The First Law of Thermodynamics — Question 6

Question 6

A falling weight calibrates a paddle-work experiment

A falling 5.00 kg mass drops 4.00 m while driving a paddle in an insulated liquid. An ideal drive delivers all lost gravitational potential energy to the liquid as work. The liquid has heat capacity 980 J/K. Take $g=9.80\,\mathrm{m/s^2}$, and neglect final macroscopic liquid motion.

1. Using the liquid alone as the system, find $Q$, work input, internal-energy increase and temperature rise.

2. Now include Earth, the mass, drive and liquid in a single isolated system. Show how its internal-energy gain and gravitational-energy change satisfy conservation without work crossing the enlarged boundary.

Question 6 - Solution

A falling weight calibrates a paddle-work experiment

(a) Method and calculation

$$Q=0,\quad W_{\mathrm{in}}=Mgh=5(9.80)(4)=196\,\mathrm J.$$$$\Delta U=Q+W_{\mathrm{in}}=196\,\mathrm J,\quad\Delta T=196/980=.200\,\mathrm K.$$

(b) Interpretation and further calculation

$$Q_{\mathrm{outer}}=0,\quad W_{\mathrm{outer}}=0,\quad\Delta E_g=-196\,\mathrm J.$$$$\Delta E_{\mathrm{total}}=\Delta U+\Delta E_g=196-196=0.$$The paddle work is an internal transfer for the enlarged system. Adding it again to its outer energy balance would double count the falling mass's energy.

Physical check

Temperature can rise in an adiabatic process through work input; heating in everyday speech need not mean thermodynamic heat transfer.