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

Irreversible Processes and Entropy — Question 3

Section 20-1 · Physics pages 3236–3237

Question 3 Equal finite bodies approach an arithmetic-mean temperature Two bodies have equal constant heat capacity C = 400 J/K and initial temperatures 400 K and 200 K. They exchange heat inside an insulated rigid enclosure; no work is done. For each body, ∆S = C ln(Tf/Ti). 400 K 200 K Q The arrow defines positive heat transfer from left to right. (a) Find the final equilibrium temperature and the total entropy change. (b) Derive the sign of total entropy for general positive Th, Tc using the arithmetic–geometric mean inequality.
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Question 3 - Solution Equal finite bodies approach an arithmetic-mean temperature 400 K 200 K Q (a) Method and calculation C(Tf −400) + C(Tf −200) = 0 ⇒Tf = 300 K. ∆S = C ln 300 400 + C ln 300 200 = 400 ln(9/8) ≈47.1132 J/K. (b) Interpretation and further calculation Tf = Th + Tc 2 , ∆S = C ln (Th + Tc)2 4ThTc . (Th + Tc)2 −4ThTc = (Th −Tc)2 ≥0 ⇒∆S ≥0. Equality requires equal starting temperatures, when there is no finite thermal relaxation. Physical check The isolated composite gains entropy even though the hotter component loses entropy.

Question 3

Equal finite bodies approach an arithmetic-mean temperature

Two bodies have equal constant heat capacity C=400C=400 J/K and initial temperatures 400 K and 200 K. They exchange heat inside an insulated rigid enclosure; no work is done. For each body, ΔS=Cln⁡(Tf/Ti)\Delta S=C\ln(T_f/T_i).

Diagram shown in the original SVG worksheet.

  1. Find the final equilibrium temperature and the total entropy change.

  2. Derive the sign of total entropy for general positive Th,TcT_h,T_c using the arithmetic–geometric mean inequality.

Question 3 - Solution

Equal finite bodies approach an arithmetic-mean temperature

Diagram shown in the original SVG worksheet.

(a) Method and calculation

C(Tf−400)+C(Tf−200)=0⇒Tf=300K.C(T_f-400)+C(T_f-200)=0\ \Rightarrow\ T_f=300\,\mathrm K.ΔS=Cln⁡300400+Cln⁡300200=400ln⁡(9/8)≈47.1132J/K.\Delta S=C\ln\frac{300}{400}+C\ln\frac{300}{200}=400\ln(9/8)\approx47.1132\,\mathrm{J/K}.

(b) Interpretation and further calculation

Tf=Th+Tc2,ΔS=Cln⁡(Th+Tc)24ThTc.T_f=\frac{T_h+T_c}{2},\quad\Delta S=C\ln\frac{(T_h+T_c)^2}{4T_hT_c}.(Th+Tc)2−4ThTc=(Th−Tc)2≥0⇒ΔS≥0.(T_h+T_c)^2-4T_hT_c=(T_h-T_c)^2\ge0\ \Rightarrow\Delta S\ge0.Equality requires equal starting temperatures, when there is no finite thermal relaxation.

Physical check

The isolated composite gains entropy even though the hotter component loses entropy.

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