The amount of ice that needs to be added into water at 0 C of temperature.

Question

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Answer 1

Question

Chapter 4.5, Problem 128RP

(a)

To determine

The amount of ice that needs to be added into water at 0 C of temperature.

(a)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

Explanation of Solution

Write the expression for the mass of the water.

mw=ρV (I)

Here, the density of the water is ρ and the volume of the water is V.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (II)

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 (II) and write energy balance relation of cold water

Qin−Wout=ΔU (III)

Here, the heat to be transfer into the system is Qin, the work to be done by the system is Wout, and the change in the internal energy is ΔU.

Conclusion:

Substitute 0 for Qin and 0 for Wout in Equation (III).

0=ΔU0=ΔUice+ΔUwater0=[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[mc(T2−T1)]water (IV)

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, and the final temperature of ice is T2

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

Substitute 1 kg/L for ρ and 0.3 L for V in Equation (I).

mw=(1 kg/L)(0.3 L)=0.3 kg

For initial temperature of ice as 0 C.

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, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water=0m=0.0546 kg×(1000 g1 kg)m=54.6 g

Thus, the amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

(b)

To determine

The amount of ice that needs to be added into water at -20 C of temperature.

The amount of water that needs to cool down the cold water at 0 C of temperature.

(b)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

The amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

Explanation of Solution

Substitute 0 for Qin and 0 for Wout in Equation (III) and write energy balance relation for cooling down the cold water.

0=ΔU0=ΔUcoldwater+ΔUwater0=[mc(T2−T1)]coldwater+[mc(T2−T1)]water (V)

Here, the mass of cold water is mcoldwater, the specific heat of cold water is ccoldwater, the initial temperature of cold water is T1,coldwater, the final temperature of cold water is T2,coldwater, the mass of water is mwater, the specific heat of water is cwater, the initial temperature of water is T1,water, and the final temperature of water is T2,water.

Conclusion:

For initial temperature of ice as -20 C instead of 0 C.

Substitute −20°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−(−20°C))solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0[m[(2.11 kJ/kg°C)(20°C)+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water]=0m=0.0487 kg×(1000 g1 kg)m=48.7 g

Thus, the amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

Substitute 4.18 kJ/kg⋅°C for c, 5°C for T2,coldwater, 0°C for T1,coldwater, 0.3 kg for mwater, 5°C for T2,water, and 20°C for T1,water in Equation (V).

[mcoldwater(4.18 kJ/kg⋅°C)(5−0)°C]+(0.3 kg)(4.18 kJ/kg⋅°C)(5−20)°C=0[mcoldwater(4.18 kJ/kg⋅°C)(5°C)]+(0.3 kg)(4.18 kJ/kg⋅°C)(−15°C)=0m=0.9 kg×(1000 g1 kg)m=900 g

Thus, the amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

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Answer 2

Question

Chapter 4.5, Problem 128RP

(a)

To determine

The amount of ice that needs to be added into water at 0 C of temperature.

(a)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

Explanation of Solution

Write the expression for the mass of the water.

mw=ρV (I)

Here, the density of the water is ρ and the volume of the water is V.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (II)

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 (II) and write energy balance relation of cold water

Qin−Wout=ΔU (III)

Here, the heat to be transfer into the system is Qin, the work to be done by the system is Wout, and the change in the internal energy is ΔU.

Conclusion:

Substitute 0 for Qin and 0 for Wout in Equation (III).

0=ΔU0=ΔUice+ΔUwater0=[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[mc(T2−T1)]water (IV)

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, and the final temperature of ice is T2

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

Substitute 1 kg/L for ρ and 0.3 L for V in Equation (I).

mw=(1 kg/L)(0.3 L)=0.3 kg

For initial temperature of ice as 0 C.

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, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water=0m=0.0546 kg×(1000 g1 kg)m=54.6 g

Thus, the amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

(b)

To determine

The amount of ice that needs to be added into water at -20 C of temperature.

The amount of water that needs to cool down the cold water at 0 C of temperature.

(b)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

The amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

Explanation of Solution

Substitute 0 for Qin and 0 for Wout in Equation (III) and write energy balance relation for cooling down the cold water.

0=ΔU0=ΔUcoldwater+ΔUwater0=[mc(T2−T1)]coldwater+[mc(T2−T1)]water (V)

Here, the mass of cold water is mcoldwater, the specific heat of cold water is ccoldwater, the initial temperature of cold water is T1,coldwater, the final temperature of cold water is T2,coldwater, the mass of water is mwater, the specific heat of water is cwater, the initial temperature of water is T1,water, and the final temperature of water is T2,water.

Conclusion:

For initial temperature of ice as -20 C instead of 0 C.

Substitute −20°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−(−20°C))solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0[m[(2.11 kJ/kg°C)(20°C)+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water]=0m=0.0487 kg×(1000 g1 kg)m=48.7 g

Thus, the amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

Substitute 4.18 kJ/kg⋅°C for c, 5°C for T2,coldwater, 0°C for T1,coldwater, 0.3 kg for mwater, 5°C for T2,water, and 20°C for T1,water in Equation (V).

[mcoldwater(4.18 kJ/kg⋅°C)(5−0)°C]+(0.3 kg)(4.18 kJ/kg⋅°C)(5−20)°C=0[mcoldwater(4.18 kJ/kg⋅°C)(5°C)]+(0.3 kg)(4.18 kJ/kg⋅°C)(−15°C)=0m=0.9 kg×(1000 g1 kg)m=900 g

Thus, the amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

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Answer 3

Question

Chapter 4.5, Problem 128RP

(a)

To determine

The amount of ice that needs to be added into water at 0 C of temperature.

(a)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

Explanation of Solution

Write the expression for the mass of the water.

mw=ρV (I)

Here, the density of the water is ρ and the volume of the water is V.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (II)

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 (II) and write energy balance relation of cold water

Qin−Wout=ΔU (III)

Here, the heat to be transfer into the system is Qin, the work to be done by the system is Wout, and the change in the internal energy is ΔU.

Conclusion:

Substitute 0 for Qin and 0 for Wout in Equation (III).

0=ΔU0=ΔUice+ΔUwater0=[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[mc(T2−T1)]water (IV)

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, and the final temperature of ice is T2

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

Substitute 1 kg/L for ρ and 0.3 L for V in Equation (I).

mw=(1 kg/L)(0.3 L)=0.3 kg

For initial temperature of ice as 0 C.

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, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water=0m=0.0546 kg×(1000 g1 kg)m=54.6 g

Thus, the amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

(b)

To determine

The amount of ice that needs to be added into water at -20 C of temperature.

The amount of water that needs to cool down the cold water at 0 C of temperature.

(b)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

The amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

Explanation of Solution

Substitute 0 for Qin and 0 for Wout in Equation (III) and write energy balance relation for cooling down the cold water.

0=ΔU0=ΔUcoldwater+ΔUwater0=[mc(T2−T1)]coldwater+[mc(T2−T1)]water (V)

Here, the mass of cold water is mcoldwater, the specific heat of cold water is ccoldwater, the initial temperature of cold water is T1,coldwater, the final temperature of cold water is T2,coldwater, the mass of water is mwater, the specific heat of water is cwater, the initial temperature of water is T1,water, and the final temperature of water is T2,water.

Conclusion:

For initial temperature of ice as -20 C instead of 0 C.

Substitute −20°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−(−20°C))solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0[m[(2.11 kJ/kg°C)(20°C)+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water]=0m=0.0487 kg×(1000 g1 kg)m=48.7 g

Thus, the amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

Substitute 4.18 kJ/kg⋅°C for c, 5°C for T2,coldwater, 0°C for T1,coldwater, 0.3 kg for mwater, 5°C for T2,water, and 20°C for T1,water in Equation (V).

[mcoldwater(4.18 kJ/kg⋅°C)(5−0)°C]+(0.3 kg)(4.18 kJ/kg⋅°C)(5−20)°C=0[mcoldwater(4.18 kJ/kg⋅°C)(5°C)]+(0.3 kg)(4.18 kJ/kg⋅°C)(−15°C)=0m=0.9 kg×(1000 g1 kg)m=900 g

Thus, the amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

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Answer 4

Question

Chapter 4.5, Problem 128RP

(a)

To determine

The amount of ice that needs to be added into water at 0 C of temperature.

(a)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

Explanation of Solution

Write the expression for the mass of the water.

mw=ρV (I)

Here, the density of the water is ρ and the volume of the water is V.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (II)

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 (II) and write energy balance relation of cold water

Qin−Wout=ΔU (III)

Here, the heat to be transfer into the system is Qin, the work to be done by the system is Wout, and the change in the internal energy is ΔU.

Conclusion:

Substitute 0 for Qin and 0 for Wout in Equation (III).

0=ΔU0=ΔUice+ΔUwater0=[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[mc(T2−T1)]water (IV)

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, and the final temperature of ice is T2

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

Substitute 1 kg/L for ρ and 0.3 L for V in Equation (I).

mw=(1 kg/L)(0.3 L)=0.3 kg

For initial temperature of ice as 0 C.

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, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water=0m=0.0546 kg×(1000 g1 kg)m=54.6 g

Thus, the amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

(b)

To determine

The amount of ice that needs to be added into water at -20 C of temperature.

The amount of water that needs to cool down the cold water at 0 C of temperature.

(b)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

The amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

Explanation of Solution

Substitute 0 for Qin and 0 for Wout in Equation (III) and write energy balance relation for cooling down the cold water.

0=ΔU0=ΔUcoldwater+ΔUwater0=[mc(T2−T1)]coldwater+[mc(T2−T1)]water (V)

Here, the mass of cold water is mcoldwater, the specific heat of cold water is ccoldwater, the initial temperature of cold water is T1,coldwater, the final temperature of cold water is T2,coldwater, the mass of water is mwater, the specific heat of water is cwater, the initial temperature of water is T1,water, and the final temperature of water is T2,water.

Conclusion:

For initial temperature of ice as -20 C instead of 0 C.

Substitute −20°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−(−20°C))solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0[m[(2.11 kJ/kg°C)(20°C)+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water]=0m=0.0487 kg×(1000 g1 kg)m=48.7 g

Thus, the amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

Substitute 4.18 kJ/kg⋅°C for c, 5°C for T2,coldwater, 0°C for T1,coldwater, 0.3 kg for mwater, 5°C for T2,water, and 20°C for T1,water in Equation (V).

[mcoldwater(4.18 kJ/kg⋅°C)(5−0)°C]+(0.3 kg)(4.18 kJ/kg⋅°C)(5−20)°C=0[mcoldwater(4.18 kJ/kg⋅°C)(5°C)]+(0.3 kg)(4.18 kJ/kg⋅°C)(−15°C)=0m=0.9 kg×(1000 g1 kg)m=900 g

Thus, the amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

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Answer 5

Chapter 4.5, Problem 128RP

(a)

To determine

The amount of ice that needs to be added into water at 0 C of temperature.

(a)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

Explanation of Solution

Write the expression for the mass of the water.

mw=ρV (I)

Here, the density of the water is ρ and the volume of the water is V.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (II)

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 (II) and write energy balance relation of cold water

Qin−Wout=ΔU (III)

Here, the heat to be transfer into the system is Qin, the work to be done by the system is Wout, and the change in the internal energy is ΔU.

Conclusion:

Substitute 0 for Qin and 0 for Wout in Equation (III).

0=ΔU0=ΔUice+ΔUwater0=[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[mc(T2−T1)]water (IV)

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, and the final temperature of ice is T2

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

Substitute 1 kg/L for ρ and 0.3 L for V in Equation (I).

mw=(1 kg/L)(0.3 L)=0.3 kg

For initial temperature of ice as 0 C.

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, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water=0m=0.0546 kg×(1000 g1 kg)m=54.6 g

Thus, the amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

(b)

To determine

The amount of ice that needs to be added into water at -20 C of temperature.

The amount of water that needs to cool down the cold water at 0 C of temperature.

(b)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

The amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

Explanation of Solution

Substitute 0 for Qin and 0 for Wout in Equation (III) and write energy balance relation for cooling down the cold water.

0=ΔU0=ΔUcoldwater+ΔUwater0=[mc(T2−T1)]coldwater+[mc(T2−T1)]water (V)

Here, the mass of cold water is mcoldwater, the specific heat of cold water is ccoldwater, the initial temperature of cold water is T1,coldwater, the final temperature of cold water is T2,coldwater, the mass of water is mwater, the specific heat of water is cwater, the initial temperature of water is T1,water, and the final temperature of water is T2,water.

Conclusion:

For initial temperature of ice as -20 C instead of 0 C.

Substitute −20°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−(−20°C))solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0[m[(2.11 kJ/kg°C)(20°C)+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water]=0m=0.0487 kg×(1000 g1 kg)m=48.7 g

Thus, the amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

Substitute 4.18 kJ/kg⋅°C for c, 5°C for T2,coldwater, 0°C for T1,coldwater, 0.3 kg for mwater, 5°C for T2,water, and 20°C for T1,water in Equation (V).

[mcoldwater(4.18 kJ/kg⋅°C)(5−0)°C]+(0.3 kg)(4.18 kJ/kg⋅°C)(5−20)°C=0[mcoldwater(4.18 kJ/kg⋅°C)(5°C)]+(0.3 kg)(4.18 kJ/kg⋅°C)(−15°C)=0m=0.9 kg×(1000 g1 kg)m=900 g

Thus, the amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

Answer 6

Chapter 4.5, Problem 128RP

(a)

To determine

The amount of ice that needs to be added into water at 0 C of temperature.

(a)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

Explanation of Solution

Write the expression for the mass of the water.

mw=ρV (I)

Here, the density of the water is ρ and the volume of the water is V.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (II)

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 (II) and write energy balance relation of cold water

Qin−Wout=ΔU (III)

Here, the heat to be transfer into the system is Qin, the work to be done by the system is Wout, and the change in the internal energy is ΔU.

Conclusion:

Substitute 0 for Qin and 0 for Wout in Equation (III).

0=ΔU0=ΔUice+ΔUwater0=[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[mc(T2−T1)]water (IV)

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, and the final temperature of ice is T2

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

Substitute 1 kg/L for ρ and 0.3 L for V in Equation (I).

mw=(1 kg/L)(0.3 L)=0.3 kg

For initial temperature of ice as 0 C.

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, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water=0m=0.0546 kg×(1000 g1 kg)m=54.6 g

Thus, the amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

Answer 7

Chapter 4.5, Problem 128RP

(a)

To determine

The amount of ice that needs to be added into water at 0 C of temperature.

Answer 8

(a)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Write the expression for the mass of the water.

mw=ρV (I)

Here, the density of the water is ρ and the volume of the water is V.

Write the expression for the energy balance equation.

Ein−Eout=ΔEsystem (II)

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 (II) and write energy balance relation of cold water

Qin−Wout=ΔU (III)

Here, the heat to be transfer into the system is Qin, the work to be done by the system is Wout, and the change in the internal energy is ΔU.

Conclusion:

Substitute 0 for Qin and 0 for Wout in Equation (III).

0=ΔU0=ΔUice+ΔUwater0=[mc(0°C−T1)solid+mhif+mc(T2−0°C)liquid]ice+[mc(T2−T1)]water (IV)

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, and the final temperature of ice is T2

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

Substitute 1 kg/L for ρ and 0.3 L for V in Equation (I).

mw=(1 kg/L)(0.3 L)=0.3 kg

For initial temperature of ice as 0 C.

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, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−0°C)solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0m[0+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water=0m=0.0546 kg×(1000 g1 kg)m=54.6 g

Thus, the amount of ice that needs to be added into water at 0 C of temperature is 54.6 g_.

Answer 13

(b)

To determine

The amount of ice that needs to be added into water at -20 C of temperature.

The amount of water that needs to cool down the cold water at 0 C of temperature.

(b)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

The amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

Explanation of Solution

Substitute 0 for Qin and 0 for Wout in Equation (III) and write energy balance relation for cooling down the cold water.

0=ΔU0=ΔUcoldwater+ΔUwater0=[mc(T2−T1)]coldwater+[mc(T2−T1)]water (V)

Here, the mass of cold water is mcoldwater, the specific heat of cold water is ccoldwater, the initial temperature of cold water is T1,coldwater, the final temperature of cold water is T2,coldwater, the mass of water is mwater, the specific heat of water is cwater, the initial temperature of water is T1,water, and the final temperature of water is T2,water.

Conclusion:

For initial temperature of ice as -20 C instead of 0 C.

Substitute −20°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−(−20°C))solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0[m[(2.11 kJ/kg°C)(20°C)+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water]=0m=0.0487 kg×(1000 g1 kg)m=48.7 g

Thus, the amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

Substitute 4.18 kJ/kg⋅°C for c, 5°C for T2,coldwater, 0°C for T1,coldwater, 0.3 kg for mwater, 5°C for T2,water, and 20°C for T1,water in Equation (V).

[mcoldwater(4.18 kJ/kg⋅°C)(5−0)°C]+(0.3 kg)(4.18 kJ/kg⋅°C)(5−20)°C=0[mcoldwater(4.18 kJ/kg⋅°C)(5°C)]+(0.3 kg)(4.18 kJ/kg⋅°C)(−15°C)=0m=0.9 kg×(1000 g1 kg)m=900 g

Thus, the amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

Answer 14

(b)

To determine

The amount of ice that needs to be added into water at -20 C of temperature.

The amount of water that needs to cool down the cold water at 0 C of temperature.

Answer 15

(b)

Expert Solution

Answer to Problem 128RP

The amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

The amount of water that needs to cool down the cold water at 0 C of temperature is 900 g_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Substitute 0 for Qin and 0 for Wout in Equation (III) and write energy balance relation for cooling down the cold water.

0=ΔU0=ΔUcoldwater+ΔUwater0=[mc(T2−T1)]coldwater+[mc(T2−T1)]water (V)

Here, the mass of cold water is mcoldwater, the specific heat of cold water is ccoldwater, the initial temperature of cold water is T1,coldwater, the final temperature of cold water is T2,coldwater, the mass of water is mwater, the specific heat of water is cwater, the initial temperature of water is T1,water, and the final temperature of water is T2,water.

Conclusion:

For initial temperature of ice as -20 C instead of 0 C.

Substitute −20°C for T1,solid, 2.11 kJ/kg°C for csolid, 333.7 kJ/kg for hif, 4.18 kJ/kg⋅°C for cliquid, 5°C for T2,liquid, 0.3 kg for mwater, and 20°C for T1,water in Equation (IV).

[[m(2.11 kJ/kg°C)(0°C−(−20°C))solid+m(333.7 kJ/kg)+m(4.18 kJ/kg⋅°C)((5°C)−0°C)liquid]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(5°C−20°C)]water]=0[m[(2.11 kJ/kg°C)(20°C)+(333.7 kJ/kg)+(4.18 kJ/kg⋅°C)(5°C)]ice+[(0.3 kJ/kg)(4.18 kJ/kg⋅°C)(−15°C)]water]=0m=0.0487 kg×(1000 g1 kg)m=48.7 g

Thus, the amount of ice that needs to be added into water at -20 C of temperature is 48.7 g_.

Substitute 4.18 kJ/kg⋅°C for c, 5°C for T2,coldwater, 0°C for T1,coldwater, 0.3 kg for mwater, 5°C for T2,water, and 20°C for T1,water in Equation (V).

[mcoldwater(4.18 kJ/kg⋅°C)(5−0)°C]+(0.3 kg)(4.18 kJ/kg⋅°C)(5−20)°C=0[mcoldwater(4.18 kJ/kg⋅°C)(5°C)]+(0.3 kg)(4.18 kJ/kg⋅°C)(−15°C)=0m=0.9 kg×(1000 g1 kg)m=900 g