#set document(title: "4.6 Entropy", author: "OpenStax / XYZ Homework")
#set page(width: 8.5in, height: auto, margin: 1in)
#import "@preview/cetz:0.5.2"
#set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en")
#show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math"))
#set par(justify: true, leading: 0.62em, spacing: 0.9em)
#set enum(spacing: 1.1em)   // room between list items so tall inline fractions don't collide
#set list(spacing: 1.1em)
#set table(stroke: 0.5pt + rgb("#c7ccd3"))

#let BLUE = rgb("#183B6F")     // brand navy — section bars + example/solution labels (white on navy 11.09:1)
#let ORANGE = rgb("#A94509")   // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text)
#let RED = rgb("#DC2626")      // brand error-600
#let GREEN = rgb("#059669")    // brand success-600 (decoration only; small green text uses green-text #007942)

#show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt,
  fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt,
  text(fill: white, weight: "bold", size: 19pt, it.body))
#show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt,
  fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt,
  text(fill: white, weight: "bold", size: 12pt, it.body))
#show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body)
#show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body)

#let examplebox(label, title, body) = block(width: 100%, breakable: true,
  fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt,
  above: 12pt, below: 12pt)[
  #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt,
    text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]]
  #body]

// rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade
#let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint,
    stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt),
    above: 11pt, below: 11pt)[
    #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)]
    #linebreak() #body]

#let solutionbox(body) = block(above: 4pt, below: 8pt)[
  #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body]

#let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"),
  stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[
  #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))]

// Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a
// consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the
// print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a
// degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero.
#let _STD_W = 3.5
#let _WIDE_W = 5.6
#let _MAX_H = 3.4
#let _ASPECT_WIDE = 2.2
#let _UPSCALE_MAX = 1.15
#let stdfig(body) = context {
  let m = measure(body)
  let w = calc.max(m.width / 1in, 0.01)
  let h = calc.max(m.height / 1in, 0.01)
  let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W }
  let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX)
  align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body)))
}

#show figure: set block(breakable: false)
#set figure(gap: 8pt)
#show figure.caption: set text(size: 8.5pt, fill: rgb("#555"))


  
== 4.6#h(0.6em)Entropy


  The second law of thermodynamics is best expressed in terms of a #emph[change] in the thermodynamic variable known as #strong[entropy], which is represented by the symbol #emph[S]. Entropy, like internal energy, is a state function. This means that when a system makes a transition from one state into another, the change in entropy #math.equation(block: false, alt: "Δ S")[$"Δ" S$] is independent of path and depends only on the thermodynamic variables of the two states.


  We first consider #math.equation(block: false, alt: "Δ S")[$"Δ" S$] for a system undergoing a reversible process at a constant temperature. In this case, the change in entropy of the system is given by


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "Δ S equals the fraction Q over T ,")[$"Δ" S = frac(Q, T) ,$]
]


  where #emph[Q] is the heat exchanged by the system kept at a temperature #emph[T] (in kelvin). If the system absorbs heat—that is, with #math.equation(block: false, alt: "Q greater than 0")[$Q > 0$]—the entropy of the system increases. As an example, suppose a gas is kept at a constant temperature of 300 K while it absorbs 10 J of heat in a reversible process. Then from , the entropy change of the gas is


  #math.equation(block: true, alt: "Δ S equals the fraction 10 J over 300 K equals 0.033 J/K .")[$"Δ" S = frac(10 #h(0.2em) "J", 300 #h(0.2em) "K") = 0.033 #h(0.2em) "J/K" .$]
  Similarly, if the gas loses 5.0 J of heat; that is, #math.equation(block: false, alt: "Q equals −5.0 J")[$Q = −5.0 #h(0.2em) "J"$], at temperature #math.equation(block: false, alt: "T equals 200 K")[$T = 200 #h(0.2em) "K"$], we have the entropy change of the system given by


  #math.equation(block: true, alt: "Δ S equals the fraction −5.0 J over 200 K equals −0.025 J/K .")[$"Δ" S = frac(−5.0 #h(0.2em) "J", 200 #h(0.2em) "K") = −0.025 #h(0.2em) "J/K" .$]
  #examplebox("Example 1")[Entropy Change of Melting Ice][
Heat is slowly added to a 50-g chunk of ice at #math.equation(block: false, alt: "0 ° C")[$0 #h(0.2em) "°" "C"$] until it completely melts into water at the same temperature. What is the entropy change of the ice?


    Strategy
Because the process is slow, we can approximate it as a reversible process. The temperature is a constant, and we can therefore use  in the calculation.


    Solution
The ice is melted by the addition of heat:


    #math.equation(block: true, alt: "Q equals m L sub f equals 50 g times 335 J/g equals 16.8 kJ .")[$Q = m L_("f") = 50 #h(0.2em) "g" #h(0.2em) × #h(0.2em) 335 #h(0.2em) "J/g" = 16.8 #h(0.2em) "kJ" .$]
    In this reversible process, the temperature of the ice-water mixture is fixed at #math.equation(block: false, alt: "0 °C")[$0 #h(0.2em) "°C"$] or 273.15 K. Now from #math.equation(block: false, alt: "Δ S equals Q / T")[$"Δ" S = Q "/" T$], the entropy change of the ice is


    #math.equation(block: true, alt: "Δ S equals the fraction 16.8 kJ over 273 K equals 61.5 J/K")[$"Δ" S = frac(16.8 #h(0.2em) "kJ", 273 #h(0.2em) "K") = 61.5 #h(0.2em) "J/K"$]
    when it melts to water at #math.equation(block: false, alt: "0 °C")[$0 #h(0.2em) "°C"$].


    Significance
During a phase change, the temperature is constant, allowing us to use  to solve this problem. The same equation could also be used if we changed from a liquid to a gas phase, since the temperature does not change during that process either.
]


  The change in entropy of a system for an arbitrary, reversible transition for which the temperature is not necessarily constant is defined by modifying #math.equation(block: false, alt: "Δ S equals Q / T")[$"Δ" S = Q "/" T$]. Imagine a system making a transition from state #emph[A] to #emph[B] in small, discrete steps. The temperatures associated with these states are #math.equation(block: false, alt: "T sub A")[$T_(A)$] and #math.equation(block: false, alt: "T sub B ,")[$T_(B) ,$] respectively. During each step of the transition, the system exchanges heat #math.equation(block: false, alt: "Δ Q sub i")[$"Δ" Q_(i)$] reversibly at a temperature #math.equation(block: false, alt: "T sub i .")[$T_(i) .$] This can be accomplished experimentally by placing the system in thermal contact with a large number of heat reservoirs of varying temperatures #math.equation(block: false, alt: "T sub i")[$T_(i)$], as illustrated in . The change in entropy for each step is #math.equation(block: false, alt: "Δ S sub i equals Q sub i / T sub i .")[$"Δ" S_(i) = Q_(i) "/" T_(i) .$] The net change in entropy of the system for the transition is


  #math.equation(block: true, alt: "Δ S equals S sub B minus S sub A equals ∑ i Δ S sub i equals ∑ i the fraction Δ Q sub i over T sub i .")[$"Δ" S = S_(B) − S_(A) = display(limits(∑)_(i) #h(0.2em) "Δ" S_(i)) = display(limits(∑)_(i) #h(0.2em) frac("Δ" Q_(i), T_(i))) .$]
  We now take the limit as #math.equation(block: false, alt: "Δ Q sub i → 0")[$"Δ" Q_(i) → 0$], and the number of steps approaches infinity. Then, replacing the summation by an integral, we obtain


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "Δ S equals S sub B minus S sub A equals ∫ A B the fraction d Q over T ,")[$"Δ" S = S_(B) − S_(A) = display(∫_(A)^(B) #h(0.2em) frac(d Q, T)) ,$]
]


  where the integral is taken between the initial state #emph[A] and the final state #emph[B]. This equation is valid only if the transition from #emph[A] to #emph[B] is reversible.


  #figure(figph[The figure shows four heat reservoirs. The first reservoir at temperature T has gas with volume V, the second reservoir at T plus delta T has gas V plus delta V, the third reservoir at T plus 2 delta T has gas V plus 2 delta V and the final reservoir at T plus n delta T has gas at V plus n delta V.], alt: "The figure shows four heat reservoirs. The first reservoir at temperature T has gas with volume V, the second reservoir at T plus delta T has gas V plus delta V, the third reservoir at T plus 2 delta T has gas V plus 2 delta V and the final reservoir at T plus n delta T has gas at V plus n delta V.", caption: [The gas expands at constant pressure as its temperature is increased in small steps through the use of a series of heat reservoirs.])


  As an example, let us determine the net entropy change of a reversible engine while it undergoes a single Carnot cycle. In the adiabatic steps 2 and 4 of the cycle shown in , no heat exchange takes place, so #math.equation(block: false, alt: "Δ S sub 2 equals Δ S sub 4 equals ∫ d Q / T equals 0 .")[$"Δ" S_(2) = "Δ" S_(4) = display(∫ #h(0.2em) d) Q "/" T = 0 .$] In step 1, the engine absorbs heat #math.equation(block: false, alt: "Q sub h")[$Q_("h")$] at a temperature #math.equation(block: false, alt: "T sub h ,")[$T_("h") ,$] so its entropy change is #math.equation(block: false, alt: "Δ S sub 1 equals Q sub h / T sub h .")[$"Δ" S_(1) = Q_("h") "/" T_("h") .$] Similarly, in step 3, #math.equation(block: false, alt: "Δ S sub 3 equals − Q sub c / T sub c .")[$"Δ" S_(3) = "−" Q_("c") "/" T_("c") .$] The net entropy change of the engine in one cycle of operation is then


  #math.equation(block: true, alt: "Δ S sub E equals Δ S sub 1 plus Δ S sub 2 plus Δ S sub 3 plus Δ S sub 4 equals the fraction Q sub h over T sub h minus the fraction Q sub c over T sub c .")[$"Δ" S_(E) = "Δ" S_(1) + "Δ" S_(2) + "Δ" S_(3) + "Δ" S_(4) = frac(Q_("h"), T_("h")) − frac(Q_("c"), T_("c")) .$]
  However, we know that for a Carnot engine,


  #math.equation(block: true, alt: "the fraction Q sub h over T sub h equals the fraction Q sub c over T sub c ,")[$frac(Q_("h"), T_("h")) = frac(Q_("c"), T_("c")) ,$]
  so


  #math.equation(block: true, alt: "Δ S sub E equals 0 .")[$"Δ" S_(E) = 0 .$]
  There is no net change in the entropy of the Carnot engine over a complete cycle. Although this result was obtained for a particular case, its validity can be shown to be far more general: There is no net change in the entropy of a system undergoing any complete reversible cyclic process. Mathematically, we write this statement as


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "∮ d S equals ∮ the fraction d Q over T equals 0")[$display(∮ d S = display(∮ frac(d Q, T))) = 0$]
]


  where #math.equation(block: false, alt: "∮")[$display(∮)$] represents the integral over a #emph[closed reversible path].


  We can use  to show that the entropy change of a system undergoing a reversible process between two given states is path independent. An arbitrary, closed path for a reversible cycle that passes through the states #emph[A] and #emph[B] is shown. From , #math.equation(block: false, alt: "∮ d S equals 0")[$display(∮ d S = 0)$] for this closed path. We may split this integral into two segments, one along I, which leads from #emph[A] to #emph[B], the other along II, which leads from #emph[B] to #emph[A]. Then


  #math.equation(block: true, alt: "[ ∫ A B d S ] sub I plus [ ∫ B A d S ] sub II equals 0 .")[$attach([ display(∫_(A)^(B) d S) ], b: "I") + attach([ display(∫_(B)^(A) d S) ], b: "II") = 0 .$]
  Since the process is reversible,


  #math.equation(block: true, alt: "[ ∫ A B d S ] sub I equals [ ∫ A B d S ] sub II .")[$attach([ display(∫_(A)^(B) d S) ], b: "I") = attach([ display(∫_(A)^(B) d S) ], b: "II") .$]
  #figure(figph[The figure shows a a pear shaped closed loop graph with x-axis V and y-axis p.], alt: "The figure shows a a pear shaped closed loop graph with x-axis V and y-axis p.", caption: [The closed loop passing through states #emph[A] and #emph[B] represents a reversible cycle.])


  Hence, the entropy change in going from #emph[A] to #emph[B] is the same for paths I and II. Since paths I and II are arbitrary, reversible paths, the entropy change in a transition between two equilibrium states is the same for all the reversible processes joining these states. Entropy, like internal energy, is therefore a state function.


  What happens if the process is irreversible? When the process is irreversible, we expect the entropy of a closed system, or the system and its environment (the universe), to increase. Therefore we can rewrite this expression as


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#math.equation(block: true, alt: "Δ S greater than or equal to 0 ,")[$"Δ" S ≥ 0 ,$]
]


  where #emph[S] is the total entropy of the closed system or the entire universe, and the equal sign is for a reversible process. The fact is the #strong[entropy statement of the second law of thermodynamics]:


  #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[Second Law of Thermodynamics (Entropy statement)]


    The entropy of a closed system and the entire universe never decreases.
]


  We can show that this statement is consistent with the Kelvin statement, the Clausius statement, and the Carnot principle.


  #examplebox("Example 2")[Entropy Change of a System during an Isobaric Process][
Determine the entropy change of an object of mass #emph[m] and specific heat #emph[c] that is cooled rapidly (and irreversibly) at constant pressure from #math.equation(block: false, alt: "T sub h")[$T_("h")$] to #math.equation(block: false, alt: "T sub c .")[$T_("c") .$]


    Strategy
The process is clearly stated as an irreversible process; therefore, we cannot simply calculate the entropy change from the actual process. However, because entropy of a system is a function of state, we can imagine a reversible process that starts from the same initial state and ends at the given final state. Then, the entropy change of the system is given by , #math.equation(block: false, alt: "Δ S equals ∫ A B d Q / T .")[$"Δ" S = display(∫_(A)^(B) d Q "/" T) .$]


    Solution
To replace this rapid cooling with a process that proceeds reversibly, we imagine that the hot object is put into thermal contact with successively cooler heat reservoirs whose temperatures range from #math.equation(block: false, alt: "T sub h")[$T_("h")$] to #math.equation(block: false, alt: "T sub c .")[$T_("c") .$] Throughout the substitute transition, the object loses infinitesimal amounts of heat #emph[dQ], so we have


    #math.equation(block: true, alt: "Δ S equals ∫ T sub h T sub c the fraction d Q over T .")[$"Δ" S = display(∫_(T_("h"))^(T_("c")) frac(d Q, T)) .$]
    From the definition of heat capacity, an infinitesimal exchange #emph[dQ] for the object is related to its temperature change #emph[dT] by


    #math.equation(block: true, alt: "d Q equals m c d T .")[$d Q = m c #h(0.2em) d T .$]
    Substituting this #emph[dQ] into the expression for #math.equation(block: false, alt: "Δ S")[$"Δ" S$], we obtain the entropy change of the object as it is cooled at constant pressure from #math.equation(block: false, alt: "T sub h to T sub c :")[$T_("h") #h(0.2em) "to" #h(0.2em) T_("c") :$]


    #math.equation(block: true, alt: "Δ S equals ∫ T sub h T sub c the fraction m c d T over T equals m c ln the fraction T sub c over T sub h .")[$"Δ" S = display(∫_(T_("h"))^(T_("c")) frac(m c #h(0.2em) d T, T)) = m c #h(0.2em) "ln" frac(T_("c"), T_("h")) .$]
    Note that #math.equation(block: false, alt: "Δ S less than 0")[$"Δ" S < 0$] here because #math.equation(block: false, alt: "T sub c less than T sub h .")[$T_("c") < T_("h") .$] In other words, the object has lost some entropy. But if we count whatever is used to remove the heat from the object, we would still end up with #math.equation(block: false, alt: "Δ S sub universe greater than 0")[$"Δ" S_("universe") > 0$] because the process is irreversible.


    Significance
If the temperature changes during the heat flow, you must keep it inside the integral to solve for the change in entropy. If, however, the temperature is constant, you can simply calculate the entropy change as the heat flow divided by the temperature.
]


  #examplebox("Example 3")[Stirling Engine][
The steps of a #strong[Stirling engine] are as follows. For this problem, we will use 0.0010 mol of a monatomic gas that starts at a temperature of #math.equation(block: false, alt: "133 ° C")[$133 #h(0.2em) "°" "C"$] and a volume of #math.equation(block: false, alt: "0.10 m cubed")[$0.10 #h(0.2em) "m"^(3)$], which will be called point #emph[A]. Then it goes through the following steps:


    + Step #emph[AB]: isothermal expansion at #math.equation(block: false, alt: "133 ° C")[$133 #h(0.2em) "°" "C"$] from #math.equation(block: false, alt: "0.10 m cubed")[$0.10 #h(0.2em) "m"^(3)$] to #math.equation(block: false, alt: "0.20 m cubed")[$0.20 #h(0.2em) "m"^(3)$]
+ Step #emph[BC]: isochoric cooling to #math.equation(block: false, alt: "33 ° C")[$33 #h(0.2em) "°" "C"$]
+ Step #emph[CD]: isothermal compression at #math.equation(block: false, alt: "33 ° C")[$33 #h(0.2em) "°" "C"$] from #math.equation(block: false, alt: "0.20 m cubed")[$0.20 #h(0.2em) "m"^(3)$] to #math.equation(block: false, alt: "0.10 m cubed")[$0.10 #h(0.2em) "m"^(3)$]
+ Step #emph[DA]: isochoric heating back to #math.equation(block: false, alt: "133 ° C")[$133 #h(0.2em) "°" "C"$] and #math.equation(block: false, alt: "0.10 m cubed")[$0.10 #h(0.2em) "m"^(3)$]


    (a) Draw the #emph[pV] diagram for the Stirling engine with proper labels.


    (b) Fill in the following table.


    #figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([Step], [#emph[W] (J)], [#emph[Q] (J)], [#math.equation(block: false, alt: "Δ S")[$"Δ" S$] (J/K)]),
  [Step #emph[AB]], [], [], [],
  [Step #emph[BC]], [], [], [],
  [Step #emph[CD]], [], [], [],
  [Step #emph[DA]], [], [], [],
  [Complete cycle], [], [], [],
))


    (c) How does the efficiency of the Stirling engine compare to the Carnot engine working within the same two heat reservoirs?


    Strategy
Using the ideal gas law, calculate the pressure at each point so that they can be labeled on the #emph[pV] diagram. Isothermal work is calculated using #math.equation(block: false, alt: "W equals n R T ln open parenthesis the fraction V sub 2 over V sub 1 close parenthesis ,")[$W = n R T #h(0.2em) "ln" ( frac(V_(2), V_(1)) ) ,$] and an isochoric process has no work done. The heat flow is calculated from the first law of thermodynamics, #math.equation(block: false, alt: "Q equals Δ E sub int minus W")[$Q = "Δ" E_("int") − W$] where #math.equation(block: false, alt: "Δ E sub int equals the fraction 3 over 2 n R Δ T")[$"Δ" E_("int") = frac(3, 2) n R "Δ" T$] for monatomic gasses. Isothermal steps have a change in entropy of #emph[Q]/#emph[T], whereas isochoric steps have #math.equation(block: false, alt: "Δ S equals the fraction 3 over 2 n R ln open parenthesis the fraction T sub 2 over T sub 1 close parenthesis .")[$"Δ" S = frac(3, 2) n R #h(0.2em) "ln" ( frac(T_(2), T_(1)) ) .$] The efficiency of a heat engine is calculated by using #math.equation(block: false, alt: "e sub Stir equals W / Q sub h .")[$e_("Stir") = W "/" Q_("h") .$]


    Solution


    + The graph is shown below. #linebreak() 
#figure(figph[The figure shows a graph with x-axis V in m superscript 3 and y-axis p in atm. The four points A (0.10, 26), B (0.20, 17), C (0.20, 13) and D (0.10, 26) are connected to form a closed loop.], alt: "The figure shows a graph with x-axis V in m superscript 3 and y-axis p in atm. The four points A (0.10, 26), B (0.20, 17), C (0.20, 13) and D (0.10, 26) are connected to form a closed loop.", caption: none)
+ The completed table is shown below. #linebreak() 
#figure(table(
  columns: 4, align: left, inset: 6pt,
  table.header([Step], [#emph[W] (J)], [#emph[Q] (J)], [#math.equation(block: false, alt: "Δ S")[$"Δ" S$] (J/K)]),
  [Step #emph[AB] Isotherm], [2.3], [2.3], [0.0057],
  [Step #emph[BC] Isochoric], [0], [–1.2], [–0.0035],
  [Step #emph[CD] Isotherm], [–1.8], [–1.8], [–0.0057],
  [Step #emph[DA] Isochoric], [0], [1.2], [0.0035],
  [Complete cycle], [0.5], [0.5], [~ 0],
))
+ The efficiency of the Stirling heat engine is
#math.equation(block: true, alt: "e sub Stir equals W / Q sub h equals open parenthesis Q sub A B plus Q sub C D close parenthesis / open parenthesis Q sub A B plus Q sub D A close parenthesis equals 0.5 / 3.5 equals 0.14 .")[$e_("Stir") = W "/" Q_("h") = ( Q_(A B) + Q_(C D) ) "/" ( Q_(A B) + Q_(D A) ) = 0.5 "/" 3.5 = 0.14 .$]


    If this were a Carnot engine operating between the same heat reservoirs, its efficiency would be


    #math.equation(block: true, alt: "e sub Car equals 1 minus open parenthesis the fraction T sub c over T sub h close parenthesis equals 0.25 .")[$e_("Car") = 1 − ( frac(T_("c"), T_("h")) ) = 0.25 .$]
    Therefore, the Carnot engine would have a greater efficiency than the Stirling engine.


    Significance
We have noted that all reversible engines operating between the same two reservoirs have the same efficiency. Since we found that the Stirling engine has a lower efficiency than the Carnot cycle, it must not be a reversible engine, and this is indeed the case. The isochoric steps are not reversible unless a regenerator is used to capture and reclaim the heat during these steps. With a perfect regenerator, the efficiency of the ideal Stirling engine would be the same as that of the Carnot engine.


    In the early days of steam engines, accidents would occur due to the high pressure of the steam in the boiler. Robert Stirling developed an engine in 1816 that did not use steam and therefore was safer. The Stirling engine was commonly used in the nineteenth century, but developments in steam and internal combustion engines have made it difficult to broaden the use of the Stirling engine.


    The Stirling engine uses compressed air as the working substance, which passes back and forth between two chambers with a porous plug, called the regenerator, which is made of material that does not conduct heat as well. In two of the steps, pistons in the two chambers move in phase.
]


  
    
=== Summary


    - The change in entropy for a reversible process at constant temperature is equal to the heat divided by the temperature. The entropy change of a system under a reversible process is given by #math.equation(block: false, alt: "Δ S equals ∫ A B d Q / T")[$"Δ" S = display(∫_(A)^(B) d) Q "/" T$].
- A system’s change in entropy between two states is independent of the reversible thermodynamic path taken by the system when it makes a transition between the states.


  
  
    
=== Conceptual Questions


    
      Does the entropy increase for a Carnot engine for each cycle?


    
    
      Is it possible for a system to have an entropy change if it neither absorbs nor emits heat during a reversible transition? What happens if the process is irreversible?


    
    #solutionbox[
Entropy will not change if it is a reversible transition but will change if the process is irreversible.
]


  
  
    
=== Problems


    
      Two hundred joules of heat are removed from a heat reservoir at a temperature of 200 K. What is the entropy change of the reservoir?


    
    #solutionbox[
–1 J/K
]


    
      In an isothermal reversible expansion at #math.equation(block: false, alt: "27 °C")[$27 #h(0.2em) "°C"$], an ideal gas does 20 J of work. What is the entropy change of the gas?


    
    
      An ideal gas at 300 K is compressed isothermally to one-fifth its original volume. Determine the entropy change per mole of the gas.


    
    #solutionbox[
–13 J(K mole)
]


    
      What is the entropy change of 10 g of steam at #math.equation(block: false, alt: "100 °C")[$100 #h(0.2em) "°C"$] when it condenses to water at the same temperature?


    
    
      A metal rod is used to conduct heat between two reservoirs at temperatures #math.equation(block: false, alt: "T sub h and T sub c ,")[$T_("h") #h(0.2em) "and" #h(0.2em) T_("c") ,$] respectively. When an amount of heat #emph[Q] flows through the rod from the hot to the cold reservoir, what is the net entropy change of the rod, the hot reservoir, the cold reservoir, and the universe?


    
    #solutionbox[
0 (no net heat remains in or leaves the rod), #math.equation(block: false, alt: "minus the fraction Q over T sub h , the fraction Q over T sub c , Q open parenthesis the fraction 1 over T sub c minus the fraction 1 over T sub h close parenthesis")[$− frac(Q, T_("h")) , frac(Q, T_("c")) , Q ( frac(1, T_("c")) − frac(1, T_("h")) )$]
]


    
      For the Carnot cycle of , what is the entropy change of the hot reservoir, the cold reservoir, and the universe?


    
    
      A 5.0-kg piece of lead at a temperature of #math.equation(block: false, alt: "600 °C")[$600 #h(0.2em) "°C"$] is placed in a lake whose temperature is #math.equation(block: false, alt: "15 °C")[$15 #h(0.2em) "°C"$]. Determine the entropy change of (a) the lead piece, (b) the lake, and (c) the universe.


    
    #solutionbox[
a. –709 J/K; b. 1300 J/K; c. 591 J/K
]


    
      One mole of an ideal gas doubles its volume in a reversible isothermal expansion. (a) What is the change in entropy of the gas? (b) If 1500 J of heat are added in this process, what is the temperature of the gas?


    
    
      An ideal monatomic gas is confined to a rigid container. When heat is added reversibly to the gas, its temperature changes from #math.equation(block: false, alt: "T sub 1 to T sub 2 .")[$T_(1) #h(0.2em) "to" #h(0.2em) T_(2) .$] (a) How much heat is added? (b) What is the change in entropy of the gas?


    
    #solutionbox[
a. #math.equation(block: false, alt: "Q equals the fraction 3 over 2 n R Δ T")[$Q = frac(3, 2) n R "Δ" T$]; b. #math.equation(block: false, alt: "S equals the fraction 3 over 2 n R ln open parenthesis T sub 2 / T sub 1 close parenthesis")[$S = frac(3, 2) n R #h(0.2em) "ln" ( T_(2) "/" T_(1) )$]
]


    
      (a) A 5.0-kg rock at a temperature of #math.equation(block: false, alt: "20 °C")[$20 #h(0.2em) "°C"$] is dropped into a shallow lake also at #math.equation(block: false, alt: "20 °C")[$20 #h(0.2em) "°C"$] from a height of #math.equation(block: false, alt: "1.0 times 10 cubed m")[$1.0 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "m"$]. What is the resulting change in entropy of the universe? (b) If the temperature of the rock is #math.equation(block: false, alt: "100 °C")[$100 #h(0.2em) "°C"$] when it is dropped, what is the change of entropy of the universe? Assume that air friction is negligible (not a good assumption) and that #math.equation(block: false, alt: "c equals 860 J/kg times K")[$c = 860 #h(0.2em) "J/kg" · "K"$] is the specific heat of the rock.