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Physics / Chapter 20: Entropy and the Second Law of Thermodynamics / Section 20-1

Irreversible Processes and Entropy — Question 1

Section 20-1 · Physics pages 3232–3233

Question 1 A reservoir pair gives a quantitative test of the direction of heat flow Two large reservoirs remain at 500 K and 300 K. Heat Q = 1200 J passes directly from the hot reservoir to the cold one. For a reservoir use ∆S = Qinto/T. 500 K 300 K Q The arrow defines positive heat transfer from left to right. (a) Find both entropy changes and the total. Check conservation of energy. (b) Compute the entropy changes for the exact reverse transfer with no other changes. Use the rule ∆Sisolated ≥0 to assess its possibility.
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Question 1 - Solution A reservoir pair gives a quantitative test of the direction of heat flow 500 K 300 K Q (a) Method and calculation ∆Sh = −1200/500 = −2.40 J/K, ∆Sc = 1200/300 = 4.00 J/K. ∆Spair = 1.60 J/K, ∆Uh + ∆Uc = −1200 + 1200 = 0. (b) Interpretation and further calculation ∆Sreverse = +2.40 −4.00 = −1.60 J/K < 0. ∆Ureverse,pair = 0. Energy conservation permits either bookkeeping direction, but the entropy condition forbids the isolated reverse transfer. An external device could drive it only while producing other changes. Physical check Zero net energy change does not imply zero entropy change.

Question 1

A reservoir pair gives a quantitative test of the direction of heat flow

Two large reservoirs remain at 500 K and 300 K. Heat Q=1200Q=1200 J passes directly from the hot reservoir to the cold one. For a reservoir use ΔS=Qinto/T\Delta S=Q_{\mathrm{into}}/T.

Diagram shown in the original SVG worksheet.

  1. Find both entropy changes and the total. Check conservation of energy.

  2. Compute the entropy changes for the exact reverse transfer with no other changes. Use the rule ΔSisolated≥0\Delta S_{\mathrm{isolated}}\ge0 to assess its possibility.

Question 1 - Solution

A reservoir pair gives a quantitative test of the direction of heat flow

Diagram shown in the original SVG worksheet.

(a) Method and calculation

ΔSh=−1200/500=−2.40J/K,ΔSc=1200/300=4.00J/K.\Delta S_h=-1200/500=-2.40\,\mathrm{J/K},\quad\Delta S_c=1200/300=4.00\,\mathrm{J/K}.ΔSpair=1.60J/K,ΔUh+ΔUc=−1200+1200=0.\Delta S_{\mathrm{pair}}=1.60\,\mathrm{J/K},\quad\Delta U_h+\Delta U_c=-1200+1200=0.

(b) Interpretation and further calculation

ΔSreverse=+2.40−4.00=−1.60J/K<0.\Delta S_{\mathrm{reverse}}=+2.40-4.00=-1.60\,\mathrm{J/K}<0.ΔUreverse,pair=0.\Delta U_{\mathrm{reverse,pair}}=0.Energy conservation permits either bookkeeping direction, but the entropy condition forbids the isolated reverse transfer. An external device could drive it only while producing other changes.

Physical check

Zero net energy change does not imply zero entropy change.

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