What Is Energy? — Question 2

Section 7-1 · Physics pages 854–855

Question 2 A zero choice cannot alter an energy change Two analysts assign energies to three states of the same system. Their conventions may differ by one constant offset. The records are below; the system makes the cycle A to B to C to A. EI (J) EII (J) A : (10, 110) B : (35, 135) C : (20, 120) The three labeled pairs are the recorded energies; coordinates use separate scales for the two axes. (a) Determine whether one constant offset relates the records. (b) Find each energy change in the cycle under both conventions. (c) A third analyst changes only the B entry of record II to 140 J. Can a new constant offset repair the whole record?
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Question 2 - Solution A zero choice cannot alter an energy change EI (J) EII (J) A : (10, 110) B : (35, 135) C : (20, 120) (a) Compare corresponding states. The governing relation is EII(s) = EI(s) + C =⇒ ∆EII = ∆EI The offset is identical in every state: 110 −10 = 135 −35 = 120 −20 = 100 J. Thus EII = EI + 100 J. (b) Subtract endpoints, not reference levels. Subtract endpoint values in either record: ∆EAB = 35 −10 = 25 J, ∆EBC = 20 −35 = −15 J. ∆ECA = 10 −20 = −10 J, ∑ cycle ∆E = 25 −15 −10 = 0. (c) Require the same offset everywhere. The proposed replacement is inconsistent with a common zero shift: CB = 140 −35 = 105 J̸ = CA = CC = 100 J. It changes energy differences and cannot be absorbed into one additive constant.

Question 2

A zero choice cannot alter an energy change

Two analysts assign energies to three states of the same system. Their conventions may differ by one constant offset. The records are below; the system makes the cycle A to B to C to A.

Diagram shown in the original SVG worksheet.

  1. Determine whether one constant offset relates the records.

  2. Find each energy change in the cycle under both conventions.

  3. A third analyst changes only the B entry of record II to 140 J. Can a new constant offset repair the whole record?

Question 2 - Solution

A zero choice cannot alter an energy change

Diagram shown in the original SVG worksheet.

(a) Compare corresponding states.

The governing relation is EII(s)=EI(s)+C⇒ΔEII=ΔEI\boxed{E_{\mathrm{II}}(s)=E_{\mathrm{I}}(s)+C\quad\Longrightarrow\quad\Delta E_{\mathrm{II}}=\Delta E_{\mathrm{I}}} The offset is identical in every state: 110−10=135−35=120−20=100J.110-10=135-35=120-20=100\ \mathrm J. Thus EII=EI+100JE_{\mathrm{II}}=E_{\mathrm{I}}+100\ \mathrm J.

(b) Subtract endpoints, not reference levels.

Subtract endpoint values in either record: ΔEAB=35−10=25J,ΔEBC=20−35=−15J.\Delta E_{AB}=35-10=25\ \mathrm J,\quad\Delta E_{BC}=20-35=-15\ \mathrm J. ΔECA=10−20=−10J,∑cycleΔE=25−15−10=0.\Delta E_{CA}=10-20=-10\ \mathrm J,\qquad\sum_{\mathrm{cycle}}\Delta E=25-15-10=0.

(c) Require the same offset everywhere.

The proposed replacement is inconsistent with a common zero shift: CB=140−35=105J≠CA=CC=100J.C_B=140-35=105\ \mathrm J\ne C_A=C_C=100\ \mathrm J. It changes energy differences and cannot be absorbed into one additive constant.