Conservation of Energy — Question 5

Question 5

A cyclic device with two useful outputs

A device completes a cycle and returns every internal store to its initial value. It receives 600 J by heat transfer, delivers 150 J of mechanical work and 90 J electrically, and releases the remaining energy as heat. No other transfers occur.

1. Find heat released to the surroundings.

2. Find the fraction of input appearing in the two useful outputs together.

3. Why is zero cycle energy change compatible with nonzero work output?

Question 5 - Solution

A cyclic device with two useful outputs

(a) Use equality of total input and output over the cycle.

The device returns to its initial state, so net storage vanishes: $$0=600-150-90-Q_{\mathrm{out}}\Rightarrow\boxed{Q_{\mathrm{out}}=360\ \mathrm J}.$$

(b) Combine the specified useful channels.

Combine the specified useful outputs: $$\boxed{\eta_{\mathrm{useful}}=\frac{W_{\mathrm{mech}}+E_{\mathrm{electric}}}{Q_{\mathrm{in}}}=\frac{150+90}{600}=.40=40\%}.$$ This is the stated device’s energy accounting, not a universal efficiency limit.

(c) Track continuing boundary transfers.

Over the cycle, $$\Delta E_{\mathrm{cycle}}=0\quad\text{while}\quad E_{\mathrm{in}}=E_{\mathrm{out}}=600\ \mathrm J.$$ The input funds nonzero outputs. Returning to the same state requires zero net storage, not zero transfer through each channel.