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 5, Problem 109RQ

(a)

To determine

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

(a)

Expert Solution

Explanation of Solution

Given:

The density of the water (ρ) is 1 kg/L.

The volume of the water (V) is 0.3 L.

The initial temperature of water (T1liquid) is 20°C.

The initial temperature of ice (T1solid) is 0°C.

The final temperature of ice (T2liquid) is 5°C.

The mass of the water (mwater) is 0.3 kg

The heat of fusion of ice (hif) is 333.7 kJ/kg.

Calculation:

Calculate the mass of the water.

mw=ρV

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

Refer the Table A-3 “Properties of common liquids, solids and foods”,

The value of specific heat of ice at 0 C(csolid) and room temperature (cliquid) as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

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 (II) and write energy balance relation of cold water.

Qin−Wout=ΔU (II)

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.

Substitute 0 for Qin and 0 for Wout in Equation (II) to obtain the mass of the ice.

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

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

(b)

Expert Solution

Explanation of Solution

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

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

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

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

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

[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 5, Problem 109RQ

(a)

To determine

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

(a)

Expert Solution

Explanation of Solution

Given:

The density of the water (ρ) is 1 kg/L.

The volume of the water (V) is 0.3 L.

The initial temperature of water (T1liquid) is 20°C.

The initial temperature of ice (T1solid) is 0°C.

The final temperature of ice (T2liquid) is 5°C.

The mass of the water (mwater) is 0.3 kg

The heat of fusion of ice (hif) is 333.7 kJ/kg.

Calculation:

Calculate the mass of the water.

mw=ρV

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

Refer the Table A-3 “Properties of common liquids, solids and foods”,

The value of specific heat of ice at 0 C(csolid) and room temperature (cliquid) as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

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 (II) and write energy balance relation of cold water.

Qin−Wout=ΔU (II)

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.

Substitute 0 for Qin and 0 for Wout in Equation (II) to obtain the mass of the ice.

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

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

(b)

Expert Solution

Explanation of Solution

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

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

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

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

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

[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 5, Problem 109RQ

(a)

To determine

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

(a)

Expert Solution

Explanation of Solution

Given:

The density of the water (ρ) is 1 kg/L.

The volume of the water (V) is 0.3 L.

The initial temperature of water (T1liquid) is 20°C.

The initial temperature of ice (T1solid) is 0°C.

The final temperature of ice (T2liquid) is 5°C.

The mass of the water (mwater) is 0.3 kg

The heat of fusion of ice (hif) is 333.7 kJ/kg.

Calculation:

Calculate the mass of the water.

mw=ρV

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

Refer the Table A-3 “Properties of common liquids, solids and foods”,

The value of specific heat of ice at 0 C(csolid) and room temperature (cliquid) as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

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 (II) and write energy balance relation of cold water.

Qin−Wout=ΔU (II)

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.

Substitute 0 for Qin and 0 for Wout in Equation (II) to obtain the mass of the ice.

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

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

(b)

Expert Solution

Explanation of Solution

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

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

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

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

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

[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 5, Problem 109RQ

(a)

To determine

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

(a)

Expert Solution

Explanation of Solution

Given:

The density of the water (ρ) is 1 kg/L.

The volume of the water (V) is 0.3 L.

The initial temperature of water (T1liquid) is 20°C.

The initial temperature of ice (T1solid) is 0°C.

The final temperature of ice (T2liquid) is 5°C.

The mass of the water (mwater) is 0.3 kg

The heat of fusion of ice (hif) is 333.7 kJ/kg.

Calculation:

Calculate the mass of the water.

mw=ρV

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

Refer the Table A-3 “Properties of common liquids, solids and foods”,

The value of specific heat of ice at 0 C(csolid) and room temperature (cliquid) as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

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 (II) and write energy balance relation of cold water.

Qin−Wout=ΔU (II)

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.

Substitute 0 for Qin and 0 for Wout in Equation (II) to obtain the mass of the ice.

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

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

(b)

Expert Solution

Explanation of Solution

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

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

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

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

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

[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 5, Problem 109RQ

(a)

To determine

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

(a)

Expert Solution

Explanation of Solution

Given:

The density of the water (ρ) is 1 kg/L.

The volume of the water (V) is 0.3 L.

The initial temperature of water (T1liquid) is 20°C.

The initial temperature of ice (T1solid) is 0°C.

The final temperature of ice (T2liquid) is 5°C.

The mass of the water (mwater) is 0.3 kg

The heat of fusion of ice (hif) is 333.7 kJ/kg.

Calculation:

Calculate the mass of the water.

mw=ρV

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

Refer the Table A-3 “Properties of common liquids, solids and foods”,

The value of specific heat of ice at 0 C(csolid) and room temperature (cliquid) as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

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 (II) and write energy balance relation of cold water.

Qin−Wout=ΔU (II)

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.

Substitute 0 for Qin and 0 for Wout in Equation (II) to obtain the mass of the ice.

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

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

(b)

Expert Solution

Explanation of Solution

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

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

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

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

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

[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 5, Problem 109RQ

(a)

To determine

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

(a)

Expert Solution

Explanation of Solution

Given:

The density of the water (ρ) is 1 kg/L.

The volume of the water (V) is 0.3 L.

The initial temperature of water (T1liquid) is 20°C.

The initial temperature of ice (T1solid) is 0°C.

The final temperature of ice (T2liquid) is 5°C.

The mass of the water (mwater) is 0.3 kg

The heat of fusion of ice (hif) is 333.7 kJ/kg.

Calculation:

Calculate the mass of the water.

mw=ρV

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

Refer the Table A-3 “Properties of common liquids, solids and foods”,

The value of specific heat of ice at 0 C(csolid) and room temperature (cliquid) as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

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 (II) and write energy balance relation of cold water.

Qin−Wout=ΔU (II)

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.

Substitute 0 for Qin and 0 for Wout in Equation (II) to obtain the mass of the ice.

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

[[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 5, Problem 109RQ

(a)

To determine

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

Answer 8

(a)

Expert Solution

Explanation of Solution

Given:

The density of the water (ρ) is 1 kg/L.

The volume of the water (V) is 0.3 L.

The initial temperature of water (T1liquid) is 20°C.

The initial temperature of ice (T1solid) is 0°C.

The final temperature of ice (T2liquid) is 5°C.

The mass of the water (mwater) is 0.3 kg

The heat of fusion of ice (hif) is 333.7 kJ/kg.

Calculation:

Calculate the mass of the water.

mw=ρV

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

Refer the Table A-3 “Properties of common liquids, solids and foods”,

The value of specific heat of ice at 0 C(csolid) and room temperature (cliquid) as 2.11 kJ/kg°C and 4.18 kJ/kg°C.

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 (II) and write energy balance relation of cold water.

Qin−Wout=ΔU (II)

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.

Substitute 0 for Qin and 0 for Wout in Equation (II) to obtain the mass of the ice.

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

[[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 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

To determine

The amount of ice that needs to be added into water at -20 C of temperature and the amount of water that needs to cool down the cold water at 0 C of temperature.

(b)

Expert Solution

Explanation of Solution

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

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

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

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

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

[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 13

(b)

To determine

The amount of ice that needs to be added into water at -20 C of temperature and the amount of water that needs to cool down the cold water at 0 C of temperature.

Answer 14

(b)

Expert Solution

Explanation of Solution

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

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

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

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

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

[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