The First Law of Thermodynamics — Question 1

Question 1

A multistage process requires one consistent sign convention

Use $\Delta U=Q-W$, where $Q$ is heat into the system and $W$ is work done by it. A stationary closed system first absorbs 800 J and does 300 J of work. Next it loses 250 J while 100 J of work is done on it. Finally it absorbs 150 J at fixed volume with no other work.

1. Find the internal-energy change in each stage and over the whole process.

2. If its initial internal energy is 2.00 kJ, find the final value. What heat would restore the initial state if the return required 200 J of work input?

Question 1 - Solution

A multistage process requires one consistent sign convention

(a) Method and calculation

$$\Delta U_1=800-300=500\,\mathrm J,\quad\Delta U_2=-250-(-100)=-150\,\mathrm J.$$$$\Delta U_3=150-0=150\,\mathrm J,\quad\Delta U_{\mathrm{total}}=500\,\mathrm J.$$

(b) Interpretation and further calculation

$$U_f=2000+500=2500\,\mathrm J.$$$$\Delta U_{\mathrm{return}}=-500=Q_{\mathrm{return}}-(-200)\ \Rightarrow\ Q_{\mathrm{return}}=-700\,\mathrm J.$$The return must reject 700 J because its work input also has to leave.

Physical check

Total heat is 700 J and total work output is 200 J on the outward process, reproducing the 500 J state change.