The molar heats of fusion and vaporization for water are 6 .0 2 kJ / mol and 4 0. 6 kJ / mol respectively, and The specific heat capacity of liquid water is 4 . 18 J / g ° C . What quantity of heat energy is required to melt 25 .0 g of ice at 0 ° C ? What quantity of heat is required for vaporize 37 . 5 g of liquid water at 1 00 ° C ? What quantity of heat is required to warm 55 . 2 g of liquid water from 0 ° C to 1 00 ° C ?

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Answer 1

Textbook Question

Chapter 14, Problem 17QAP

The molar heats of fusion and vaporization for water are 6 .0 2 kJ / mol and 4 0. 6 kJ / mol respectively, and The specific heat capacity of liquid water is 4 . 18 J / g ° C . What quantity of heat energy is required to melt 25 .0 g of ice at 0 ° C ? What quantity of heat is required for vaporize 37 . 5 g of liquid water at 1 00 ° C ? What quantity of heat is required to warm 55 . 2 g of liquid water from 0 ° C to 1 00 ° C ?

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The heat required to melt 25 g of ice at 00C, heat required to vaporize 37.5 g liquid water at 1000 C and heat required to warm 55.2 g water from 00 C to 1000 C must be calculated.

Concept Introduction:

For melting and vaporization latent heat is used but for increasing temperature specific heat is used.

Answer to Problem 17QAP

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00 C to 1000 C is 23.073 kJ.

Explanation of Solution

Given information:

Molar heat of fusion and vaporization of water is 6.02 kJ/mol and 40.6 kJ/mol respectively.

The specific heat capacity of water is 4.18 J/g0C.

Heat of fusion of ice is 6.02 kJ/mol

So fusion of 18 g ice needs 6.02 kJ heat energy.

Fusion of 1 g ice needs 6.02 kJ18 heat energy.

Fusion of 25 g ice needs (6.02×25) kJ18 heat energy = 150.5 kJ18 heat energy = 8.36 kJ

Heat of vaporization of water is 40.6 kJ/mol

So vaporization of 18 g water needs 40.6 kJ heat energy.

Vaporization of 1 g water needs 40.6 kJ18 heat energy.

Vaporization of 37.5 g water needs (40.6 kJ×37.5)18 heat energy = 1522.5 kJ18

=84.583 kJ

Heat required to warm 55.2 g water from 00 C to 1000 C is H=m×s×t where m, s, t are mass of water, specific heat capacity of water and temperature rise of water respectively.

Now putting all the values we get H=55.2 g×4.18 J/g0C×(100−0)0C=23.073 kJ

Conclusion

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00C to 1000C is 23.073 kJ.

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Answer 2

Textbook Question

Chapter 14, Problem 17QAP

The molar heats of fusion and vaporization for water are 6 .0 2 kJ / mol and 4 0. 6 kJ / mol respectively, and The specific heat capacity of liquid water is 4 . 18 J / g ° C . What quantity of heat energy is required to melt 25 .0 g of ice at 0 ° C ? What quantity of heat is required for vaporize 37 . 5 g of liquid water at 1 00 ° C ? What quantity of heat is required to warm 55 . 2 g of liquid water from 0 ° C to 1 00 ° C ?

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The heat required to melt 25 g of ice at 00C, heat required to vaporize 37.5 g liquid water at 1000 C and heat required to warm 55.2 g water from 00 C to 1000 C must be calculated.

Concept Introduction:

For melting and vaporization latent heat is used but for increasing temperature specific heat is used.

Answer to Problem 17QAP

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00 C to 1000 C is 23.073 kJ.

Explanation of Solution

Given information:

Molar heat of fusion and vaporization of water is 6.02 kJ/mol and 40.6 kJ/mol respectively.

The specific heat capacity of water is 4.18 J/g0C.

Heat of fusion of ice is 6.02 kJ/mol

So fusion of 18 g ice needs 6.02 kJ heat energy.

Fusion of 1 g ice needs 6.02 kJ18 heat energy.

Fusion of 25 g ice needs (6.02×25) kJ18 heat energy = 150.5 kJ18 heat energy = 8.36 kJ

Heat of vaporization of water is 40.6 kJ/mol

So vaporization of 18 g water needs 40.6 kJ heat energy.

Vaporization of 1 g water needs 40.6 kJ18 heat energy.

Vaporization of 37.5 g water needs (40.6 kJ×37.5)18 heat energy = 1522.5 kJ18

=84.583 kJ

Heat required to warm 55.2 g water from 00 C to 1000 C is H=m×s×t where m, s, t are mass of water, specific heat capacity of water and temperature rise of water respectively.

Now putting all the values we get H=55.2 g×4.18 J/g0C×(100−0)0C=23.073 kJ

Conclusion

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00C to 1000C is 23.073 kJ.

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Answer 3

Textbook Question

Chapter 14, Problem 17QAP

The molar heats of fusion and vaporization for water are 6 .0 2 kJ / mol and 4 0. 6 kJ / mol respectively, and The specific heat capacity of liquid water is 4 . 18 J / g ° C . What quantity of heat energy is required to melt 25 .0 g of ice at 0 ° C ? What quantity of heat is required for vaporize 37 . 5 g of liquid water at 1 00 ° C ? What quantity of heat is required to warm 55 . 2 g of liquid water from 0 ° C to 1 00 ° C ?

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The heat required to melt 25 g of ice at 00C, heat required to vaporize 37.5 g liquid water at 1000 C and heat required to warm 55.2 g water from 00 C to 1000 C must be calculated.

Concept Introduction:

For melting and vaporization latent heat is used but for increasing temperature specific heat is used.

Answer to Problem 17QAP

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00 C to 1000 C is 23.073 kJ.

Explanation of Solution

Given information:

Molar heat of fusion and vaporization of water is 6.02 kJ/mol and 40.6 kJ/mol respectively.

The specific heat capacity of water is 4.18 J/g0C.

Heat of fusion of ice is 6.02 kJ/mol

So fusion of 18 g ice needs 6.02 kJ heat energy.

Fusion of 1 g ice needs 6.02 kJ18 heat energy.

Fusion of 25 g ice needs (6.02×25) kJ18 heat energy = 150.5 kJ18 heat energy = 8.36 kJ

Heat of vaporization of water is 40.6 kJ/mol

So vaporization of 18 g water needs 40.6 kJ heat energy.

Vaporization of 1 g water needs 40.6 kJ18 heat energy.

Vaporization of 37.5 g water needs (40.6 kJ×37.5)18 heat energy = 1522.5 kJ18

=84.583 kJ

Heat required to warm 55.2 g water from 00 C to 1000 C is H=m×s×t where m, s, t are mass of water, specific heat capacity of water and temperature rise of water respectively.

Now putting all the values we get H=55.2 g×4.18 J/g0C×(100−0)0C=23.073 kJ

Conclusion

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00C to 1000C is 23.073 kJ.

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Answer 4

Textbook Question

Chapter 14, Problem 17QAP

The molar heats of fusion and vaporization for water are 6 .0 2 kJ / mol and 4 0. 6 kJ / mol respectively, and The specific heat capacity of liquid water is 4 . 18 J / g ° C . What quantity of heat energy is required to melt 25 .0 g of ice at 0 ° C ? What quantity of heat is required for vaporize 37 . 5 g of liquid water at 1 00 ° C ? What quantity of heat is required to warm 55 . 2 g of liquid water from 0 ° C to 1 00 ° C ?

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The heat required to melt 25 g of ice at 00C, heat required to vaporize 37.5 g liquid water at 1000 C and heat required to warm 55.2 g water from 00 C to 1000 C must be calculated.

Concept Introduction:

For melting and vaporization latent heat is used but for increasing temperature specific heat is used.

Answer to Problem 17QAP

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00 C to 1000 C is 23.073 kJ.

Explanation of Solution

Given information:

Molar heat of fusion and vaporization of water is 6.02 kJ/mol and 40.6 kJ/mol respectively.

The specific heat capacity of water is 4.18 J/g0C.

Heat of fusion of ice is 6.02 kJ/mol

So fusion of 18 g ice needs 6.02 kJ heat energy.

Fusion of 1 g ice needs 6.02 kJ18 heat energy.

Fusion of 25 g ice needs (6.02×25) kJ18 heat energy = 150.5 kJ18 heat energy = 8.36 kJ

Heat of vaporization of water is 40.6 kJ/mol

So vaporization of 18 g water needs 40.6 kJ heat energy.

Vaporization of 1 g water needs 40.6 kJ18 heat energy.

Vaporization of 37.5 g water needs (40.6 kJ×37.5)18 heat energy = 1522.5 kJ18

=84.583 kJ

Heat required to warm 55.2 g water from 00 C to 1000 C is H=m×s×t where m, s, t are mass of water, specific heat capacity of water and temperature rise of water respectively.

Now putting all the values we get H=55.2 g×4.18 J/g0C×(100−0)0C=23.073 kJ

Conclusion

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00C to 1000C is 23.073 kJ.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The heat required to melt 25 g of ice at 00C, heat required to vaporize 37.5 g liquid water at 1000 C and heat required to warm 55.2 g water from 00 C to 1000 C must be calculated.

Concept Introduction:

For melting and vaporization latent heat is used but for increasing temperature specific heat is used.

Answer to Problem 17QAP

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00 C to 1000 C is 23.073 kJ.

Explanation of Solution

Given information:

Molar heat of fusion and vaporization of water is 6.02 kJ/mol and 40.6 kJ/mol respectively.

The specific heat capacity of water is 4.18 J/g0C.

Heat of fusion of ice is 6.02 kJ/mol

So fusion of 18 g ice needs 6.02 kJ heat energy.

Fusion of 1 g ice needs 6.02 kJ18 heat energy.

Fusion of 25 g ice needs (6.02×25) kJ18 heat energy = 150.5 kJ18 heat energy = 8.36 kJ

Heat of vaporization of water is 40.6 kJ/mol

So vaporization of 18 g water needs 40.6 kJ heat energy.

Vaporization of 1 g water needs 40.6 kJ18 heat energy.

Vaporization of 37.5 g water needs (40.6 kJ×37.5)18 heat energy = 1522.5 kJ18

=84.583 kJ

Heat required to warm 55.2 g water from 00 C to 1000 C is H=m×s×t where m, s, t are mass of water, specific heat capacity of water and temperature rise of water respectively.

Now putting all the values we get H=55.2 g×4.18 J/g0C×(100−0)0C=23.073 kJ

Conclusion

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00C to 1000C is 23.073 kJ.

Answer 8

Expert Solution & Answer

Interpretation Introduction

Interpretation:

The heat required to melt 25 g of ice at 00C, heat required to vaporize 37.5 g liquid water at 1000 C and heat required to warm 55.2 g water from 00 C to 1000 C must be calculated.

Concept Introduction:

For melting and vaporization latent heat is used but for increasing temperature specific heat is used.

Answer to Problem 17QAP

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00 C to 1000 C is 23.073 kJ.

Explanation of Solution

Given information:

Molar heat of fusion and vaporization of water is 6.02 kJ/mol and 40.6 kJ/mol respectively.

The specific heat capacity of water is 4.18 J/g0C.

Heat of fusion of ice is 6.02 kJ/mol

So fusion of 18 g ice needs 6.02 kJ heat energy.

Fusion of 1 g ice needs 6.02 kJ18 heat energy.

Fusion of 25 g ice needs (6.02×25) kJ18 heat energy = 150.5 kJ18 heat energy = 8.36 kJ

Heat of vaporization of water is 40.6 kJ/mol

So vaporization of 18 g water needs 40.6 kJ heat energy.

Vaporization of 1 g water needs 40.6 kJ18 heat energy.

Vaporization of 37.5 g water needs (40.6 kJ×37.5)18 heat energy = 1522.5 kJ18

=84.583 kJ

Heat required to warm 55.2 g water from 00 C to 1000 C is H=m×s×t where m, s, t are mass of water, specific heat capacity of water and temperature rise of water respectively.

Now putting all the values we get H=55.2 g×4.18 J/g0C×(100−0)0C=23.073 kJ

Conclusion

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00C to 1000C is 23.073 kJ.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Interpretation Introduction

Interpretation:

The heat required to melt 25 g of ice at 00C, heat required to vaporize 37.5 g liquid water at 1000 C and heat required to warm 55.2 g water from 00 C to 1000 C must be calculated.

Concept Introduction:

For melting and vaporization latent heat is used but for increasing temperature specific heat is used.

Answer 12

Answer to Problem 17QAP

The heat required to melt 25 g of ice at 00 C is 8.36 kJ.

The heat required to vaporize 37.5 g liquid water at 1000 C is 84.583 kJ.

The heat required to warm 55.2 g water from 00 C to 1000 C is 23.073 kJ.

Answer 13

Explanation of Solution

Given information:

Molar heat of fusion and vaporization of water is 6.02 kJ/mol and 40.6 kJ/mol respectively.

The specific heat capacity of water is 4.18 J/g0C.

Heat of fusion of ice is 6.02 kJ/mol

So fusion of 18 g ice needs 6.02 kJ heat energy.

Fusion of 1 g ice needs 6.02 kJ18 heat energy.

Fusion of 25 g ice needs (6.02×25) kJ18 heat energy = 150.5 kJ18 heat energy = 8.36 kJ

Heat of vaporization of water is 40.6 kJ/mol

So vaporization of 18 g water needs 40.6 kJ heat energy.

Vaporization of 1 g water needs 40.6 kJ18 heat energy.

Vaporization of 37.5 g water needs (40.6 kJ×37.5)18 heat energy = 1522.5 kJ18

=84.583 kJ

Heat required to warm 55.2 g water from 00 C to 1000 C is H=m×s×t where m, s, t are mass of water, specific heat capacity of water and temperature rise of water respectively.