The final equilibrium temperature.

Question

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Answer 1

Question

Chapter 8, Problem 97P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Explanation of Solution

Given:

The mass of the water (m) is 45 kg.

The initial temperature of the water (T1) is 95°C.

The volume of the room (v) is 90 m3.

The initial temperature of the room (T1) is 12°C.

Calculation;

Refer the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Calculate the mass of the air.

mair=P1vRT1mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

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

The specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Explanation of Solution

Calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air)Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

c)

To determine

The entropy generation.

c)

Expert Solution

Explanation of Solution

Calculate the final pressure (P2).

mair=P2vRT2

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)P2=121.98 kPa=122 kPa

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSwater in Equation (II).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water))

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/K.

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Answer 2

Question

Chapter 8, Problem 97P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Explanation of Solution

Given:

The mass of the water (m) is 45 kg.

The initial temperature of the water (T1) is 95°C.

The volume of the room (v) is 90 m3.

The initial temperature of the room (T1) is 12°C.

Calculation;

Refer the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Calculate the mass of the air.

mair=P1vRT1mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

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

The specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Explanation of Solution

Calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air)Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

c)

To determine

The entropy generation.

c)

Expert Solution

Explanation of Solution

Calculate the final pressure (P2).

mair=P2vRT2

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)P2=121.98 kPa=122 kPa

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSwater in Equation (II).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water))

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/K.

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Answer 3

Question

Chapter 8, Problem 97P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Explanation of Solution

Given:

The mass of the water (m) is 45 kg.

The initial temperature of the water (T1) is 95°C.

The volume of the room (v) is 90 m3.

The initial temperature of the room (T1) is 12°C.

Calculation;

Refer the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Calculate the mass of the air.

mair=P1vRT1mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

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

The specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Explanation of Solution

Calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air)Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

c)

To determine

The entropy generation.

c)

Expert Solution

Explanation of Solution

Calculate the final pressure (P2).

mair=P2vRT2

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)P2=121.98 kPa=122 kPa

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSwater in Equation (II).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water))

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/K.

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Answer 4

Question

Chapter 8, Problem 97P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Explanation of Solution

Given:

The mass of the water (m) is 45 kg.

The initial temperature of the water (T1) is 95°C.

The volume of the room (v) is 90 m3.

The initial temperature of the room (T1) is 12°C.

Calculation;

Refer the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Calculate the mass of the air.

mair=P1vRT1mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

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

The specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Explanation of Solution

Calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air)Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

c)

To determine

The entropy generation.

c)

Expert Solution

Explanation of Solution

Calculate the final pressure (P2).

mair=P2vRT2

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)P2=121.98 kPa=122 kPa

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSwater in Equation (II).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water))

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/K.

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Answer 5

Chapter 8, Problem 97P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Explanation of Solution

Given:

The mass of the water (m) is 45 kg.

The initial temperature of the water (T1) is 95°C.

The volume of the room (v) is 90 m3.

The initial temperature of the room (T1) is 12°C.

Calculation;

Refer the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Calculate the mass of the air.

mair=P1vRT1mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

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

The specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Explanation of Solution

Calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air)Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

c)

To determine

The entropy generation.

c)

Expert Solution

Explanation of Solution

Calculate the final pressure (P2).

mair=P2vRT2

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)P2=121.98 kPa=122 kPa

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSwater in Equation (II).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water))

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/K.

Answer 6

Chapter 8, Problem 97P

a)

To determine

The final equilibrium temperature.

a)

Expert Solution

Explanation of Solution

Given:

The mass of the water (m) is 45 kg.

The initial temperature of the water (T1) is 95°C.

The volume of the room (v) is 90 m3.

The initial temperature of the room (T1) is 12°C.

Calculation;

Refer the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Calculate the mass of the air.

mair=P1vRT1mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

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

The specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

Answer 7

Chapter 8, Problem 97P

a)

To determine

The final equilibrium temperature.

Answer 8

a)

Expert Solution

Explanation of Solution

Given:

The mass of the water (m) is 45 kg.

The initial temperature of the water (T1) is 95°C.

The volume of the room (v) is 90 m3.

The initial temperature of the room (T1) is 12°C.

Calculation;

Refer the Table A-2, “Ideal-gas specific heats of various common gases”, obtain the properties for air.

R=0.2870 kJ/kg⋅Kcp=1.005 kJ/kg⋅Kcv=0.718 kJ/kg⋅K

Calculate the mass of the air.

mair=P1vRT1mair=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)12 °C=(101.3 kPa)(90 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(12+273)K=111.46 kg

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

The specific heat of water cw at room temperature is 4.18 kJ/kg⋅K

Write the expression for the energy balance equation for closed system.

Ein−Eout=ΔEsystem (I)

Here, energy transfer into the control volume is Ein, energy transfer out of control volume is Eout and change in internal energy of system is ΔEsystem.

Substitute 0 for Ein, 0 for Eout and ΔUair+ΔUwater for ΔEsystem in Equation (I).

0−0=ΔUwater+ΔUair[mcw(T2−T1)]water+[mcv(T2−T1)]air=0mwatercw(T2−(T1)water)+maircv(T2−(T1)air)=0

{[(45 kg)(4.18 kJ/kg⋅K)(T2−95°C)]+[(111.46 kg)(0.718 kJ/kg⋅K)(T2−12°C)]}=0{[(188.1 kJ/K)(T2−95°C)]+[(80.02 kJ/K)(T2−12°C)]}=0T2=70.2 °C

Thus, the final equilibrium temperature is 70.2 °C.

Answer 9

a)

Expert Solution

Answer 10

a)

Expert Solution

Answer 11

Expert Solution

Answer 12

b)

To determine

The amount of heat transfer to the air.

b)

Expert Solution

Explanation of Solution

Calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air)Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

Answer 13

b)

To determine

The amount of heat transfer to the air.

Answer 14

b)

Expert Solution

Explanation of Solution

Calculate the heat transfer (Q) to the air.

Q=maircv(T2−(T1)air)Q=(111.46 kg)(0.718 kJ/kg⋅K)(70.2 °C−12 °C)=(111.46 kg)(0.718 kJ/kg⋅°K)[(70.2+273)K−(12 +273)K]=4660 kJ

Thus, the amount of heat transfer to the air is 4660 kJ.

Answer 15

b)

Expert Solution

Answer 16

b)

Expert Solution

Answer 17

Expert Solution

Answer 18

c)

To determine

The entropy generation.

c)

Expert Solution

Explanation of Solution

Calculate the final pressure (P2).

mair=P2vRT2

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)P2=121.98 kPa=122 kPa

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSwater in Equation (II).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water))

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K

Thus, the entropy generation is 1.77 kJ/K.

Answer 19

c)

To determine

The entropy generation.

Answer 20

c)

Expert Solution

Explanation of Solution

Calculate the final pressure (P2).

mair=P2vRT2

111.46 kg=P2(90 m3)(0.2870 kJ/kg⋅K)70.2°CP2=(111.46 kg)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(70.2+273) K(90 m3)P2=121.98 kPa=122 kPa

Write the expression for the entropy balance equation of the system.

Sin−Sout+Sgen=ΔSsystem (II)

Here, rate of net entropy in is Sin, rate of net entropy out is Sout, rate of entropy generation is Sgen and change of entropy of the system is ΔSsystem

Substitute Sin=0, Sout=0, and ΔSsystem=ΔSair+ΔSwater in Equation (II).

0−0+Sgen=ΔSair+ΔSwaterSgen=mair(cpln(T2(T1)air)−Rln(P2P1))+(mwatercwln(T2(T1)water))

Sgen=[(111.46 kg)((1.005 kJ/kg⋅K)ln(70.2 °C12 °C)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln(70.2 °C95 °C))]=[(111.46 kg)((1.005 kJ/kg⋅K)ln((70.2+273) K(12+273) K)−(0.2870 kJ/kg⋅K)ln(122 kPa101.3 kPa))+(45 kg(4.18 kJ/kg⋅K)ln((70.2+273) K(95+273) K))]=1.488 kJ/kg+(−13.12 kJ/kg)=1.77 kJ/K