(a) The boiling point of acetylene is − 28.1 ° 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 ° F . What is this temperature on the Celsius and Kelvin scales?

Question

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Answer 1

Textbook Question

Chapter 1, Problem 9PP

(a) The boiling point of acetylene is $− 28.1 ° C .$ 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 $98.6 ° F .$ What is this temperature on the Celsius and Kelvin scales?

(a)

Expert Solution

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:

$T°F=95T°C+32$ $1$

Here, $T°F$ is the temperature in degree Fahrenheit and $T°C$ is the temperature in degree Celsius.

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

$TK=T°C+273.15$ $2$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

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

$T°C=T°F−32×59$ $3$

Here, $T°C$ is the temperature in degree Celsius and $T°F$ is the temperature in degree Fahrenheit.

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, the value of $T°C$ as $−28.1 °C$ in the equation $2$ :

$TK=−28.1+273.15=245.05 K$

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

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, $T°C$ as $−28.1 °C$ in the equation $1$ :

$T°F=95−28.1 °C+32=−50.58+32=−18.58 °F$

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

(b)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

$T°C=TK−273.15$ $4$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

The boiling point of helium is given to be $4 K$ . Substitute $T°K$ as $4 K$ in the equation $4$ :

$T°C=4 K−273.15=(−269.15 °C)$

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

(c)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The temperature (in $°C$ ) of the human body can be calculated using the equation $3$ as follows:

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $3$ :

$T°C=98.6 °F−32×59=37 °C$

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

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

$TK=T°F−32×59+273.15$ $5$

Here, $TK$ is the temperature in kelvin and $T°F$ is the temperature in Fahrenheit.

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $5$ .

$TK=98.6 °F−32×59+273.5=37 °C+273.15=310.15 K$

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

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Answer 2

Textbook Question

Chapter 1, Problem 9PP

(a) The boiling point of acetylene is $− 28.1 ° C .$ 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 $98.6 ° F .$ What is this temperature on the Celsius and Kelvin scales?

(a)

Expert Solution

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:

$T°F=95T°C+32$ $1$

Here, $T°F$ is the temperature in degree Fahrenheit and $T°C$ is the temperature in degree Celsius.

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

$TK=T°C+273.15$ $2$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

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

$T°C=T°F−32×59$ $3$

Here, $T°C$ is the temperature in degree Celsius and $T°F$ is the temperature in degree Fahrenheit.

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, the value of $T°C$ as $−28.1 °C$ in the equation $2$ :

$TK=−28.1+273.15=245.05 K$

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

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, $T°C$ as $−28.1 °C$ in the equation $1$ :

$T°F=95−28.1 °C+32=−50.58+32=−18.58 °F$

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

(b)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

$T°C=TK−273.15$ $4$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

The boiling point of helium is given to be $4 K$ . Substitute $T°K$ as $4 K$ in the equation $4$ :

$T°C=4 K−273.15=(−269.15 °C)$

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

(c)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The temperature (in $°C$ ) of the human body can be calculated using the equation $3$ as follows:

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $3$ :

$T°C=98.6 °F−32×59=37 °C$

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

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

$TK=T°F−32×59+273.15$ $5$

Here, $TK$ is the temperature in kelvin and $T°F$ is the temperature in Fahrenheit.

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $5$ .

$TK=98.6 °F−32×59+273.5=37 °C+273.15=310.15 K$

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

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Answer 3

Textbook Question

Chapter 1, Problem 9PP

(a) The boiling point of acetylene is $− 28.1 ° C .$ 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 $98.6 ° F .$ What is this temperature on the Celsius and Kelvin scales?

(a)

Expert Solution

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:

$T°F=95T°C+32$ $1$

Here, $T°F$ is the temperature in degree Fahrenheit and $T°C$ is the temperature in degree Celsius.

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

$TK=T°C+273.15$ $2$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

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

$T°C=T°F−32×59$ $3$

Here, $T°C$ is the temperature in degree Celsius and $T°F$ is the temperature in degree Fahrenheit.

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, the value of $T°C$ as $−28.1 °C$ in the equation $2$ :

$TK=−28.1+273.15=245.05 K$

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

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, $T°C$ as $−28.1 °C$ in the equation $1$ :

$T°F=95−28.1 °C+32=−50.58+32=−18.58 °F$

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

(b)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

$T°C=TK−273.15$ $4$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

The boiling point of helium is given to be $4 K$ . Substitute $T°K$ as $4 K$ in the equation $4$ :

$T°C=4 K−273.15=(−269.15 °C)$

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

(c)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The temperature (in $°C$ ) of the human body can be calculated using the equation $3$ as follows:

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $3$ :

$T°C=98.6 °F−32×59=37 °C$

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

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

$TK=T°F−32×59+273.15$ $5$

Here, $TK$ is the temperature in kelvin and $T°F$ is the temperature in Fahrenheit.

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $5$ .

$TK=98.6 °F−32×59+273.5=37 °C+273.15=310.15 K$

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

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Answer 4

Textbook Question

Chapter 1, Problem 9PP

(a) The boiling point of acetylene is $− 28.1 ° C .$ 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 $98.6 ° F .$ What is this temperature on the Celsius and Kelvin scales?

(a)

Expert Solution

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:

$T°F=95T°C+32$ $1$

Here, $T°F$ is the temperature in degree Fahrenheit and $T°C$ is the temperature in degree Celsius.

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

$TK=T°C+273.15$ $2$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

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

$T°C=T°F−32×59$ $3$

Here, $T°C$ is the temperature in degree Celsius and $T°F$ is the temperature in degree Fahrenheit.

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, the value of $T°C$ as $−28.1 °C$ in the equation $2$ :

$TK=−28.1+273.15=245.05 K$

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

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, $T°C$ as $−28.1 °C$ in the equation $1$ :

$T°F=95−28.1 °C+32=−50.58+32=−18.58 °F$

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

(b)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

$T°C=TK−273.15$ $4$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

The boiling point of helium is given to be $4 K$ . Substitute $T°K$ as $4 K$ in the equation $4$ :

$T°C=4 K−273.15=(−269.15 °C)$

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

(c)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The temperature (in $°C$ ) of the human body can be calculated using the equation $3$ as follows:

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $3$ :

$T°C=98.6 °F−32×59=37 °C$

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

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

$TK=T°F−32×59+273.15$ $5$

Here, $TK$ is the temperature in kelvin and $T°F$ is the temperature in Fahrenheit.

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $5$ .

$TK=98.6 °F−32×59+273.5=37 °C+273.15=310.15 K$

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

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

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:

$T°F=95T°C+32$ $1$

Here, $T°F$ is the temperature in degree Fahrenheit and $T°C$ is the temperature in degree Celsius.

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

$TK=T°C+273.15$ $2$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

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

$T°C=T°F−32×59$ $3$

Here, $T°C$ is the temperature in degree Celsius and $T°F$ is the temperature in degree Fahrenheit.

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, the value of $T°C$ as $−28.1 °C$ in the equation $2$ :

$TK=−28.1+273.15=245.05 K$

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

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, $T°C$ as $−28.1 °C$ in the equation $1$ :

$T°F=95−28.1 °C+32=−50.58+32=−18.58 °F$

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

(b)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

$T°C=TK−273.15$ $4$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

The boiling point of helium is given to be $4 K$ . Substitute $T°K$ as $4 K$ in the equation $4$ :

$T°C=4 K−273.15=(−269.15 °C)$

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

(c)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The temperature (in $°C$ ) of the human body can be calculated using the equation $3$ as follows:

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $3$ :

$T°C=98.6 °F−32×59=37 °C$

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

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

$TK=T°F−32×59+273.15$ $5$

Here, $TK$ is the temperature in kelvin and $T°F$ is the temperature in Fahrenheit.

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $5$ .

$TK=98.6 °F−32×59+273.5=37 °C+273.15=310.15 K$

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

Answer 8

(a)

Expert Solution

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:

$T°F=95T°C+32$ $1$

Here, $T°F$ is the temperature in degree Fahrenheit and $T°C$ is the temperature in degree Celsius.

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

$TK=T°C+273.15$ $2$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

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

$T°C=T°F−32×59$ $3$

Here, $T°C$ is the temperature in degree Celsius and $T°F$ is the temperature in degree Fahrenheit.

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, the value of $T°C$ as $−28.1 °C$ in the equation $2$ :

$TK=−28.1+273.15=245.05 K$

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

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, $T°C$ as $−28.1 °C$ in the equation $1$ :

$T°F=95−28.1 °C+32=−50.58+32=−18.58 °F$

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

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation:

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

Answer 13

Explanation of Solution

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

$T°F=95T°C+32$ $1$

Here, $T°F$ is the temperature in degree Fahrenheit and $T°C$ is the temperature in degree Celsius.

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

$TK=T°C+273.15$ $2$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

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

$T°C=T°F−32×59$ $3$

Here, $T°C$ is the temperature in degree Celsius and $T°F$ is the temperature in degree Fahrenheit.

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, the value of $T°C$ as $−28.1 °C$ in the equation $2$ :

$TK=−28.1+273.15=245.05 K$

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

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

The boiling point of acetylene is given to be $−28.1 °C$ . Substitute, $T°C$ as $−28.1 °C$ in the equation $1$ :

$T°F=95−28.1 °C+32=−50.58+32=−18.58 °F$

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

Answer 14

(b)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

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

$T°C=TK−273.15$ $4$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

The boiling point of helium is given to be $4 K$ . Substitute $T°K$ as $4 K$ in the equation $4$ :

$T°C=4 K−273.15=(−269.15 °C)$

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

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Interpretation Introduction

Interpretation:

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

Answer 19

Explanation of Solution

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

$T°C=TK−273.15$ $4$

Here, $TK$ is the temperature in Kelvin and $T°C$ is the temperature in degree Celsius.

The boiling point of helium is given to be $4 K$ . Substitute $T°K$ as $4 K$ in the equation $4$ :

$T°C=4 K−273.15=(−269.15 °C)$

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

Answer 20

(c)

Expert Solution

Interpretation Introduction

Interpretation:

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

Explanation of Solution

The temperature (in $°C$ ) of the human body can be calculated using the equation $3$ as follows:

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $3$ :

$T°C=98.6 °F−32×59=37 °C$

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

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

$TK=T°F−32×59+273.15$ $5$

Here, $TK$ is the temperature in kelvin and $T°F$ is the temperature in Fahrenheit.

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $5$ .

$TK=98.6 °F−32×59+273.5=37 °C+273.15=310.15 K$

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

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Interpretation Introduction

Interpretation:

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

Answer 25

Explanation of Solution

The temperature (in $°C$ ) of the human body can be calculated using the equation $3$ as follows:

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $3$ :

$T°C=98.6 °F−32×59=37 °C$

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

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

$TK=T°F−32×59+273.15$ $5$

Here, $TK$ is the temperature in kelvin and $T°F$ is the temperature in Fahrenheit.

The normal body temperature of the human body is $98.6 °F$ . Substitute $T°F$ as $98.6 °F$ in equation $5$ .

$TK=98.6 °F−32×59+273.5=37 °C+273.15=310.15 K$

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

Answer 26

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