The amount of ice that needs to be added in an insulated cylinder.

Question

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Answer 1

Question

Chapter 5, Problem 100RQ

To determine

The amount of ice that needs to be added in an insulated cylinder.

Expert Solution & Answer

Explanation of Solution

Given:

The volume of the saturated mixture (V1) is 0.01 m3.

The quality of the mixture is 0.2.

The initial temperature of the mixture (T1) is 120°C.

The heat of fusion is 333.7 kJ/kg.

Calculation:

From the Table A-3 “Properties of common liquids, solids and foods”, obtain the value of specific heat of ice at 0 C and 120 C temperature as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

From the Table A-4 “Saturated water”, obtain the value of specific volume and enthalpy of saturated liquid, vapour and change upon vaporization for insulated cylinder device.

At final temperature of water as 120°C.

The specific volume of ice for insulated cylinder device (vf) is 0.001060

The specific volume of ice for insulated cylinder device (vg) is 0.89133 m3/kg

The specific enthalpy of ice (hf) is 503.81 kJ/kg.

The specific enthalpy of the mixture (hfg) is 2202.1 kJ/kg

Write the expression of initial specific volume of a two-phase system for insulated cylinder device.

v1=vf+xvfg=vf+x(vg−vf)

v1=(0.001060)+(0.2)×(0.89133 m3/kg−0.001060)=(0.001060)+(0.2)×(0.89027 m3/kg)=0.179114 m3/kg

Write the expression of initial specific enthalpy of a two-phase system for insulated cylinder device.

h1=hf+xhfg

h1=(503.81)+(0.2)×(2202.1 kJ/kg)=944.23 kJ/kg

Here the value final specific enthalpy is equal to specific enthalpy of saturated liquid as 503.81 kJ/kg.

Determine the initial specific volume of the refrigerant.

v1=V1mm=V1v1

m=0.01 m30.179114 m3/kg=0.05583 kg

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (I) and write energy balance relation of an insulated cylinder.

Wb,in=ΔUWb,in−ΔU=0 (II)

Here, the boundary work to be done into the system is Wb,in and the change in the internal energy is ΔU.

Substitute ΔH for Wb,in−ΔU in Equation (II).

ΔH=0ΔHice+ΔHwater=0[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[m(h2−h1)]water=0 (III)

Here, the mass of the ice is m, the heat of fusion of ice is hif, the specific heat of ice at liquid state is c, the initial temperature of ice is T1, the final temperature of ice is T2, the initial specific enthalpy of ice is h1, and the final specific enthalpy of ice is h2.

Substitute 0°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 120°C for T2,liquid, 0.05583 kg for mwater, 503.81 kJ/kg for h2, and 944.24 kJ/kg for h1 in Equation (III).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((120°C)−0°C)liquid]ice+[(0.05583 kg)((503.81 kJ/kg)−(944.24 kJ/kg))]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(120°C)]ice+[(0.05583 kg)(−440.43 kJ/kg)]=0m=0.0294 kg×(1000 g1 kg)m=29.4 g

Thus, the amount of ice that needs to be added in an insulated cylinder is 29.4 g_.

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Answer 2

Question

Chapter 5, Problem 100RQ

To determine

The amount of ice that needs to be added in an insulated cylinder.

Expert Solution & Answer

Explanation of Solution

Given:

The volume of the saturated mixture (V1) is 0.01 m3.

The quality of the mixture is 0.2.

The initial temperature of the mixture (T1) is 120°C.

The heat of fusion is 333.7 kJ/kg.

Calculation:

From the Table A-3 “Properties of common liquids, solids and foods”, obtain the value of specific heat of ice at 0 C and 120 C temperature as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

From the Table A-4 “Saturated water”, obtain the value of specific volume and enthalpy of saturated liquid, vapour and change upon vaporization for insulated cylinder device.

At final temperature of water as 120°C.

The specific volume of ice for insulated cylinder device (vf) is 0.001060

The specific volume of ice for insulated cylinder device (vg) is 0.89133 m3/kg

The specific enthalpy of ice (hf) is 503.81 kJ/kg.

The specific enthalpy of the mixture (hfg) is 2202.1 kJ/kg

Write the expression of initial specific volume of a two-phase system for insulated cylinder device.

v1=vf+xvfg=vf+x(vg−vf)

v1=(0.001060)+(0.2)×(0.89133 m3/kg−0.001060)=(0.001060)+(0.2)×(0.89027 m3/kg)=0.179114 m3/kg

Write the expression of initial specific enthalpy of a two-phase system for insulated cylinder device.

h1=hf+xhfg

h1=(503.81)+(0.2)×(2202.1 kJ/kg)=944.23 kJ/kg

Here the value final specific enthalpy is equal to specific enthalpy of saturated liquid as 503.81 kJ/kg.

Determine the initial specific volume of the refrigerant.

v1=V1mm=V1v1

m=0.01 m30.179114 m3/kg=0.05583 kg

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (I) and write energy balance relation of an insulated cylinder.

Wb,in=ΔUWb,in−ΔU=0 (II)

Here, the boundary work to be done into the system is Wb,in and the change in the internal energy is ΔU.

Substitute ΔH for Wb,in−ΔU in Equation (II).

ΔH=0ΔHice+ΔHwater=0[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[m(h2−h1)]water=0 (III)

Here, the mass of the ice is m, the heat of fusion of ice is hif, the specific heat of ice at liquid state is c, the initial temperature of ice is T1, the final temperature of ice is T2, the initial specific enthalpy of ice is h1, and the final specific enthalpy of ice is h2.

Substitute 0°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 120°C for T2,liquid, 0.05583 kg for mwater, 503.81 kJ/kg for h2, and 944.24 kJ/kg for h1 in Equation (III).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((120°C)−0°C)liquid]ice+[(0.05583 kg)((503.81 kJ/kg)−(944.24 kJ/kg))]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(120°C)]ice+[(0.05583 kg)(−440.43 kJ/kg)]=0m=0.0294 kg×(1000 g1 kg)m=29.4 g

Thus, the amount of ice that needs to be added in an insulated cylinder is 29.4 g_.

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Answer 3

Question

Chapter 5, Problem 100RQ

To determine

The amount of ice that needs to be added in an insulated cylinder.

Expert Solution & Answer

Explanation of Solution

Given:

The volume of the saturated mixture (V1) is 0.01 m3.

The quality of the mixture is 0.2.

The initial temperature of the mixture (T1) is 120°C.

The heat of fusion is 333.7 kJ/kg.

Calculation:

From the Table A-3 “Properties of common liquids, solids and foods”, obtain the value of specific heat of ice at 0 C and 120 C temperature as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

From the Table A-4 “Saturated water”, obtain the value of specific volume and enthalpy of saturated liquid, vapour and change upon vaporization for insulated cylinder device.

At final temperature of water as 120°C.

The specific volume of ice for insulated cylinder device (vf) is 0.001060

The specific volume of ice for insulated cylinder device (vg) is 0.89133 m3/kg

The specific enthalpy of ice (hf) is 503.81 kJ/kg.

The specific enthalpy of the mixture (hfg) is 2202.1 kJ/kg

Write the expression of initial specific volume of a two-phase system for insulated cylinder device.

v1=vf+xvfg=vf+x(vg−vf)

v1=(0.001060)+(0.2)×(0.89133 m3/kg−0.001060)=(0.001060)+(0.2)×(0.89027 m3/kg)=0.179114 m3/kg

Write the expression of initial specific enthalpy of a two-phase system for insulated cylinder device.

h1=hf+xhfg

h1=(503.81)+(0.2)×(2202.1 kJ/kg)=944.23 kJ/kg

Here the value final specific enthalpy is equal to specific enthalpy of saturated liquid as 503.81 kJ/kg.

Determine the initial specific volume of the refrigerant.

v1=V1mm=V1v1

m=0.01 m30.179114 m3/kg=0.05583 kg

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (I) and write energy balance relation of an insulated cylinder.

Wb,in=ΔUWb,in−ΔU=0 (II)

Here, the boundary work to be done into the system is Wb,in and the change in the internal energy is ΔU.

Substitute ΔH for Wb,in−ΔU in Equation (II).

ΔH=0ΔHice+ΔHwater=0[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[m(h2−h1)]water=0 (III)

Here, the mass of the ice is m, the heat of fusion of ice is hif, the specific heat of ice at liquid state is c, the initial temperature of ice is T1, the final temperature of ice is T2, the initial specific enthalpy of ice is h1, and the final specific enthalpy of ice is h2.

Substitute 0°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 120°C for T2,liquid, 0.05583 kg for mwater, 503.81 kJ/kg for h2, and 944.24 kJ/kg for h1 in Equation (III).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((120°C)−0°C)liquid]ice+[(0.05583 kg)((503.81 kJ/kg)−(944.24 kJ/kg))]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(120°C)]ice+[(0.05583 kg)(−440.43 kJ/kg)]=0m=0.0294 kg×(1000 g1 kg)m=29.4 g

Thus, the amount of ice that needs to be added in an insulated cylinder is 29.4 g_.

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Answer 4

Question

Chapter 5, Problem 100RQ

To determine

The amount of ice that needs to be added in an insulated cylinder.

Expert Solution & Answer

Explanation of Solution

Given:

The volume of the saturated mixture (V1) is 0.01 m3.

The quality of the mixture is 0.2.

The initial temperature of the mixture (T1) is 120°C.

The heat of fusion is 333.7 kJ/kg.

Calculation:

From the Table A-3 “Properties of common liquids, solids and foods”, obtain the value of specific heat of ice at 0 C and 120 C temperature as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

From the Table A-4 “Saturated water”, obtain the value of specific volume and enthalpy of saturated liquid, vapour and change upon vaporization for insulated cylinder device.

At final temperature of water as 120°C.

The specific volume of ice for insulated cylinder device (vf) is 0.001060

The specific volume of ice for insulated cylinder device (vg) is 0.89133 m3/kg

The specific enthalpy of ice (hf) is 503.81 kJ/kg.

The specific enthalpy of the mixture (hfg) is 2202.1 kJ/kg

Write the expression of initial specific volume of a two-phase system for insulated cylinder device.

v1=vf+xvfg=vf+x(vg−vf)

v1=(0.001060)+(0.2)×(0.89133 m3/kg−0.001060)=(0.001060)+(0.2)×(0.89027 m3/kg)=0.179114 m3/kg

Write the expression of initial specific enthalpy of a two-phase system for insulated cylinder device.

h1=hf+xhfg

h1=(503.81)+(0.2)×(2202.1 kJ/kg)=944.23 kJ/kg

Here the value final specific enthalpy is equal to specific enthalpy of saturated liquid as 503.81 kJ/kg.

Determine the initial specific volume of the refrigerant.

v1=V1mm=V1v1

m=0.01 m30.179114 m3/kg=0.05583 kg

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (I) and write energy balance relation of an insulated cylinder.

Wb,in=ΔUWb,in−ΔU=0 (II)

Here, the boundary work to be done into the system is Wb,in and the change in the internal energy is ΔU.

Substitute ΔH for Wb,in−ΔU in Equation (II).

ΔH=0ΔHice+ΔHwater=0[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[m(h2−h1)]water=0 (III)

Here, the mass of the ice is m, the heat of fusion of ice is hif, the specific heat of ice at liquid state is c, the initial temperature of ice is T1, the final temperature of ice is T2, the initial specific enthalpy of ice is h1, and the final specific enthalpy of ice is h2.

Substitute 0°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 120°C for T2,liquid, 0.05583 kg for mwater, 503.81 kJ/kg for h2, and 944.24 kJ/kg for h1 in Equation (III).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((120°C)−0°C)liquid]ice+[(0.05583 kg)((503.81 kJ/kg)−(944.24 kJ/kg))]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(120°C)]ice+[(0.05583 kg)(−440.43 kJ/kg)]=0m=0.0294 kg×(1000 g1 kg)m=29.4 g

Thus, the amount of ice that needs to be added in an insulated cylinder is 29.4 g_.

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Answer 5

Chapter 5, Problem 100RQ

To determine

The amount of ice that needs to be added in an insulated cylinder.

Expert Solution & Answer

Explanation of Solution

Given:

The volume of the saturated mixture (V1) is 0.01 m3.

The quality of the mixture is 0.2.

The initial temperature of the mixture (T1) is 120°C.

The heat of fusion is 333.7 kJ/kg.

Calculation:

From the Table A-3 “Properties of common liquids, solids and foods”, obtain the value of specific heat of ice at 0 C and 120 C temperature as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

From the Table A-4 “Saturated water”, obtain the value of specific volume and enthalpy of saturated liquid, vapour and change upon vaporization for insulated cylinder device.

At final temperature of water as 120°C.

The specific volume of ice for insulated cylinder device (vf) is 0.001060

The specific volume of ice for insulated cylinder device (vg) is 0.89133 m3/kg

The specific enthalpy of ice (hf) is 503.81 kJ/kg.

The specific enthalpy of the mixture (hfg) is 2202.1 kJ/kg

Write the expression of initial specific volume of a two-phase system for insulated cylinder device.

v1=vf+xvfg=vf+x(vg−vf)

v1=(0.001060)+(0.2)×(0.89133 m3/kg−0.001060)=(0.001060)+(0.2)×(0.89027 m3/kg)=0.179114 m3/kg

Write the expression of initial specific enthalpy of a two-phase system for insulated cylinder device.

h1=hf+xhfg

h1=(503.81)+(0.2)×(2202.1 kJ/kg)=944.23 kJ/kg

Here the value final specific enthalpy is equal to specific enthalpy of saturated liquid as 503.81 kJ/kg.

Determine the initial specific volume of the refrigerant.

v1=V1mm=V1v1

m=0.01 m30.179114 m3/kg=0.05583 kg

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (I) and write energy balance relation of an insulated cylinder.

Wb,in=ΔUWb,in−ΔU=0 (II)

Here, the boundary work to be done into the system is Wb,in and the change in the internal energy is ΔU.

Substitute ΔH for Wb,in−ΔU in Equation (II).

ΔH=0ΔHice+ΔHwater=0[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[m(h2−h1)]water=0 (III)

Here, the mass of the ice is m, the heat of fusion of ice is hif, the specific heat of ice at liquid state is c, the initial temperature of ice is T1, the final temperature of ice is T2, the initial specific enthalpy of ice is h1, and the final specific enthalpy of ice is h2.

Substitute 0°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 120°C for T2,liquid, 0.05583 kg for mwater, 503.81 kJ/kg for h2, and 944.24 kJ/kg for h1 in Equation (III).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((120°C)−0°C)liquid]ice+[(0.05583 kg)((503.81 kJ/kg)−(944.24 kJ/kg))]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(120°C)]ice+[(0.05583 kg)(−440.43 kJ/kg)]=0m=0.0294 kg×(1000 g1 kg)m=29.4 g

Thus, the amount of ice that needs to be added in an insulated cylinder is 29.4 g_.

Answer 6

Chapter 5, Problem 100RQ

To determine

The amount of ice that needs to be added in an insulated cylinder.

Expert Solution & Answer

Explanation of Solution

Given:

The volume of the saturated mixture (V1) is 0.01 m3.

The quality of the mixture is 0.2.

The initial temperature of the mixture (T1) is 120°C.

The heat of fusion is 333.7 kJ/kg.

Calculation:

From the Table A-3 “Properties of common liquids, solids and foods”, obtain the value of specific heat of ice at 0 C and 120 C temperature as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

From the Table A-4 “Saturated water”, obtain the value of specific volume and enthalpy of saturated liquid, vapour and change upon vaporization for insulated cylinder device.

At final temperature of water as 120°C.

The specific volume of ice for insulated cylinder device (vf) is 0.001060

The specific volume of ice for insulated cylinder device (vg) is 0.89133 m3/kg

The specific enthalpy of ice (hf) is 503.81 kJ/kg.

The specific enthalpy of the mixture (hfg) is 2202.1 kJ/kg

Write the expression of initial specific volume of a two-phase system for insulated cylinder device.

v1=vf+xvfg=vf+x(vg−vf)

v1=(0.001060)+(0.2)×(0.89133 m3/kg−0.001060)=(0.001060)+(0.2)×(0.89027 m3/kg)=0.179114 m3/kg

Write the expression of initial specific enthalpy of a two-phase system for insulated cylinder device.

h1=hf+xhfg

h1=(503.81)+(0.2)×(2202.1 kJ/kg)=944.23 kJ/kg

Here the value final specific enthalpy is equal to specific enthalpy of saturated liquid as 503.81 kJ/kg.

Determine the initial specific volume of the refrigerant.

v1=V1mm=V1v1

m=0.01 m30.179114 m3/kg=0.05583 kg

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (I) and write energy balance relation of an insulated cylinder.

Wb,in=ΔUWb,in−ΔU=0 (II)

Here, the boundary work to be done into the system is Wb,in and the change in the internal energy is ΔU.

Substitute ΔH for Wb,in−ΔU in Equation (II).

ΔH=0ΔHice+ΔHwater=0[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[m(h2−h1)]water=0 (III)

Here, the mass of the ice is m, the heat of fusion of ice is hif, the specific heat of ice at liquid state is c, the initial temperature of ice is T1, the final temperature of ice is T2, the initial specific enthalpy of ice is h1, and the final specific enthalpy of ice is h2.

Substitute 0°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 120°C for T2,liquid, 0.05583 kg for mwater, 503.81 kJ/kg for h2, and 944.24 kJ/kg for h1 in Equation (III).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((120°C)−0°C)liquid]ice+[(0.05583 kg)((503.81 kJ/kg)−(944.24 kJ/kg))]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(120°C)]ice+[(0.05583 kg)(−440.43 kJ/kg)]=0m=0.0294 kg×(1000 g1 kg)m=29.4 g

Thus, the amount of ice that needs to be added in an insulated cylinder is 29.4 g_.

Answer 7

Chapter 5, Problem 100RQ

To determine

The amount of ice that needs to be added in an insulated cylinder.

Answer 8

Expert Solution & Answer

Explanation of Solution

Given:

The volume of the saturated mixture (V1) is 0.01 m3.

The quality of the mixture is 0.2.

The initial temperature of the mixture (T1) is 120°C.

The heat of fusion is 333.7 kJ/kg.

Calculation:

From the Table A-3 “Properties of common liquids, solids and foods”, obtain the value of specific heat of ice at 0 C and 120 C temperature as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

From the Table A-4 “Saturated water”, obtain the value of specific volume and enthalpy of saturated liquid, vapour and change upon vaporization for insulated cylinder device.

At final temperature of water as 120°C.

The specific volume of ice for insulated cylinder device (vf) is 0.001060

The specific volume of ice for insulated cylinder device (vg) is 0.89133 m3/kg

The specific enthalpy of ice (hf) is 503.81 kJ/kg.

The specific enthalpy of the mixture (hfg) is 2202.1 kJ/kg

Write the expression of initial specific volume of a two-phase system for insulated cylinder device.

v1=vf+xvfg=vf+x(vg−vf)

v1=(0.001060)+(0.2)×(0.89133 m3/kg−0.001060)=(0.001060)+(0.2)×(0.89027 m3/kg)=0.179114 m3/kg

Write the expression of initial specific enthalpy of a two-phase system for insulated cylinder device.

h1=hf+xhfg

h1=(503.81)+(0.2)×(2202.1 kJ/kg)=944.23 kJ/kg

Here the value final specific enthalpy is equal to specific enthalpy of saturated liquid as 503.81 kJ/kg.

Determine the initial specific volume of the refrigerant.

v1=V1mm=V1v1

m=0.01 m30.179114 m3/kg=0.05583 kg

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (I)

Here, the total energy entering the system is Ein, the total energy leaving the system is Eout, and the change in the total energy of the system is ΔEsystem.

Simplify Equation (I) and write energy balance relation of an insulated cylinder.

Wb,in=ΔUWb,in−ΔU=0 (II)

Here, the boundary work to be done into the system is Wb,in and the change in the internal energy is ΔU.

Substitute ΔH for Wb,in−ΔU in Equation (II).

ΔH=0ΔHice+ΔHwater=0[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[m(h2−h1)]water=0 (III)

Here, the mass of the ice is m, the heat of fusion of ice is hif, the specific heat of ice at liquid state is c, the initial temperature of ice is T1, the final temperature of ice is T2, the initial specific enthalpy of ice is h1, and the final specific enthalpy of ice is h2.

Substitute 0°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 120°C for T2,liquid, 0.05583 kg for mwater, 503.81 kJ/kg for h2, and 944.24 kJ/kg for h1 in Equation (III).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((120°C)−0°C)liquid]ice+[(0.05583 kg)((503.81 kJ/kg)−(944.24 kJ/kg))]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(120°C)]ice+[(0.05583 kg)(−440.43 kJ/kg)]=0m=0.0294 kg×(1000 g1 kg)m=29.4 g