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A closed system undergoes a process in which work is done on the system and the heat transfer Q occurs only at temperature Tb. For each case, determine whether the entropy change of the system is positive, negative, zero, or indeterminate.(a) internally reversible process, Q > 0.(b) internally reversible process, Q = 0.(c) internally reversible process, Q < 0.d) internal irreversibilities present, Q > 0.(e) internal irreversibilities present, Q = 0.(f) internal irreversibilities present, Q < 0.

Respuesta :

ogorwyne

For the system the enthropy change in

  • a. positive
  • b. zero
  • c. negative
  • d. positive
  • e. positive
  • f. indeterminate

a. The internally reversible process Q<0

Heat is added to the process,

Snet outflow = [tex]\frac{-Q}{Tb} \\[/tex]

ΔS = S₂ - S₁

= [tex]0-\frac{-Q}{Tb}[/tex]

Q/Tb is positive

b. internally reversible process, Q = 0

S g.i.r = 0

Snet = 0

S2 - S1 = 0

The enthropy change in the system is zero here

c. Internally reversible process Q < 0

S g.i.r = 0

Snet outflow = [tex]\frac{-Q}{Tb} \\[/tex]

0 [tex]\frac{-Q}{Tb} \\[/tex]

The system is negative here

d. internally irreversible present, Q < 0

S g.i.r > 0

Snet outflow = [tex]\frac{-Q}{Tb} \\[/tex]

S2 - S1 = S g.i,r - [tex]\frac{-Q}{Tb} \\[/tex]

= S g.i.r + [tex]\frac{Q}{Tb} \\[/tex]

The system is positive

e. Internally irreversibilities present, Q = 0

Q = 0

S g.i.r >  0

Snet = 0

S2 - S1 = S g.i.r

ΔS = positive

f. internal irreversibilities present, Q < 0

S g.i.r >  0

S2 - S1 = S g.i.r [tex]\frac{-Q}{Tb} \\[/tex]

The system is indeterminate here

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