What Is Energy? — Question 10

Question 10

A complete state must include hidden internal stores

For this introductory problem use the supplied accounting rule $\Delta E=E_{\mathrm{in}}-E_{\mathrm{out}}$, with all quantities in joules. Two otherwise identical sealed devices have the same measured motion before and after a process. Device X receives 40 J and releases 10 J; Y receives 25 J and releases 5 J. No matter crosses either boundary. The visible-motion energy does not change in either device.

1. Find the total stored-energy change in each.

2. What must be true of the combined unmeasured internal stores?

3. Can one infer which device has the larger final total energy without initial-energy information?

Question 10 - Solution

A complete state must include hidden internal stores

(a) Use the full energy balance.

The governing relation is $$\boxed{\Delta E_{\mathrm{total}}=\Delta E_{\mathrm{visible}}+\Delta E_{\mathrm{hidden}}=I-O}$$ X gains $$40-10=30\ \mathrm J;$$ Y gains $$25-5=20\ \mathrm J.$$

(b) Account for the unchanged visible contribution.

The balance includes all internal stores: $$\Delta E_{\mathrm{hidden}}=E_{\mathrm{in}}-E_{\mathrm{out}}-\Delta E_{\mathrm{visible}}.$$ Thus $\boxed{\Delta E_{{\mathrm{hidden}},X}=30\ \mathrm J}$ and $\boxed{\Delta E_{{\mathrm{hidden}},Y}=20\ \mathrm J}$. Unchanged visible motion does not mean unchanged total energy.

(c) Distinguish changes from final levels.

No. $$E_{X,f}-E_{Y,f}=(E_{X,i}-E_{Y,i})+10\ \mathrm J.$$ X ends 10 J higher only if their initial total energies were equal. For an initial difference of $-15\ \mathrm J$, X ends 5 J lower.