature from degree Fahrenheit to degree Celsius can be done using the following relation:

$T_{°\text{C}}=\left(T_{°\text{F}}-32\right)\times \frac{5}{9}$ $\left(3\right)$

Here, $T_{°\text{C}}$ is the temperature in degree Celsius and $T_{°\text{F}}$ is the temperature in degree Fahrenheit.

The temperature ( in $\text{K}$ ) at which the acetylene exists as a liquid can be calculated as follows:

The boiling point of acetylene is given to be $-28.1°\text{C}$ . Substitute, the value of $T_{°\text{C}}$ as $-28.1°\text{C}$ in the equation $\left(2\right)$ :

$\begin{array}{c}{T}_{\text{K}}=-28.1+273.15\\ =245.05\text{ K}\end{array}$

Therefore, the temperature at which the acetylene exists as a liquid, in terms of kelvin is found to be $245.05\text{ K}$ .

The temperature (in $°\text{F}$ ) at which acetylene exists as a liquid is calculated as follows:

The boiling point of acetylene is given to be $-28.1°\text{C}$ . Substitute, $T_{°\text{C}}$ as $-28.1°\text{C}$ in the equation $\left(1\right)$ :

$\begin{array}{c}{T}_{°\text{F}}=\frac{9}{5}\left(-28.1°\text{C}\right)+32\\ =-50.58+32\\ =-18.58°\text{F}\end{array}$

Therefore, the temperature at which the acetylene exists as a liquid, in terms of degree Fahrenheit is found to be $-18.58°\text{F}$ .

Interpretation Introduction

Interpretation:

The temperature in terms of degree Celcius below which helium exits in a liquid state.

Explanation of Solution

The temperature ( in $\text{K}$ ) at which helium exists as a liquid can be calculated using the following relation.

$T_{°\text{C}}=T_{\text{K}}-273.15$ $\left(4\right)$

Here, $T_{\text{K}}$ is the temperature in Kelvin and $T_{°\text{C}}$ is the temperature in degree Celsius.

The boiling point of helium is given to be $\text{4 K}$ . Substitute $T_{°\text{K}}$ as $\text{4 K}$ in the equation $\left(4\right)$ :

$\begin{array}{c}{T}_{°\text{C}}=4\text{ K}-273.15\\ =(-269.15°\text{C)}\end{array}$

Therefore, the temperature at which helium exists as a liquid, in terms of Kelvin, is found to be $-269.15°\text{C}$ .

Interpretation Introduction

Interpretation:

The normal temperature of the human body in terms of degree Celsius and kelvin.

Explanation of Solution

The temperature (in $°\text{C}$ ) of the human body can be calculated using the equation $\left(3\right)$ as follows:

The normal body temperature of the human body is $98.6°\text{F}$ . Substitute $T_{°\text{F}}$ as $98.6°\text{F}$ in equation $\left(3\right)$ :

$\begin{array}{c}{T}_{°\text{C}}=\left(98.6°\text{F}-32\right)\times \frac{5}{9}\\ =37°\text{C}\end{array}$

Therefore, the normal temperature of the human body in degree Celsius is found to be $37°\text{C}$ .

The normal body temperature (in $\text{K}$ ) can be calculated using the following relation.

$T_{\text{K}}=\left(T_{°\text{F}}-32\right)\times \frac{5}{9}+273.15$ $\left(5\right)$

Here, $T_{\text{K}}$ is the temperature in kelvin and $T_{°\text{F}}$ is the temperature in Fahrenheit.

The normal body temperature of the human body is $98.6°\text{F}$ . Substitute $T_{°\text{F}}$ as $98.6°\text{F}$ in equation $\left(5\right)$ .

$\begin{array}{c}{T}_{\text{K}}=\left(98.6°\text{F}-32\right)\times \frac{5}{9}+273.5\\ =37°\text{C}+273.15\\ =310.15\text{ K}\end{array}$

Therefore, the normal temperature of the human body in kelvin is found to be $310.15\text{ K}$ .