The result of losing 3 MJ heat from 1 kg of steam at 200 °C .

Question

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Answer 1

Question

Chapter 5, Problem 35MC

To determine

The result of losing 3 MJ heat from 1 kg of steam at 200°C.

Expert Solution & Answer

Answer to Problem 35MC

Option (b) the result of losing 3 MJ heat from 1 kg of steam at 200°C is water and ice.

Explanation of Solution

Write the expression for the expression for the heat liberated when temperature of a substance reduces.

Q=mcΔT=mc(T1−T2) (1)

Here,

Q is the amount of heat added to the system

m is the mass of substance

c is the specific heat of substance

ΔT is the change in temperature

T2 is the final temperature

T1 is the initial temperature

The specific heat of steam is 1.996 kJ/kg⋅°C.

Substitute 1 kg for m, 1.996 kJ/kg⋅°C for c, 200°C for T1 and 100°C for T2 to find the heat liberated when 1 kg of steam at 200°C becomes steam at 100°C.

Q=(1 kg)(1.996 kJ/kg⋅°C)(200°C−100°C)=199 kJ

Write the expression for the heat liberated when steam becomes water at the boiling point of water.

Q=mLv

Here,

Lv is the latent heat of vaporization

The heat of vaporization for water is 2260 kJ/kg.

Substitute 1 kg for m and 2260 kJ/kg for Lv to find the heat liberated when 1 kg of steam at 100°C becomes water at 100°C.

Q=(1 kg)(2260 kJ/kg)=2260 kJ

The specific heat of water is 4.186 kJ/kg⋅°C.

Substitute 1 kg for m, 4.186 kJ/kg⋅°C for c, 100°C for T1 and 0°C for T2 in equation (1) to find the heat energy liberated when 1 kg of water at 100°C becomes water at 0°C.

Q=(1 kg)(4.186 kJ/kg⋅°C)(100°C−0°C)=418.6 kJ

Thus, the total heat liberated when 1 kg of steam at 200°C becomes water at 0°C is,

Qtotal=199 kJ+2260 kJ+418.6 kJ=2878 kJ=2878 kJ×10−3 MJ1 kJ=2.9MJ

When the steam liberate 2.9MJ heat, it will completely turn into water at 0°C. Sill there is 0.1MJ of energy remaining in 3MJ.

Write the expression for the heat corresponding to the phase change of water at 0°C to ice at 0°C.

Q=mLf

Here,

Lf is the latent heat of fusion

The heat of fusion of water is 335 kJ/kg.

Substitute 0.1MJ for Q and 335 kJ/kg for Lf to find the amount of water at 0°C becomes ice at 0°C by liberating heat equal to 0.1MJ.

0.1MJ=m(335 kJ/kg)m=0.1MJ335 kJ/kg=0.1MJ×103 kJ1 MJ335 kJ/kg=0.3 kg

So, liberation the 0.1MJ heat will cause 0.3 kg of water at 0°C to turn into ice at 0°C. Thus, the result of losing 3 MJ heat from 1 kg of steam at 200°C will be a mixture of water and ice.

Conclusion:

Since, the liberated heat energy of 3 MJ is enough to result the 1 kg steam at 200°C to become water and ice, the final state of the system will be a mixture of water and ice. Thus, option (b) is correct.

The liberated heat energy of 3 MJ is not enough to result the whole 1 kg steam at 200°C to become ice. Thus, option (a) is incorrect.

Since there is energy lost from the system even after it become completely water, the final state is not only consist of water but also ice. Thus, option (c) is incorrect.

Losing 3 MJ heat from 1 kg of steam at 200°C will definitely convert all the steam into other forms. Thus, option (d) is incorrect.

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Answer 2

Question

Chapter 5, Problem 35MC

To determine

The result of losing 3 MJ heat from 1 kg of steam at 200°C.

Expert Solution & Answer

Answer to Problem 35MC

Option (b) the result of losing 3 MJ heat from 1 kg of steam at 200°C is water and ice.

Explanation of Solution

Write the expression for the expression for the heat liberated when temperature of a substance reduces.

Q=mcΔT=mc(T1−T2) (1)

Here,

Q is the amount of heat added to the system

m is the mass of substance

c is the specific heat of substance

ΔT is the change in temperature

T2 is the final temperature

T1 is the initial temperature

The specific heat of steam is 1.996 kJ/kg⋅°C.

Substitute 1 kg for m, 1.996 kJ/kg⋅°C for c, 200°C for T1 and 100°C for T2 to find the heat liberated when 1 kg of steam at 200°C becomes steam at 100°C.

Q=(1 kg)(1.996 kJ/kg⋅°C)(200°C−100°C)=199 kJ

Write the expression for the heat liberated when steam becomes water at the boiling point of water.

Q=mLv

Here,

Lv is the latent heat of vaporization

The heat of vaporization for water is 2260 kJ/kg.

Substitute 1 kg for m and 2260 kJ/kg for Lv to find the heat liberated when 1 kg of steam at 100°C becomes water at 100°C.

Q=(1 kg)(2260 kJ/kg)=2260 kJ

The specific heat of water is 4.186 kJ/kg⋅°C.

Substitute 1 kg for m, 4.186 kJ/kg⋅°C for c, 100°C for T1 and 0°C for T2 in equation (1) to find the heat energy liberated when 1 kg of water at 100°C becomes water at 0°C.

Q=(1 kg)(4.186 kJ/kg⋅°C)(100°C−0°C)=418.6 kJ

Thus, the total heat liberated when 1 kg of steam at 200°C becomes water at 0°C is,

Qtotal=199 kJ+2260 kJ+418.6 kJ=2878 kJ=2878 kJ×10−3 MJ1 kJ=2.9MJ

When the steam liberate 2.9MJ heat, it will completely turn into water at 0°C. Sill there is 0.1MJ of energy remaining in 3MJ.

Write the expression for the heat corresponding to the phase change of water at 0°C to ice at 0°C.

Q=mLf

Here,

Lf is the latent heat of fusion

The heat of fusion of water is 335 kJ/kg.

Substitute 0.1MJ for Q and 335 kJ/kg for Lf to find the amount of water at 0°C becomes ice at 0°C by liberating heat equal to 0.1MJ.

0.1MJ=m(335 kJ/kg)m=0.1MJ335 kJ/kg=0.1MJ×103 kJ1 MJ335 kJ/kg=0.3 kg

So, liberation the 0.1MJ heat will cause 0.3 kg of water at 0°C to turn into ice at 0°C. Thus, the result of losing 3 MJ heat from 1 kg of steam at 200°C will be a mixture of water and ice.

Conclusion:

Since, the liberated heat energy of 3 MJ is enough to result the 1 kg steam at 200°C to become water and ice, the final state of the system will be a mixture of water and ice. Thus, option (b) is correct.

The liberated heat energy of 3 MJ is not enough to result the whole 1 kg steam at 200°C to become ice. Thus, option (a) is incorrect.

Since there is energy lost from the system even after it become completely water, the final state is not only consist of water but also ice. Thus, option (c) is incorrect.

Losing 3 MJ heat from 1 kg of steam at 200°C will definitely convert all the steam into other forms. Thus, option (d) is incorrect.

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Answer 3

Question

Chapter 5, Problem 35MC

To determine

The result of losing 3 MJ heat from 1 kg of steam at 200°C.

Expert Solution & Answer

Answer to Problem 35MC

Option (b) the result of losing 3 MJ heat from 1 kg of steam at 200°C is water and ice.

Explanation of Solution

Write the expression for the expression for the heat liberated when temperature of a substance reduces.

Q=mcΔT=mc(T1−T2) (1)

Here,

Q is the amount of heat added to the system

m is the mass of substance

c is the specific heat of substance

ΔT is the change in temperature

T2 is the final temperature

T1 is the initial temperature

The specific heat of steam is 1.996 kJ/kg⋅°C.

Substitute 1 kg for m, 1.996 kJ/kg⋅°C for c, 200°C for T1 and 100°C for T2 to find the heat liberated when 1 kg of steam at 200°C becomes steam at 100°C.

Q=(1 kg)(1.996 kJ/kg⋅°C)(200°C−100°C)=199 kJ

Write the expression for the heat liberated when steam becomes water at the boiling point of water.

Q=mLv

Here,

Lv is the latent heat of vaporization

The heat of vaporization for water is 2260 kJ/kg.

Substitute 1 kg for m and 2260 kJ/kg for Lv to find the heat liberated when 1 kg of steam at 100°C becomes water at 100°C.

Q=(1 kg)(2260 kJ/kg)=2260 kJ

The specific heat of water is 4.186 kJ/kg⋅°C.

Substitute 1 kg for m, 4.186 kJ/kg⋅°C for c, 100°C for T1 and 0°C for T2 in equation (1) to find the heat energy liberated when 1 kg of water at 100°C becomes water at 0°C.

Q=(1 kg)(4.186 kJ/kg⋅°C)(100°C−0°C)=418.6 kJ

Thus, the total heat liberated when 1 kg of steam at 200°C becomes water at 0°C is,

Qtotal=199 kJ+2260 kJ+418.6 kJ=2878 kJ=2878 kJ×10−3 MJ1 kJ=2.9MJ

When the steam liberate 2.9MJ heat, it will completely turn into water at 0°C. Sill there is 0.1MJ of energy remaining in 3MJ.

Write the expression for the heat corresponding to the phase change of water at 0°C to ice at 0°C.

Q=mLf

Here,

Lf is the latent heat of fusion

The heat of fusion of water is 335 kJ/kg.

Substitute 0.1MJ for Q and 335 kJ/kg for Lf to find the amount of water at 0°C becomes ice at 0°C by liberating heat equal to 0.1MJ.

0.1MJ=m(335 kJ/kg)m=0.1MJ335 kJ/kg=0.1MJ×103 kJ1 MJ335 kJ/kg=0.3 kg

So, liberation the 0.1MJ heat will cause 0.3 kg of water at 0°C to turn into ice at 0°C. Thus, the result of losing 3 MJ heat from 1 kg of steam at 200°C will be a mixture of water and ice.

Conclusion:

Since, the liberated heat energy of 3 MJ is enough to result the 1 kg steam at 200°C to become water and ice, the final state of the system will be a mixture of water and ice. Thus, option (b) is correct.

The liberated heat energy of 3 MJ is not enough to result the whole 1 kg steam at 200°C to become ice. Thus, option (a) is incorrect.

Since there is energy lost from the system even after it become completely water, the final state is not only consist of water but also ice. Thus, option (c) is incorrect.

Losing 3 MJ heat from 1 kg of steam at 200°C will definitely convert all the steam into other forms. Thus, option (d) is incorrect.

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Answer 4

Question

Chapter 5, Problem 35MC

To determine

The result of losing 3 MJ heat from 1 kg of steam at 200°C.

Expert Solution & Answer

Answer to Problem 35MC

Option (b) the result of losing 3 MJ heat from 1 kg of steam at 200°C is water and ice.

Explanation of Solution

Write the expression for the expression for the heat liberated when temperature of a substance reduces.

Q=mcΔT=mc(T1−T2) (1)

Here,

Q is the amount of heat added to the system

m is the mass of substance

c is the specific heat of substance

ΔT is the change in temperature

T2 is the final temperature

T1 is the initial temperature

The specific heat of steam is 1.996 kJ/kg⋅°C.

Substitute 1 kg for m, 1.996 kJ/kg⋅°C for c, 200°C for T1 and 100°C for T2 to find the heat liberated when 1 kg of steam at 200°C becomes steam at 100°C.

Q=(1 kg)(1.996 kJ/kg⋅°C)(200°C−100°C)=199 kJ

Write the expression for the heat liberated when steam becomes water at the boiling point of water.

Q=mLv

Here,

Lv is the latent heat of vaporization

The heat of vaporization for water is 2260 kJ/kg.

Substitute 1 kg for m and 2260 kJ/kg for Lv to find the heat liberated when 1 kg of steam at 100°C becomes water at 100°C.

Q=(1 kg)(2260 kJ/kg)=2260 kJ

The specific heat of water is 4.186 kJ/kg⋅°C.

Substitute 1 kg for m, 4.186 kJ/kg⋅°C for c, 100°C for T1 and 0°C for T2 in equation (1) to find the heat energy liberated when 1 kg of water at 100°C becomes water at 0°C.

Q=(1 kg)(4.186 kJ/kg⋅°C)(100°C−0°C)=418.6 kJ

Thus, the total heat liberated when 1 kg of steam at 200°C becomes water at 0°C is,

Qtotal=199 kJ+2260 kJ+418.6 kJ=2878 kJ=2878 kJ×10−3 MJ1 kJ=2.9MJ

When the steam liberate 2.9MJ heat, it will completely turn into water at 0°C. Sill there is 0.1MJ of energy remaining in 3MJ.

Write the expression for the heat corresponding to the phase change of water at 0°C to ice at 0°C.

Q=mLf

Here,

Lf is the latent heat of fusion

The heat of fusion of water is 335 kJ/kg.

Substitute 0.1MJ for Q and 335 kJ/kg for Lf to find the amount of water at 0°C becomes ice at 0°C by liberating heat equal to 0.1MJ.

0.1MJ=m(335 kJ/kg)m=0.1MJ335 kJ/kg=0.1MJ×103 kJ1 MJ335 kJ/kg=0.3 kg

So, liberation the 0.1MJ heat will cause 0.3 kg of water at 0°C to turn into ice at 0°C. Thus, the result of losing 3 MJ heat from 1 kg of steam at 200°C will be a mixture of water and ice.

Conclusion:

Since, the liberated heat energy of 3 MJ is enough to result the 1 kg steam at 200°C to become water and ice, the final state of the system will be a mixture of water and ice. Thus, option (b) is correct.

The liberated heat energy of 3 MJ is not enough to result the whole 1 kg steam at 200°C to become ice. Thus, option (a) is incorrect.

Since there is energy lost from the system even after it become completely water, the final state is not only consist of water but also ice. Thus, option (c) is incorrect.

Losing 3 MJ heat from 1 kg of steam at 200°C will definitely convert all the steam into other forms. Thus, option (d) is incorrect.

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

Answer 5

Chapter 5, Problem 35MC

To determine

The result of losing 3 MJ heat from 1 kg of steam at 200°C.

Expert Solution & Answer

Answer to Problem 35MC

Option (b) the result of losing 3 MJ heat from 1 kg of steam at 200°C is water and ice.

Explanation of Solution

Write the expression for the expression for the heat liberated when temperature of a substance reduces.

Q=mcΔT=mc(T1−T2) (1)

Here,

Q is the amount of heat added to the system

m is the mass of substance

c is the specific heat of substance

ΔT is the change in temperature

T2 is the final temperature

T1 is the initial temperature

The specific heat of steam is 1.996 kJ/kg⋅°C.

Substitute 1 kg for m, 1.996 kJ/kg⋅°C for c, 200°C for T1 and 100°C for T2 to find the heat liberated when 1 kg of steam at 200°C becomes steam at 100°C.

Q=(1 kg)(1.996 kJ/kg⋅°C)(200°C−100°C)=199 kJ

Write the expression for the heat liberated when steam becomes water at the boiling point of water.

Q=mLv

Here,

Lv is the latent heat of vaporization

The heat of vaporization for water is 2260 kJ/kg.

Substitute 1 kg for m and 2260 kJ/kg for Lv to find the heat liberated when 1 kg of steam at 100°C becomes water at 100°C.

Q=(1 kg)(2260 kJ/kg)=2260 kJ

The specific heat of water is 4.186 kJ/kg⋅°C.

Substitute 1 kg for m, 4.186 kJ/kg⋅°C for c, 100°C for T1 and 0°C for T2 in equation (1) to find the heat energy liberated when 1 kg of water at 100°C becomes water at 0°C.

Q=(1 kg)(4.186 kJ/kg⋅°C)(100°C−0°C)=418.6 kJ

Thus, the total heat liberated when 1 kg of steam at 200°C becomes water at 0°C is,

Qtotal=199 kJ+2260 kJ+418.6 kJ=2878 kJ=2878 kJ×10−3 MJ1 kJ=2.9MJ

When the steam liberate 2.9MJ heat, it will completely turn into water at 0°C. Sill there is 0.1MJ of energy remaining in 3MJ.

Write the expression for the heat corresponding to the phase change of water at 0°C to ice at 0°C.

Q=mLf

Here,

Lf is the latent heat of fusion

The heat of fusion of water is 335 kJ/kg.

Substitute 0.1MJ for Q and 335 kJ/kg for Lf to find the amount of water at 0°C becomes ice at 0°C by liberating heat equal to 0.1MJ.

0.1MJ=m(335 kJ/kg)m=0.1MJ335 kJ/kg=0.1MJ×103 kJ1 MJ335 kJ/kg=0.3 kg

So, liberation the 0.1MJ heat will cause 0.3 kg of water at 0°C to turn into ice at 0°C. Thus, the result of losing 3 MJ heat from 1 kg of steam at 200°C will be a mixture of water and ice.

Conclusion:

Since, the liberated heat energy of 3 MJ is enough to result the 1 kg steam at 200°C to become water and ice, the final state of the system will be a mixture of water and ice. Thus, option (b) is correct.

The liberated heat energy of 3 MJ is not enough to result the whole 1 kg steam at 200°C to become ice. Thus, option (a) is incorrect.

Since there is energy lost from the system even after it become completely water, the final state is not only consist of water but also ice. Thus, option (c) is incorrect.

Losing 3 MJ heat from 1 kg of steam at 200°C will definitely convert all the steam into other forms. Thus, option (d) is incorrect.

Answer 6

Chapter 5, Problem 35MC

To determine

The result of losing 3 MJ heat from 1 kg of steam at 200°C.

Expert Solution & Answer

Answer to Problem 35MC

Option (b) the result of losing 3 MJ heat from 1 kg of steam at 200°C is water and ice.

Explanation of Solution

Write the expression for the expression for the heat liberated when temperature of a substance reduces.

Q=mcΔT=mc(T1−T2) (1)

Here,

Q is the amount of heat added to the system

m is the mass of substance

c is the specific heat of substance

ΔT is the change in temperature

T2 is the final temperature

T1 is the initial temperature

The specific heat of steam is 1.996 kJ/kg⋅°C.

Substitute 1 kg for m, 1.996 kJ/kg⋅°C for c, 200°C for T1 and 100°C for T2 to find the heat liberated when 1 kg of steam at 200°C becomes steam at 100°C.

Q=(1 kg)(1.996 kJ/kg⋅°C)(200°C−100°C)=199 kJ

Write the expression for the heat liberated when steam becomes water at the boiling point of water.

Q=mLv

Here,

Lv is the latent heat of vaporization

The heat of vaporization for water is 2260 kJ/kg.

Substitute 1 kg for m and 2260 kJ/kg for Lv to find the heat liberated when 1 kg of steam at 100°C becomes water at 100°C.

Q=(1 kg)(2260 kJ/kg)=2260 kJ

The specific heat of water is 4.186 kJ/kg⋅°C.

Substitute 1 kg for m, 4.186 kJ/kg⋅°C for c, 100°C for T1 and 0°C for T2 in equation (1) to find the heat energy liberated when 1 kg of water at 100°C becomes water at 0°C.

Q=(1 kg)(4.186 kJ/kg⋅°C)(100°C−0°C)=418.6 kJ

Thus, the total heat liberated when 1 kg of steam at 200°C becomes water at 0°C is,

Qtotal=199 kJ+2260 kJ+418.6 kJ=2878 kJ=2878 kJ×10−3 MJ1 kJ=2.9MJ

When the steam liberate 2.9MJ heat, it will completely turn into water at 0°C. Sill there is 0.1MJ of energy remaining in 3MJ.

Write the expression for the heat corresponding to the phase change of water at 0°C to ice at 0°C.

Q=mLf

Here,

Lf is the latent heat of fusion

The heat of fusion of water is 335 kJ/kg.

Substitute 0.1MJ for Q and 335 kJ/kg for Lf to find the amount of water at 0°C becomes ice at 0°C by liberating heat equal to 0.1MJ.

0.1MJ=m(335 kJ/kg)m=0.1MJ335 kJ/kg=0.1MJ×103 kJ1 MJ335 kJ/kg=0.3 kg

So, liberation the 0.1MJ heat will cause 0.3 kg of water at 0°C to turn into ice at 0°C. Thus, the result of losing 3 MJ heat from 1 kg of steam at 200°C will be a mixture of water and ice.

Conclusion:

Since, the liberated heat energy of 3 MJ is enough to result the 1 kg steam at 200°C to become water and ice, the final state of the system will be a mixture of water and ice. Thus, option (b) is correct.

The liberated heat energy of 3 MJ is not enough to result the whole 1 kg steam at 200°C to become ice. Thus, option (a) is incorrect.

Since there is energy lost from the system even after it become completely water, the final state is not only consist of water but also ice. Thus, option (c) is incorrect.

Losing 3 MJ heat from 1 kg of steam at 200°C will definitely convert all the steam into other forms. Thus, option (d) is incorrect.

Answer 7

Chapter 5, Problem 35MC

To determine

The result of losing 3 MJ heat from 1 kg of steam at 200°C.

Answer 8

Expert Solution & Answer

Answer to Problem 35MC

Option (b) the result of losing 3 MJ heat from 1 kg of steam at 200°C is water and ice.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Write the expression for the expression for the heat liberated when temperature of a substance reduces.

Q=mcΔT=mc(T1−T2) (1)

Here,

Q is the amount of heat added to the system

m is the mass of substance

c is the specific heat of substance

ΔT is the change in temperature

T2 is the final temperature

T1 is the initial temperature

The specific heat of steam is 1.996 kJ/kg⋅°C.

Substitute 1 kg for m, 1.996 kJ/kg⋅°C for c, 200°C for T1 and 100°C for T2 to find the heat liberated when 1 kg of steam at 200°C becomes steam at 100°C.

Q=(1 kg)(1.996 kJ/kg⋅°C)(200°C−100°C)=199 kJ

Write the expression for the heat liberated when steam becomes water at the boiling point of water.

Q=mLv

Here,

Lv is the latent heat of vaporization

The heat of vaporization for water is 2260 kJ/kg.

Substitute 1 kg for m and 2260 kJ/kg for Lv to find the heat liberated when 1 kg of steam at 100°C becomes water at 100°C.

Q=(1 kg)(2260 kJ/kg)=2260 kJ

The specific heat of water is 4.186 kJ/kg⋅°C.

Substitute 1 kg for m, 4.186 kJ/kg⋅°C for c, 100°C for T1 and 0°C for T2 in equation (1) to find the heat energy liberated when 1 kg of water at 100°C becomes water at 0°C.

Q=(1 kg)(4.186 kJ/kg⋅°C)(100°C−0°C)=418.6 kJ

Thus, the total heat liberated when 1 kg of steam at 200°C becomes water at 0°C is,

Qtotal=199 kJ+2260 kJ+418.6 kJ=2878 kJ=2878 kJ×10−3 MJ1 kJ=2.9MJ

When the steam liberate 2.9MJ heat, it will completely turn into water at 0°C. Sill there is 0.1MJ of energy remaining in 3MJ.

Write the expression for the heat corresponding to the phase change of water at 0°C to ice at 0°C.

Q=mLf

Here,

Lf is the latent heat of fusion

The heat of fusion of water is 335 kJ/kg.

Substitute 0.1MJ for Q and 335 kJ/kg for Lf to find the amount of water at 0°C becomes ice at 0°C by liberating heat equal to 0.1MJ.

0.1MJ=m(335 kJ/kg)m=0.1MJ335 kJ/kg=0.1MJ×103 kJ1 MJ335 kJ/kg=0.3 kg

So, liberation the 0.1MJ heat will cause 0.3 kg of water at 0°C to turn into ice at 0°C. Thus, the result of losing 3 MJ heat from 1 kg of steam at 200°C will be a mixture of water and ice.

Conclusion:

Since, the liberated heat energy of 3 MJ is enough to result the 1 kg steam at 200°C to become water and ice, the final state of the system will be a mixture of water and ice. Thus, option (b) is correct.

The liberated heat energy of 3 MJ is not enough to result the whole 1 kg steam at 200°C to become ice. Thus, option (a) is incorrect.

Since there is energy lost from the system even after it become completely water, the final state is not only consist of water but also ice. Thus, option (c) is incorrect.

Losing 3 MJ heat from 1 kg of steam at 200°C will definitely convert all the steam into other forms. Thus, option (d) is incorrect.