The First Law of Thermodynamics — Question 5

Question 5

A cycle must balance net heat and net work

Use $\Delta U=Q-W$, where $Q$ is heat into the system and $W$ is work done by it. A cyclic device absorbs 1.80 kJ from one thermal reservoir and releases 1.10 kJ to another during every cycle. No material enters or leaves; there are no other heat contacts. The device returns to its complete initial state.

1. Find net work output per cycle and average output power at 20.0 cycles/s.

2. An instrument instead reports 900 J of output per cycle with the same heat readings. Find the energy-balance residual and the additional heat input required if the work reading is correct.

Question 5 - Solution

A cycle must balance net heat and net work

(a) Method and calculation

$$\Delta U_{\mathrm{cycle}}=0\ \Rightarrow\ W_{\mathrm{cycle}}=Q_{\mathrm{net}}=1800-1100=700\,\mathrm J.$$$$P=700(20.0)=14000\,\mathrm W=14.0\,\mathrm{kW}.$$

(b) Interpretation and further calculation

$$Q_{\mathrm{net}}-W_{\mathrm{reported}}=700-900=-200\,\mathrm J.$$$$Q_{\mathrm{missing}}=900-700=200\,\mathrm J\quad\text{per cycle}.$$Without that input, the reported operation would reduce internal energy by 200 J per cycle and could not be a repeatable cycle.

Physical check

This is a first-law consistency test; it does not establish whether a proposed device satisfies the second law.