EBK THERMODYNAMICS: AN ENGINEERING APPR
EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 9780100257054
Author: CENGEL
Publisher: YUZU
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Entropy
In physics, entropy means the amount of disorder or the degree of randomness associated with a thermodynamic system. The system is not self-organized when the entropy of the system is high. For instance, when a system is at an elevated temperature, its m…
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Chapter 7.13, Problem 61P

(a)

To determine

The entropy change of the computer chips.

(a)

Expert Solution
Check Mark

Answer to Problem 61P

The entropy change of the computer chips is 0.000772 kJ/K_.

Explanation of Solution

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.

Substitute 0 for Ein, 0 for Eout, and ΔU for ΔEsystem in Equation (I)

(0)−(0)=ΔU(0)=[m(u2−u1)]Chips+[m(u2−u1)]R-134a[m(u2−u1)]Chips=[m(u2−u1)]R-134a (II)

Here, the mass is m, the initial specific internal energy is u1, and the final specific internal energy is u2.

Determine the heat released by the computer chips.

Qchips=mc(T1−T2) (III)

Here, the mass of the computer chips is m, the specific heat of the computer chips is c, the initial temperature of the computer chips is T1, and the final temperature of the computer chips is T2.

Determine the mass of the refrigerant vaporized during this heat exchange process.

mg,2=Qchipshg−hf=Qchipshfg@−40°C (IV)

Here, the saturated specific enthalpy change upon vaporization at −40°C is hfg@−40°C.

Determine the change in the entropy of the R-134a.

ΔSR-134a=mg,2sg,2+mf,2sf,2−mf,1sf,1 (V)

Here, the mass of the refrigerant vaporized at state 2 is mg,2, the entropy of the refrigerant vaporized at state 2 is sg,2, the mass of the refrigerant liquid at state 2 is mf,2, the entropy of the refrigerant liquid at state 2 is sf,2, the mass of the refrigerant liquid at state 1 is mf,1, and the entropy of the refrigerant liquid at state 1 is sf,1.

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C  for T1, and −40°C for T2 in Equation (III).

Qchips=(10 g)(0.3 kJ/kg⋅K)((20°C)−(−40°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)((20°C+273)−(−40°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)((293 K)−(233 K))=0.18 kJ

From the Table A-11, to obtain the value of the specific enthalpy change upon vaporization, entropy of the refrigerant vaporized at state 2, entropy of the refrigerant liquid at state 2, entropy of the refrigerant liquid at state 1 at final temperature of −40°C as

hfg=225.86 kJ/kgsg,2=0.96866 kJ/kg⋅Ksf,1=0.00000 kJ/kg⋅K

Substitute 0.18 kJ for Qchips, 225.86 kJ/kg for hfg@−40°C in Equation (IV).

mg,2=(0.18 kJ)(225.86 kJ/kg)=0.000797 kg

Substitute 0.000797 kg for mg,2, 0.96866 kJ/kg⋅K for sg,2, 5 g for mf,2, 5 g for mf,1, and 0.00000 kJ/kg⋅K for sf,1 in Equation (V).

ΔSR-134a=[(0.000797 kg)×(0.96866 kJ/kg⋅K)+(5 g)×(0.00000 kJ/kg⋅K)−(5 g)×(0.00000 kJ/kg⋅K)]=0.00772 kJ/K

Thus, the entropy change of the computer chips is 0.000772 kJ/K_.

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C  for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

(b)

To determine

The entropy change of the R-134.

(b)

Expert Solution
Check Mark

Answer to Problem 61P

The entropy change of the R-134 is −0.00069 kJ/K_.

Explanation of Solution

Determine the entropy change of the computer chips.

ΔSchips=mclnT2T1 (VI)

Conclusion:

Substitute 10 g for mchips, 0.3 kJ/kg⋅K for c, 20°C  for T1, and −40°C for T2 in Equation (VI).

ΔSchips=(10 g)(0.3 kJ/kg⋅K)ln((−40°C)(20°C))=(10 g×(10−3kg1 g))(0.3 kJ/kg⋅K)ln((−40°C+273)(20°C+273))=(0.010 kg)(0.3 kJ/kg⋅K)(−0.22913)=−0.00069 kJ/K

Thus, the entropy change of the R-134 is −0.00069 kJ/K_.

(c)

To determine

The entropy change of the entire system.

(c)

Expert Solution
Check Mark

Answer to Problem 61P

The entropy change of the entire system is 0.00008457 kJ/K_.

Explanation of Solution

Determine the total entropy change of the entire system.

ΔStotal=ΔSR-134a+ΔSchips (VII)

Conclusion:

Substitute 0.000772 kJ/K for ΔSR-134a and −0.00069 kJ/K for ΔSchips in Equation (VII).

ΔStotal=(0.000772 kJ/K)+(−0.00069 kJ/K)=0.00008457 kJ/K

Thus, the entropy change of the entire system is 0.00008457 kJ/K_.

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Chapter 7 Solutions

EBK THERMODYNAMICS: AN ENGINEERING APPR

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