You have a l.00-mole sample of water at −30.°C and you heat it until you have gaseous water at 140.°C. Calculate q for the entire process. Use the following data. Specific heat capacity of ice = 2.03 J/ ° C ⋅ g Specific heat capacity of water = 4.18 J/ ° C ⋅ g Specific heat capacity of steam = 2.02 J/ ° C ⋅ g H 2 O ( s ) → H 2 O ( l ) Δ H fusion = 6.02 KJ/mol ( at 0 ° C ) H 2 O ( l ) → H 2 O ( g ) Δ H vaporization = 40.7 KJ/mol ( at 100 . ° C )

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 6, Problem 131CP

You have a l.00-mole sample of water at −30.°C and you heat it until you have gaseous water at 140.°C. Calculate q for the entire process. Use the following data.

$Specific heat capacity of ice = 2.03 J/ ° C ⋅ g Specific heat capacity of water = 4.18 J/ ° C ⋅ g Specific heat capacity of steam = 2.02 J/ ° C ⋅ g$

$H 2 O ( s ) → H 2 O ( l ) Δ H fusion = 6.02 KJ/mol ( at 0 ° C ) H 2 O ( l ) → H 2 O ( g ) Δ H vaporization = 40.7 KJ/mol ( at 100 . ° C )$

Expert Solution & Answer

Interpretation Introduction

Interpretation: The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ should be determined.

Concept Introduction:

The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

$C = Heat absorbedTemperature change......(1)$

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

$Absorbed heat (J) = Specific heat capacity×Temperature change(c)×mass ofsubstance (g)......(2)$

For the above equation heat is:

$S = q×M×T.....(1)$

q is heat (J)

M is mass of sample $(g)$

S is specific heat capacity $(J/°C·g.)$

T is temperature change $(C)$

For the process no heat loss to the surroundings means then the heat is

$(absorbed)-q×M×ΔT=-q×M×ΔT (released)......(3)$ '

Answer to Problem 131CP

Answer

The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ is $56.9 kJ$ .

Explanation of Solution

Explanation

Record the given data:

Specific heat capacity of ice $= 2.03J/°C· g$

Specific heat capacity of water $= 4.18 J/°C · g$

Specific heat capacity of steam $= 2.02J/°C· g$

Sample water is l.00 mole

Initial temperature is $- 30°C$

Final temperature is $- 140°C$ .

$H2O(s) →H2O(l) ΔΗfusion= 6.02kJ/mol (at 0°C)H2O(l)→H2O(g) ΔΗvaporization= 40.7kJ/mol (at 100°C)$

To calculate the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .

$q(1) = 2.30J°C.g×18.2g×(0-(-30))°C = 1.1 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .
Required heat for vaporization of 1 mole of water is $1.1 ×103 J$ .

To calculate the required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into 1 mol $H2O(l)$ at $0°C$ .

$q(1) = 1.00 mol ×6.02×103 J/mol = 6.02×103 J$

The given enthalpy of fusion is multiplied by the mole of water to give required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ .
Required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ is $6.02×103 J$ .

To calculate the required to heat $H2O(l)$ from $0°C$ to $100°C$ .

$q(3) = 4.18J°C.g×18.2g×(100-0)°C = 7.53 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(l)$ from $0°C$ to $100°C$ .
Required to heat $H2O(l)$ from $0°C$ to $100°C$ is $7.53 ×103 J$ .

To calculate the required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

$q(4) = 1.00 mol ×40.7×103 J/mol = 4.07 ×104 J$

The given enthalpy of vaporization is multiplied by the mole of water to give required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

To calculate the required to heat $H2O(g)$ from $100°C$ to $140°C$ .

$q(5) = 2.30J°C.g×18.2g×(140−100)°C = 1.5 ×103 J$

The given values are plugging in equation (1) to give required to heat $H2O(g)$ from $100°C$ to $140°C$ .
Required to heat $H2O(g)$ from $100°C$ to $140°C$ is $1.5 ×103 J$ .

To calculate the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ .

$qtotal= q1+q2+q3+q4+q5qtotal= 1.1 ×103 J+6.02×103 J+ 7.53 ×103 J+4.07 ×104 J+1.5 ×103 J = 5.69 ×104 J = 56.9 kJ$

Conclusion

Conclusion

The heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was calculated by using given specific heat capacities of water and temperatures and enthalpies and the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was found to be $56.9 kJ$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Highlight the chirality (or stereogenic) center(s) in the given compound. A compound may have one or more stereogenic centers. OH OH OH OH OH OH

Using wedge-and-dash bonds, modify the bonds on the chiral carbon in the molecule below so the molecule has R stereochemical configuration. NH H Br X टे

Provide photos of models of the following molecules. (Include a key for identification of the atoms) 1,2-dichloropropane 2,3,3-trimethylhexane 2-bromo-3-methybutane

Answer 2

Textbook Question

Chapter 6, Problem 131CP

You have a l.00-mole sample of water at −30.°C and you heat it until you have gaseous water at 140.°C. Calculate q for the entire process. Use the following data.

$Specific heat capacity of ice = 2.03 J/ ° C ⋅ g Specific heat capacity of water = 4.18 J/ ° C ⋅ g Specific heat capacity of steam = 2.02 J/ ° C ⋅ g$

$H 2 O ( s ) → H 2 O ( l ) Δ H fusion = 6.02 KJ/mol ( at 0 ° C ) H 2 O ( l ) → H 2 O ( g ) Δ H vaporization = 40.7 KJ/mol ( at 100 . ° C )$

Expert Solution & Answer

Interpretation Introduction

Interpretation: The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ should be determined.

Concept Introduction:

The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

$C = Heat absorbedTemperature change......(1)$

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

$Absorbed heat (J) = Specific heat capacity×Temperature change(c)×mass ofsubstance (g)......(2)$

For the above equation heat is:

$S = q×M×T.....(1)$

q is heat (J)

M is mass of sample $(g)$

S is specific heat capacity $(J/°C·g.)$

T is temperature change $(C)$

For the process no heat loss to the surroundings means then the heat is

$(absorbed)-q×M×ΔT=-q×M×ΔT (released)......(3)$ '

Answer to Problem 131CP

Answer

The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ is $56.9 kJ$ .

Explanation of Solution

Explanation

Record the given data:

Specific heat capacity of ice $= 2.03J/°C· g$

Specific heat capacity of water $= 4.18 J/°C · g$

Specific heat capacity of steam $= 2.02J/°C· g$

Sample water is l.00 mole

Initial temperature is $- 30°C$

Final temperature is $- 140°C$ .

$H2O(s) →H2O(l) ΔΗfusion= 6.02kJ/mol (at 0°C)H2O(l)→H2O(g) ΔΗvaporization= 40.7kJ/mol (at 100°C)$

To calculate the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .

$q(1) = 2.30J°C.g×18.2g×(0-(-30))°C = 1.1 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .
Required heat for vaporization of 1 mole of water is $1.1 ×103 J$ .

To calculate the required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into 1 mol $H2O(l)$ at $0°C$ .

$q(1) = 1.00 mol ×6.02×103 J/mol = 6.02×103 J$

The given enthalpy of fusion is multiplied by the mole of water to give required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ .
Required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ is $6.02×103 J$ .

To calculate the required to heat $H2O(l)$ from $0°C$ to $100°C$ .

$q(3) = 4.18J°C.g×18.2g×(100-0)°C = 7.53 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(l)$ from $0°C$ to $100°C$ .
Required to heat $H2O(l)$ from $0°C$ to $100°C$ is $7.53 ×103 J$ .

To calculate the required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

$q(4) = 1.00 mol ×40.7×103 J/mol = 4.07 ×104 J$

The given enthalpy of vaporization is multiplied by the mole of water to give required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

To calculate the required to heat $H2O(g)$ from $100°C$ to $140°C$ .

$q(5) = 2.30J°C.g×18.2g×(140−100)°C = 1.5 ×103 J$

The given values are plugging in equation (1) to give required to heat $H2O(g)$ from $100°C$ to $140°C$ .
Required to heat $H2O(g)$ from $100°C$ to $140°C$ is $1.5 ×103 J$ .

To calculate the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ .

$qtotal= q1+q2+q3+q4+q5qtotal= 1.1 ×103 J+6.02×103 J+ 7.53 ×103 J+4.07 ×104 J+1.5 ×103 J = 5.69 ×104 J = 56.9 kJ$

Conclusion

Conclusion

The heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was calculated by using given specific heat capacities of water and temperatures and enthalpies and the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was found to be $56.9 kJ$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Textbook Question

Chapter 6, Problem 131CP

You have a l.00-mole sample of water at −30.°C and you heat it until you have gaseous water at 140.°C. Calculate q for the entire process. Use the following data.

$Specific heat capacity of ice = 2.03 J/ ° C ⋅ g Specific heat capacity of water = 4.18 J/ ° C ⋅ g Specific heat capacity of steam = 2.02 J/ ° C ⋅ g$

$H 2 O ( s ) → H 2 O ( l ) Δ H fusion = 6.02 KJ/mol ( at 0 ° C ) H 2 O ( l ) → H 2 O ( g ) Δ H vaporization = 40.7 KJ/mol ( at 100 . ° C )$

Expert Solution & Answer

Interpretation Introduction

Interpretation: The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ should be determined.

Concept Introduction:

The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

$C = Heat absorbedTemperature change......(1)$

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

$Absorbed heat (J) = Specific heat capacity×Temperature change(c)×mass ofsubstance (g)......(2)$

For the above equation heat is:

$S = q×M×T.....(1)$

q is heat (J)

M is mass of sample $(g)$

S is specific heat capacity $(J/°C·g.)$

T is temperature change $(C)$

For the process no heat loss to the surroundings means then the heat is

$(absorbed)-q×M×ΔT=-q×M×ΔT (released)......(3)$ '

Answer to Problem 131CP

Answer

The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ is $56.9 kJ$ .

Explanation of Solution

Explanation

Record the given data:

Specific heat capacity of ice $= 2.03J/°C· g$

Specific heat capacity of water $= 4.18 J/°C · g$

Specific heat capacity of steam $= 2.02J/°C· g$

Sample water is l.00 mole

Initial temperature is $- 30°C$

Final temperature is $- 140°C$ .

$H2O(s) →H2O(l) ΔΗfusion= 6.02kJ/mol (at 0°C)H2O(l)→H2O(g) ΔΗvaporization= 40.7kJ/mol (at 100°C)$

To calculate the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .

$q(1) = 2.30J°C.g×18.2g×(0-(-30))°C = 1.1 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .
Required heat for vaporization of 1 mole of water is $1.1 ×103 J$ .

To calculate the required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into 1 mol $H2O(l)$ at $0°C$ .

$q(1) = 1.00 mol ×6.02×103 J/mol = 6.02×103 J$

The given enthalpy of fusion is multiplied by the mole of water to give required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ .
Required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ is $6.02×103 J$ .

To calculate the required to heat $H2O(l)$ from $0°C$ to $100°C$ .

$q(3) = 4.18J°C.g×18.2g×(100-0)°C = 7.53 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(l)$ from $0°C$ to $100°C$ .
Required to heat $H2O(l)$ from $0°C$ to $100°C$ is $7.53 ×103 J$ .

To calculate the required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

$q(4) = 1.00 mol ×40.7×103 J/mol = 4.07 ×104 J$

The given enthalpy of vaporization is multiplied by the mole of water to give required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

To calculate the required to heat $H2O(g)$ from $100°C$ to $140°C$ .

$q(5) = 2.30J°C.g×18.2g×(140−100)°C = 1.5 ×103 J$

The given values are plugging in equation (1) to give required to heat $H2O(g)$ from $100°C$ to $140°C$ .
Required to heat $H2O(g)$ from $100°C$ to $140°C$ is $1.5 ×103 J$ .

To calculate the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ .

$qtotal= q1+q2+q3+q4+q5qtotal= 1.1 ×103 J+6.02×103 J+ 7.53 ×103 J+4.07 ×104 J+1.5 ×103 J = 5.69 ×104 J = 56.9 kJ$

Conclusion

Conclusion

The heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was calculated by using given specific heat capacities of water and temperatures and enthalpies and the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was found to be $56.9 kJ$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Textbook Question

Chapter 6, Problem 131CP

You have a l.00-mole sample of water at −30.°C and you heat it until you have gaseous water at 140.°C. Calculate q for the entire process. Use the following data.

$Specific heat capacity of ice = 2.03 J/ ° C ⋅ g Specific heat capacity of water = 4.18 J/ ° C ⋅ g Specific heat capacity of steam = 2.02 J/ ° C ⋅ g$

$H 2 O ( s ) → H 2 O ( l ) Δ H fusion = 6.02 KJ/mol ( at 0 ° C ) H 2 O ( l ) → H 2 O ( g ) Δ H vaporization = 40.7 KJ/mol ( at 100 . ° C )$

Expert Solution & Answer

Interpretation Introduction

Interpretation: The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ should be determined.

Concept Introduction:

The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

$C = Heat absorbedTemperature change......(1)$

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

$Absorbed heat (J) = Specific heat capacity×Temperature change(c)×mass ofsubstance (g)......(2)$

For the above equation heat is:

$S = q×M×T.....(1)$

q is heat (J)

M is mass of sample $(g)$

S is specific heat capacity $(J/°C·g.)$

T is temperature change $(C)$

For the process no heat loss to the surroundings means then the heat is

$(absorbed)-q×M×ΔT=-q×M×ΔT (released)......(3)$ '

Answer to Problem 131CP

Answer

The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ is $56.9 kJ$ .

Explanation of Solution

Explanation

Record the given data:

Specific heat capacity of ice $= 2.03J/°C· g$

Specific heat capacity of water $= 4.18 J/°C · g$

Specific heat capacity of steam $= 2.02J/°C· g$

Sample water is l.00 mole

Initial temperature is $- 30°C$

Final temperature is $- 140°C$ .

$H2O(s) →H2O(l) ΔΗfusion= 6.02kJ/mol (at 0°C)H2O(l)→H2O(g) ΔΗvaporization= 40.7kJ/mol (at 100°C)$

To calculate the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .

$q(1) = 2.30J°C.g×18.2g×(0-(-30))°C = 1.1 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .
Required heat for vaporization of 1 mole of water is $1.1 ×103 J$ .

To calculate the required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into 1 mol $H2O(l)$ at $0°C$ .

$q(1) = 1.00 mol ×6.02×103 J/mol = 6.02×103 J$

The given enthalpy of fusion is multiplied by the mole of water to give required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ .
Required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ is $6.02×103 J$ .

To calculate the required to heat $H2O(l)$ from $0°C$ to $100°C$ .

$q(3) = 4.18J°C.g×18.2g×(100-0)°C = 7.53 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(l)$ from $0°C$ to $100°C$ .
Required to heat $H2O(l)$ from $0°C$ to $100°C$ is $7.53 ×103 J$ .

To calculate the required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

$q(4) = 1.00 mol ×40.7×103 J/mol = 4.07 ×104 J$

The given enthalpy of vaporization is multiplied by the mole of water to give required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

To calculate the required to heat $H2O(g)$ from $100°C$ to $140°C$ .

$q(5) = 2.30J°C.g×18.2g×(140−100)°C = 1.5 ×103 J$

The given values are plugging in equation (1) to give required to heat $H2O(g)$ from $100°C$ to $140°C$ .
Required to heat $H2O(g)$ from $100°C$ to $140°C$ is $1.5 ×103 J$ .

To calculate the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ .

$qtotal= q1+q2+q3+q4+q5qtotal= 1.1 ×103 J+6.02×103 J+ 7.53 ×103 J+4.07 ×104 J+1.5 ×103 J = 5.69 ×104 J = 56.9 kJ$

Conclusion

Conclusion

The heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was calculated by using given specific heat capacities of water and temperatures and enthalpies and the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was found to be $56.9 kJ$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation: The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ should be determined.

Concept Introduction:

The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

$C = Heat absorbedTemperature change......(1)$

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

$Absorbed heat (J) = Specific heat capacity×Temperature change(c)×mass ofsubstance (g)......(2)$

For the above equation heat is:

$S = q×M×T.....(1)$

q is heat (J)

M is mass of sample $(g)$

S is specific heat capacity $(J/°C·g.)$

T is temperature change $(C)$

For the process no heat loss to the surroundings means then the heat is

$(absorbed)-q×M×ΔT=-q×M×ΔT (released)......(3)$ '

Answer to Problem 131CP

Answer

The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ is $56.9 kJ$ .

Explanation of Solution

Explanation

Record the given data:

Specific heat capacity of ice $= 2.03J/°C· g$

Specific heat capacity of water $= 4.18 J/°C · g$

Specific heat capacity of steam $= 2.02J/°C· g$

Sample water is l.00 mole

Initial temperature is $- 30°C$

Final temperature is $- 140°C$ .

$H2O(s) →H2O(l) ΔΗfusion= 6.02kJ/mol (at 0°C)H2O(l)→H2O(g) ΔΗvaporization= 40.7kJ/mol (at 100°C)$

To calculate the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .

$q(1) = 2.30J°C.g×18.2g×(0-(-30))°C = 1.1 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .
Required heat for vaporization of 1 mole of water is $1.1 ×103 J$ .

To calculate the required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into 1 mol $H2O(l)$ at $0°C$ .

$q(1) = 1.00 mol ×6.02×103 J/mol = 6.02×103 J$

The given enthalpy of fusion is multiplied by the mole of water to give required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ .
Required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ is $6.02×103 J$ .

To calculate the required to heat $H2O(l)$ from $0°C$ to $100°C$ .

$q(3) = 4.18J°C.g×18.2g×(100-0)°C = 7.53 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(l)$ from $0°C$ to $100°C$ .
Required to heat $H2O(l)$ from $0°C$ to $100°C$ is $7.53 ×103 J$ .

To calculate the required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

$q(4) = 1.00 mol ×40.7×103 J/mol = 4.07 ×104 J$

The given enthalpy of vaporization is multiplied by the mole of water to give required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

To calculate the required to heat $H2O(g)$ from $100°C$ to $140°C$ .

$q(5) = 2.30J°C.g×18.2g×(140−100)°C = 1.5 ×103 J$

The given values are plugging in equation (1) to give required to heat $H2O(g)$ from $100°C$ to $140°C$ .
Required to heat $H2O(g)$ from $100°C$ to $140°C$ is $1.5 ×103 J$ .

To calculate the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ .

$qtotal= q1+q2+q3+q4+q5qtotal= 1.1 ×103 J+6.02×103 J+ 7.53 ×103 J+4.07 ×104 J+1.5 ×103 J = 5.69 ×104 J = 56.9 kJ$

Conclusion

Conclusion

The heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was calculated by using given specific heat capacities of water and temperatures and enthalpies and the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was found to be $56.9 kJ$ .

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation: The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ should be determined.

Concept Introduction:

The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

$C = Heat absorbedTemperature change......(1)$

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

$Absorbed heat (J) = Specific heat capacity×Temperature change(c)×mass ofsubstance (g)......(2)$

For the above equation heat is:

$S = q×M×T.....(1)$

q is heat (J)

M is mass of sample $(g)$

S is specific heat capacity $(J/°C·g.)$

T is temperature change $(C)$

For the process no heat loss to the surroundings means then the heat is

$(absorbed)-q×M×ΔT=-q×M×ΔT (released)......(3)$ '

Answer to Problem 131CP

Answer

The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ is $56.9 kJ$ .

Explanation of Solution

Explanation

Record the given data:

Specific heat capacity of ice $= 2.03J/°C· g$

Specific heat capacity of water $= 4.18 J/°C · g$

Specific heat capacity of steam $= 2.02J/°C· g$

Sample water is l.00 mole

Initial temperature is $- 30°C$

Final temperature is $- 140°C$ .

$H2O(s) →H2O(l) ΔΗfusion= 6.02kJ/mol (at 0°C)H2O(l)→H2O(g) ΔΗvaporization= 40.7kJ/mol (at 100°C)$

To calculate the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .

$q(1) = 2.30J°C.g×18.2g×(0-(-30))°C = 1.1 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .
Required heat for vaporization of 1 mole of water is $1.1 ×103 J$ .

To calculate the required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into 1 mol $H2O(l)$ at $0°C$ .

$q(1) = 1.00 mol ×6.02×103 J/mol = 6.02×103 J$

The given enthalpy of fusion is multiplied by the mole of water to give required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ .
Required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ is $6.02×103 J$ .

To calculate the required to heat $H2O(l)$ from $0°C$ to $100°C$ .

$q(3) = 4.18J°C.g×18.2g×(100-0)°C = 7.53 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(l)$ from $0°C$ to $100°C$ .
Required to heat $H2O(l)$ from $0°C$ to $100°C$ is $7.53 ×103 J$ .

To calculate the required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

$q(4) = 1.00 mol ×40.7×103 J/mol = 4.07 ×104 J$

The given enthalpy of vaporization is multiplied by the mole of water to give required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

To calculate the required to heat $H2O(g)$ from $100°C$ to $140°C$ .

$q(5) = 2.30J°C.g×18.2g×(140−100)°C = 1.5 ×103 J$

The given values are plugging in equation (1) to give required to heat $H2O(g)$ from $100°C$ to $140°C$ .
Required to heat $H2O(g)$ from $100°C$ to $140°C$ is $1.5 ×103 J$ .

To calculate the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ .

$qtotal= q1+q2+q3+q4+q5qtotal= 1.1 ×103 J+6.02×103 J+ 7.53 ×103 J+4.07 ×104 J+1.5 ×103 J = 5.69 ×104 J = 56.9 kJ$

Conclusion

Conclusion

The heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was calculated by using given specific heat capacities of water and temperatures and enthalpies and the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was found to be $56.9 kJ$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation: The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ should be determined.

Concept Introduction:

The heat capacity C is defined as the relation of heat absorbed to the temperature change. It can be given by,

$C = Heat absorbedTemperature change......(1)$

Require heat for an one gram of substance raise to its temperature by one degree Celsius is called specific heat capacity.

$Absorbed heat (J) = Specific heat capacity×Temperature change(c)×mass ofsubstance (g)......(2)$

For the above equation heat is:

$S = q×M×T.....(1)$

q is heat (J)

M is mass of sample $(g)$

S is specific heat capacity $(J/°C·g.)$

T is temperature change $(C)$

For the process no heat loss to the surroundings means then the heat is

$(absorbed)-q×M×ΔT=-q×M×ΔT (released)......(3)$ '

Answer 12

Answer to Problem 131CP

Answer

The amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ is $56.9 kJ$ .

Answer 13

Explanation of Solution

Explanation

Record the given data:

Specific heat capacity of ice $= 2.03J/°C· g$

Specific heat capacity of water $= 4.18 J/°C · g$

Specific heat capacity of steam $= 2.02J/°C· g$

Sample water is l.00 mole

Initial temperature is $- 30°C$

Final temperature is $- 140°C$ .

$H2O(s) →H2O(l) ΔΗfusion= 6.02kJ/mol (at 0°C)H2O(l)→H2O(g) ΔΗvaporization= 40.7kJ/mol (at 100°C)$

To calculate the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .

$q(1) = 2.30J°C.g×18.2g×(0-(-30))°C = 1.1 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(s)$ from $-30°C$ to $0°C$ .
Required heat for vaporization of 1 mole of water is $1.1 ×103 J$ .

To calculate the required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into 1 mol $H2O(l)$ at $0°C$ .

$q(1) = 1.00 mol ×6.02×103 J/mol = 6.02×103 J$

The given enthalpy of fusion is multiplied by the mole of water to give required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ .
Required heat (q) to convert 1 mol $H2O(s)$ at $0°C$ into $H2O(l)$ at $0°C$ is $6.02×103 J$ .

To calculate the required to heat $H2O(l)$ from $0°C$ to $100°C$ .

$q(3) = 4.18J°C.g×18.2g×(100-0)°C = 7.53 ×103 J$

The given values are plugging in equation (1) to give the required to heat $H2O(l)$ from $0°C$ to $100°C$ .
Required to heat $H2O(l)$ from $0°C$ to $100°C$ is $7.53 ×103 J$ .

To calculate the required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

$q(4) = 1.00 mol ×40.7×103 J/mol = 4.07 ×104 J$

The given enthalpy of vaporization is multiplied by the mole of water to give required to convert 1 mol $H2O(l)$ at $100°C$ into 1 mol $H2O(g)$ at $100°C$ .

To calculate the required to heat $H2O(g)$ from $100°C$ to $140°C$ .

$q(5) = 2.30J°C.g×18.2g×(140−100)°C = 1.5 ×103 J$

The given values are plugging in equation (1) to give required to heat $H2O(g)$ from $100°C$ to $140°C$ .
Required to heat $H2O(g)$ from $100°C$ to $140°C$ is $1.5 ×103 J$ .

To calculate the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ .

$qtotal= q1+q2+q3+q4+q5qtotal= 1.1 ×103 J+6.02×103 J+ 7.53 ×103 J+4.07 ×104 J+1.5 ×103 J = 5.69 ×104 J = 56.9 kJ$

Answer 14

Conclusion

Conclusion

The heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was calculated by using given specific heat capacities of water and temperatures and enthalpies and the amount of heat required to convert 1.00 mol of $H2O(s)$ at $-30°C$ to $H2O(g)$ at $140°C$ was found to be $56.9 kJ$ .

Answer 15

Ch. 6 - Define the following terms: potential energy,...Ch. 6 - Consider the following potential energy diagrams...Ch. 6 - What is the first law of thermodynamics? How can a...Ch. 6 - When a gas expands, what is the sign of w? Why?...Ch. 6 - What is the heat gained/released at constant...Ch. 6 - High-quality audio amplifiers generate large...Ch. 6 - Explain how calorimetry works to calculate H or E...Ch. 6 - What is Hesss law? When a reaction is reversed,...Ch. 6 - Define the standard enthalpy of formation. What...Ch. 6 - Objects placed together eventually reach the same...

Concept explainers

Videos

Answer to Problem 131CP

Explanation of Solution

Want to see more full solutions like this?

Chapter 6 Solutions