The final equilibrium temperature in the room.

Question

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Answer 1

Question

Chapter 7.13, Problem 205RP

a)

To determine

The final equilibrium temperature in the room.

a)

Expert Solution

Answer to Problem 205RP

The final equilibrium temperature in the room is 78.4°C.

Explanation of Solution

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.

Write the expression to calculate the ideal gas equation, to find the mass of the air.

mair=P1(ν1)airR(T1)air (II)

Here, mass of the air is mair, initial pressure is P1, initial volume of air is (ν1)air, gas constant is R and initial temperature of air is (T1)air.

Conclusion:

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

0−0=ΔUwater+ΔUair[mwatercp,water(T2−(T1)water)]+[maircv,air(T2−(T1)air)]=0 (III)

Here, change in internal energy in air is ΔUair, change in internal energy in water is ΔUwater, mass of the air is mair, mass of the water is mwater, initial temperature of water is (T1)water, specific heat at constant pressure is cp,water , specific heat at constant volume is cv,air , and initial temperature of air is (T1)air.

From the Table A-1, “the molar mass, gas constant and critical point properties table”, select the gas constant (R) of air as 0.2870 kJ/kg⋅K.

Substitute 100 kPa for P1, (4m×5m×7m) for (ν1)air, 0.2870 kJ/kg⋅K for R and 22 °C for (T1)air in Equation (II).

mair=(100 kPa)(4m×5m×7m)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)22 °C=(100 kPa)(140 m3)(0.2870 kJ/kg⋅K)(1 kPa⋅m31 kJ)(22+273)K=165.4 kg

From the Table A-3, “the properties of common liquids, solids and foods table”, select the specific heat at constant pressure (cp) and select the specific heat at constant volume (cv) of water at room temperature as 4.18 kJ/kg⋅K and 0.718 kJ/kg⋅K respectively.

Substitute 1000 kg for mwater, 4.18 kJ/kg⋅K for cp,water, 80 °C for (T1)water, 165.4 kg for mair, 0.718 kJ/kg⋅K for cv and 22 °C for (T1)air in Equation (III).

{[(1000 kg)(4.18 kJ/kg⋅K)(T2−80°C)]+[(165.4 kg)(0.718 kJ/kg⋅K)(T2−22°C)]}=04298.75T2=337012.6T2=337012.64298.75T2=78.4°C

Thus, the final equilibrium temperature in the room is 78.4°C.

b)

To determine

The entropy generation during the process.

b)

Expert Solution

Answer to Problem 205RP

The entropy generation during the process is 1.79 kJ/K.

Explanation of Solution

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

Sin−Sout+Sgen=ΔSsystem (IV)

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

Conclusion:

Substitute 0 for Sin, 0 for Sout and ΔSair+ΔSwater for ΔSsystem in Equation (V).

0−0+Sgen=ΔSair+ΔSwaterSgen=[(maircv,airln(T2(T1)air)+mairRln((ν2)air(ν1)air))+(mwatercp,waterln(T2(T1)water))] (V)

Here, final volume of air is (ν2)air.

Substitute 165.4 kg for mair, 0.718 kJ/kg⋅K for cv,air, 22 °C for (T1)air, 78.4 °C for T2, 1000 kg for mwater, 4.18 kJ/kg⋅K for cp,water, 80 °C for (T1)water, (4m×5m×7m) for (ν2)air, (4m×5m×7m) for (ν1)air, and 0.2870 kJ/kg⋅K for R in Equation (V).

Sgen={((165.4 kg)(0.718 kJ/kg⋅K)ln(78.4 °C22°C)+(165.4 kg)(0.2870 kJ/kg⋅K)ln((4m×5m×7m)(4m×5m×7m)))+((1000 kg)(4.18 kJ/kg⋅K)ln(78.4 °C22°C))}

Sgen={((165.4 kg)(0.718 kJ/kg⋅K)ln((78.4+273)K(22+273)K)+(165.4 kg)(0.2870 kJ/kg⋅K)ln(140 m3140 m3))+((1000 kg)(4.18 kJ/kg⋅K)ln((78.4+273)K(80+273)K))}=20.78+(−18.99)=1.79 kJ/K

Thus, the entropy generation during the process is 1.79 kJ/K.

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