The heat released when steam is converted into ice is to be calculated. Concept Introduction: Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process, q . The heat absorbed or released by a substance is calculated by the following equation: q = m × C × Δ T â€¦â€¦(1) Here, m is the mass, Δ T is the temperature change, and C is the specific heat capacity. The expression to calculate the value of Δ T : Δ T = T 2 − T 1 â€¦..(2) Here T 1 is the initial temperature and T 2 is the final temperature. Substitute (2) in (1) q = m × C × T 2 − T 1 â€¦â€¦(3)

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Answer 1

Question

Chapter 10, Problem 38QP

Interpretation Introduction

Interpretation:

The heat released when steam is converted into ice is to be calculated.

Concept Introduction:

Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process, q .

The heat absorbed or released by a substance is calculated by the following equation:

q=m×C×ΔT â€¦â€¦(1)

Here, m is the mass, ΔT is the temperature change, and C is the specific heat capacity.

The expression to calculate the value of ΔT :

ΔT=T2−T1 â€¦..(2)

Here T1 is the initial temperature and T2 is the final temperature.

Substitute (2) in (1)

q=m×C×T2−T1 â€¦â€¦(3)

Expert Solution & Answer

Answer to Problem 38QP

Solution:

The heat released in the process of converting steam to ice is −98.27 kJ .

Explanation of Solution

Given Information: Mass of steam is 32.0 g , the temperature of steam is 115.0°C , the

specific heat of ice is 2.03 J g−1°C−1 , the heat of fusion of ice is 6.01×103 J mol−1 , the

the specific heat of water is 4.18 J g−1°C−1 , the heat of vaporization of water is 4.07×104 J mol−1 and the specific heat of steam is 2.02 J g−1°C−1 .

Calculate the heat lost on cooling steam from 115.0°C to 100.0°C .

Using equation (3) to calculate the heat required to cool the temperature of ice from 115°C to 100°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, 100°C for T2 and 115°C for T1 in equation (3):

q1=32 g×2.03 J g−1°C−1×100°C−115°C=−969.6 J

Calculate the heat lost to condense steam to water by multiplying the molar heat of condensation by the number of moles, n .

q2=n×− molar heat of vaporization=32 g18.02 g mol−1 ×−4.07×104J mol−1=−72275.25 J

Using equation (3) to calculate the heat required to cool the temperature of water from 100°C to 0°C .

Substitute 32 g for m , 4.18 J g−1°C−1 for C as the specific heat of ice, 0.0°C for T2 and 100°C for T1 in equation (3):

q3=32 g×4.18 J g−1°C−1×0.0°C−100.0°C =−13376 J

Now, calculate the heat lost on the freezing of water by multiplying the molar heat of freezing by the number of moles.

q4=n×− molar heat of fusion=32 g18.02 g mol−1 ×−6.01×103 J mol−1=−10672.59 J

Using equation (3) to calculate the heat required to cool the temperature of ice from 0°C to −15°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, −15°C for T2 and 0°C for T1 in equation (3):

q5=32 g×2.03 J g−1°C−1×−15.0°C−0.0°C=− 974.4 J

Calculate the total heat released in the entire process by adding the heats lost for the individual steps as follows.

qtotal=q1+q2+q3+q4+q5= −969.6 J − 72275.25 J − 13376 J − 10672.59 J − 974.4 J= −98267.84 J= −98.27 kJ

The total heat released in the process is −98.27 kJ .

Conclusion

The total heat released in the process is −98.27 kJ .

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Answer 2

Question

Chapter 10, Problem 38QP

Interpretation Introduction

Interpretation:

The heat released when steam is converted into ice is to be calculated.

Concept Introduction:

Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process, q .

The heat absorbed or released by a substance is calculated by the following equation:

q=m×C×ΔT â€¦â€¦(1)

Here, m is the mass, ΔT is the temperature change, and C is the specific heat capacity.

The expression to calculate the value of ΔT :

ΔT=T2−T1 â€¦..(2)

Here T1 is the initial temperature and T2 is the final temperature.

Substitute (2) in (1)

q=m×C×T2−T1 â€¦â€¦(3)

Expert Solution & Answer

Answer to Problem 38QP

Solution:

The heat released in the process of converting steam to ice is −98.27 kJ .

Explanation of Solution

Given Information: Mass of steam is 32.0 g , the temperature of steam is 115.0°C , the

specific heat of ice is 2.03 J g−1°C−1 , the heat of fusion of ice is 6.01×103 J mol−1 , the

the specific heat of water is 4.18 J g−1°C−1 , the heat of vaporization of water is 4.07×104 J mol−1 and the specific heat of steam is 2.02 J g−1°C−1 .

Calculate the heat lost on cooling steam from 115.0°C to 100.0°C .

Using equation (3) to calculate the heat required to cool the temperature of ice from 115°C to 100°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, 100°C for T2 and 115°C for T1 in equation (3):

q1=32 g×2.03 J g−1°C−1×100°C−115°C=−969.6 J

Calculate the heat lost to condense steam to water by multiplying the molar heat of condensation by the number of moles, n .

q2=n×− molar heat of vaporization=32 g18.02 g mol−1 ×−4.07×104J mol−1=−72275.25 J

Using equation (3) to calculate the heat required to cool the temperature of water from 100°C to 0°C .

Substitute 32 g for m , 4.18 J g−1°C−1 for C as the specific heat of ice, 0.0°C for T2 and 100°C for T1 in equation (3):

q3=32 g×4.18 J g−1°C−1×0.0°C−100.0°C =−13376 J

Now, calculate the heat lost on the freezing of water by multiplying the molar heat of freezing by the number of moles.

q4=n×− molar heat of fusion=32 g18.02 g mol−1 ×−6.01×103 J mol−1=−10672.59 J

Using equation (3) to calculate the heat required to cool the temperature of ice from 0°C to −15°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, −15°C for T2 and 0°C for T1 in equation (3):

q5=32 g×2.03 J g−1°C−1×−15.0°C−0.0°C=− 974.4 J

Calculate the total heat released in the entire process by adding the heats lost for the individual steps as follows.

qtotal=q1+q2+q3+q4+q5= −969.6 J − 72275.25 J − 13376 J − 10672.59 J − 974.4 J= −98267.84 J= −98.27 kJ

The total heat released in the process is −98.27 kJ .

Conclusion

The total heat released in the process is −98.27 kJ .

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

Answer 3

Question

Chapter 10, Problem 38QP

Interpretation Introduction

Interpretation:

The heat released when steam is converted into ice is to be calculated.

Concept Introduction:

Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process, q .

The heat absorbed or released by a substance is calculated by the following equation:

q=m×C×ΔT â€¦â€¦(1)

Here, m is the mass, ΔT is the temperature change, and C is the specific heat capacity.

The expression to calculate the value of ΔT :

ΔT=T2−T1 â€¦..(2)

Here T1 is the initial temperature and T2 is the final temperature.

Substitute (2) in (1)

q=m×C×T2−T1 â€¦â€¦(3)

Expert Solution & Answer

Answer to Problem 38QP

Solution:

The heat released in the process of converting steam to ice is −98.27 kJ .

Explanation of Solution

Given Information: Mass of steam is 32.0 g , the temperature of steam is 115.0°C , the

specific heat of ice is 2.03 J g−1°C−1 , the heat of fusion of ice is 6.01×103 J mol−1 , the

the specific heat of water is 4.18 J g−1°C−1 , the heat of vaporization of water is 4.07×104 J mol−1 and the specific heat of steam is 2.02 J g−1°C−1 .

Calculate the heat lost on cooling steam from 115.0°C to 100.0°C .

Using equation (3) to calculate the heat required to cool the temperature of ice from 115°C to 100°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, 100°C for T2 and 115°C for T1 in equation (3):

q1=32 g×2.03 J g−1°C−1×100°C−115°C=−969.6 J

Calculate the heat lost to condense steam to water by multiplying the molar heat of condensation by the number of moles, n .

q2=n×− molar heat of vaporization=32 g18.02 g mol−1 ×−4.07×104J mol−1=−72275.25 J

Using equation (3) to calculate the heat required to cool the temperature of water from 100°C to 0°C .

Substitute 32 g for m , 4.18 J g−1°C−1 for C as the specific heat of ice, 0.0°C for T2 and 100°C for T1 in equation (3):

q3=32 g×4.18 J g−1°C−1×0.0°C−100.0°C =−13376 J

Now, calculate the heat lost on the freezing of water by multiplying the molar heat of freezing by the number of moles.

q4=n×− molar heat of fusion=32 g18.02 g mol−1 ×−6.01×103 J mol−1=−10672.59 J

Using equation (3) to calculate the heat required to cool the temperature of ice from 0°C to −15°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, −15°C for T2 and 0°C for T1 in equation (3):

q5=32 g×2.03 J g−1°C−1×−15.0°C−0.0°C=− 974.4 J

Calculate the total heat released in the entire process by adding the heats lost for the individual steps as follows.

qtotal=q1+q2+q3+q4+q5= −969.6 J − 72275.25 J − 13376 J − 10672.59 J − 974.4 J= −98267.84 J= −98.27 kJ

The total heat released in the process is −98.27 kJ .

Conclusion

The total heat released in the process is −98.27 kJ .

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Answer 4

Question

Chapter 10, Problem 38QP

Interpretation Introduction

Interpretation:

The heat released when steam is converted into ice is to be calculated.

Concept Introduction:

Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process, q .

The heat absorbed or released by a substance is calculated by the following equation:

q=m×C×ΔT â€¦â€¦(1)

Here, m is the mass, ΔT is the temperature change, and C is the specific heat capacity.

The expression to calculate the value of ΔT :

ΔT=T2−T1 â€¦..(2)

Here T1 is the initial temperature and T2 is the final temperature.

Substitute (2) in (1)

q=m×C×T2−T1 â€¦â€¦(3)

Expert Solution & Answer

Answer to Problem 38QP

Solution:

The heat released in the process of converting steam to ice is −98.27 kJ .

Explanation of Solution

Given Information: Mass of steam is 32.0 g , the temperature of steam is 115.0°C , the

specific heat of ice is 2.03 J g−1°C−1 , the heat of fusion of ice is 6.01×103 J mol−1 , the

the specific heat of water is 4.18 J g−1°C−1 , the heat of vaporization of water is 4.07×104 J mol−1 and the specific heat of steam is 2.02 J g−1°C−1 .

Calculate the heat lost on cooling steam from 115.0°C to 100.0°C .

Using equation (3) to calculate the heat required to cool the temperature of ice from 115°C to 100°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, 100°C for T2 and 115°C for T1 in equation (3):

q1=32 g×2.03 J g−1°C−1×100°C−115°C=−969.6 J

Calculate the heat lost to condense steam to water by multiplying the molar heat of condensation by the number of moles, n .

q2=n×− molar heat of vaporization=32 g18.02 g mol−1 ×−4.07×104J mol−1=−72275.25 J

Using equation (3) to calculate the heat required to cool the temperature of water from 100°C to 0°C .

Substitute 32 g for m , 4.18 J g−1°C−1 for C as the specific heat of ice, 0.0°C for T2 and 100°C for T1 in equation (3):

q3=32 g×4.18 J g−1°C−1×0.0°C−100.0°C =−13376 J

Now, calculate the heat lost on the freezing of water by multiplying the molar heat of freezing by the number of moles.

q4=n×− molar heat of fusion=32 g18.02 g mol−1 ×−6.01×103 J mol−1=−10672.59 J

Using equation (3) to calculate the heat required to cool the temperature of ice from 0°C to −15°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, −15°C for T2 and 0°C for T1 in equation (3):

q5=32 g×2.03 J g−1°C−1×−15.0°C−0.0°C=− 974.4 J

Calculate the total heat released in the entire process by adding the heats lost for the individual steps as follows.

qtotal=q1+q2+q3+q4+q5= −969.6 J − 72275.25 J − 13376 J − 10672.59 J − 974.4 J= −98267.84 J= −98.27 kJ

The total heat released in the process is −98.27 kJ .

Conclusion

The total heat released in the process is −98.27 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

Chapter 10, Problem 38QP

Interpretation Introduction

Interpretation:

The heat released when steam is converted into ice is to be calculated.

Concept Introduction:

Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process, q .

The heat absorbed or released by a substance is calculated by the following equation:

q=m×C×ΔT â€¦â€¦(1)

Here, m is the mass, ΔT is the temperature change, and C is the specific heat capacity.

The expression to calculate the value of ΔT :

ΔT=T2−T1 â€¦..(2)

Here T1 is the initial temperature and T2 is the final temperature.

Substitute (2) in (1)

q=m×C×T2−T1 â€¦â€¦(3)

Expert Solution & Answer

Answer to Problem 38QP

Solution:

The heat released in the process of converting steam to ice is −98.27 kJ .

Explanation of Solution

Given Information: Mass of steam is 32.0 g , the temperature of steam is 115.0°C , the

specific heat of ice is 2.03 J g−1°C−1 , the heat of fusion of ice is 6.01×103 J mol−1 , the

the specific heat of water is 4.18 J g−1°C−1 , the heat of vaporization of water is 4.07×104 J mol−1 and the specific heat of steam is 2.02 J g−1°C−1 .

Calculate the heat lost on cooling steam from 115.0°C to 100.0°C .

Using equation (3) to calculate the heat required to cool the temperature of ice from 115°C to 100°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, 100°C for T2 and 115°C for T1 in equation (3):

q1=32 g×2.03 J g−1°C−1×100°C−115°C=−969.6 J

Calculate the heat lost to condense steam to water by multiplying the molar heat of condensation by the number of moles, n .

q2=n×− molar heat of vaporization=32 g18.02 g mol−1 ×−4.07×104J mol−1=−72275.25 J

Using equation (3) to calculate the heat required to cool the temperature of water from 100°C to 0°C .

Substitute 32 g for m , 4.18 J g−1°C−1 for C as the specific heat of ice, 0.0°C for T2 and 100°C for T1 in equation (3):

q3=32 g×4.18 J g−1°C−1×0.0°C−100.0°C =−13376 J

Now, calculate the heat lost on the freezing of water by multiplying the molar heat of freezing by the number of moles.

q4=n×− molar heat of fusion=32 g18.02 g mol−1 ×−6.01×103 J mol−1=−10672.59 J

Using equation (3) to calculate the heat required to cool the temperature of ice from 0°C to −15°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, −15°C for T2 and 0°C for T1 in equation (3):

q5=32 g×2.03 J g−1°C−1×−15.0°C−0.0°C=− 974.4 J

Calculate the total heat released in the entire process by adding the heats lost for the individual steps as follows.

qtotal=q1+q2+q3+q4+q5= −969.6 J − 72275.25 J − 13376 J − 10672.59 J − 974.4 J= −98267.84 J= −98.27 kJ

The total heat released in the process is −98.27 kJ .

Conclusion

The total heat released in the process is −98.27 kJ .

Answer 6

Chapter 10, Problem 38QP

Interpretation Introduction

Interpretation:

The heat released when steam is converted into ice is to be calculated.

Concept Introduction:

Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process, q .

The heat absorbed or released by a substance is calculated by the following equation:

q=m×C×ΔT â€¦â€¦(1)

Here, m is the mass, ΔT is the temperature change, and C is the specific heat capacity.

The expression to calculate the value of ΔT :

ΔT=T2−T1 â€¦..(2)

Here T1 is the initial temperature and T2 is the final temperature.

Substitute (2) in (1)

q=m×C×T2−T1 â€¦â€¦(3)

Expert Solution & Answer

Answer to Problem 38QP

Solution:

The heat released in the process of converting steam to ice is −98.27 kJ .

Explanation of Solution

Given Information: Mass of steam is 32.0 g , the temperature of steam is 115.0°C , the

specific heat of ice is 2.03 J g−1°C−1 , the heat of fusion of ice is 6.01×103 J mol−1 , the

the specific heat of water is 4.18 J g−1°C−1 , the heat of vaporization of water is 4.07×104 J mol−1 and the specific heat of steam is 2.02 J g−1°C−1 .

Calculate the heat lost on cooling steam from 115.0°C to 100.0°C .

Using equation (3) to calculate the heat required to cool the temperature of ice from 115°C to 100°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, 100°C for T2 and 115°C for T1 in equation (3):

q1=32 g×2.03 J g−1°C−1×100°C−115°C=−969.6 J

Calculate the heat lost to condense steam to water by multiplying the molar heat of condensation by the number of moles, n .

q2=n×− molar heat of vaporization=32 g18.02 g mol−1 ×−4.07×104J mol−1=−72275.25 J

Using equation (3) to calculate the heat required to cool the temperature of water from 100°C to 0°C .

Substitute 32 g for m , 4.18 J g−1°C−1 for C as the specific heat of ice, 0.0°C for T2 and 100°C for T1 in equation (3):

q3=32 g×4.18 J g−1°C−1×0.0°C−100.0°C =−13376 J

Now, calculate the heat lost on the freezing of water by multiplying the molar heat of freezing by the number of moles.

q4=n×− molar heat of fusion=32 g18.02 g mol−1 ×−6.01×103 J mol−1=−10672.59 J

Using equation (3) to calculate the heat required to cool the temperature of ice from 0°C to −15°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, −15°C for T2 and 0°C for T1 in equation (3):

q5=32 g×2.03 J g−1°C−1×−15.0°C−0.0°C=− 974.4 J

Calculate the total heat released in the entire process by adding the heats lost for the individual steps as follows.

qtotal=q1+q2+q3+q4+q5= −969.6 J − 72275.25 J − 13376 J − 10672.59 J − 974.4 J= −98267.84 J= −98.27 kJ

The total heat released in the process is −98.27 kJ .

Conclusion

The total heat released in the process is −98.27 kJ .

Answer 7

Chapter 10, Problem 38QP

Interpretation Introduction

Interpretation:

The heat released when steam is converted into ice is to be calculated.

Concept Introduction:

Phase change is the process by which ice is converted into steam. Each phase change occurs at a constant temperature along with a change in energy. This change in energy is called the heat of that process, q .

The heat absorbed or released by a substance is calculated by the following equation:

q=m×C×ΔT â€¦â€¦(1)

Here, m is the mass, ΔT is the temperature change, and C is the specific heat capacity.

The expression to calculate the value of ΔT :

ΔT=T2−T1 â€¦..(2)

Here T1 is the initial temperature and T2 is the final temperature.

Substitute (2) in (1)

q=m×C×T2−T1 â€¦â€¦(3)

Answer 8

Expert Solution & Answer

Answer to Problem 38QP

Solution:

The heat released in the process of converting steam to ice is −98.27 kJ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given Information: Mass of steam is 32.0 g , the temperature of steam is 115.0°C , the

specific heat of ice is 2.03 J g−1°C−1 , the heat of fusion of ice is 6.01×103 J mol−1 , the

the specific heat of water is 4.18 J g−1°C−1 , the heat of vaporization of water is 4.07×104 J mol−1 and the specific heat of steam is 2.02 J g−1°C−1 .

Calculate the heat lost on cooling steam from 115.0°C to 100.0°C .

Using equation (3) to calculate the heat required to cool the temperature of ice from 115°C to 100°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, 100°C for T2 and 115°C for T1 in equation (3):

q1=32 g×2.03 J g−1°C−1×100°C−115°C=−969.6 J

Calculate the heat lost to condense steam to water by multiplying the molar heat of condensation by the number of moles, n .

q2=n×− molar heat of vaporization=32 g18.02 g mol−1 ×−4.07×104J mol−1=−72275.25 J

Using equation (3) to calculate the heat required to cool the temperature of water from 100°C to 0°C .

Substitute 32 g for m , 4.18 J g−1°C−1 for C as the specific heat of ice, 0.0°C for T2 and 100°C for T1 in equation (3):

q3=32 g×4.18 J g−1°C−1×0.0°C−100.0°C =−13376 J

Now, calculate the heat lost on the freezing of water by multiplying the molar heat of freezing by the number of moles.

q4=n×− molar heat of fusion=32 g18.02 g mol−1 ×−6.01×103 J mol−1=−10672.59 J

Using equation (3) to calculate the heat required to cool the temperature of ice from 0°C to −15°C .

Substitute 32 g for m , 2.03 J g−1°C−1 for C as the specific heat of ice, −15°C for T2 and 0°C for T1 in equation (3):

q5=32 g×2.03 J g−1°C−1×−15.0°C−0.0°C=− 974.4 J

Calculate the total heat released in the entire process by adding the heats lost for the individual steps as follows.

qtotal=q1+q2+q3+q4+q5= −969.6 J − 72275.25 J − 13376 J − 10672.59 J − 974.4 J= −98267.84 J= −98.27 kJ

The total heat released in the process is −98.27 kJ .

Answer 12

Conclusion

The total heat released in the process is −98.27 kJ .

Answer 13

Ch. 10 - How do the properties of liquids and solid differ,...Ch. 10 - Prob. 2QC Ch. 10 - Prob. 3QC Ch. 10 - Prob. 4QC Ch. 10 - Prob. 1PP Ch. 10 - Prob. 2PP Ch. 10 - Prob. 3PP Ch. 10 - Prob. 4PP Ch. 10 - Which has the stronger London dispersion forces,...Ch. 10 - Prob. 6PP

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Answer to Problem 38QP

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