Temperature and Heat — Question 3

Question 3

Zero net heat can hide a large throughput

A plate remains at a steady temperature while receiving heat from a hot reservoir at 80 W and releasing heat to a cold reservoir at the same rate. The observation lasts 5.00 min. There is no work transfer.

1. Calculate inward, outward and net heat over the interval.

2. For a second run, the outward rate is 65 W while the inward rate stays 80 W. If the plate has constant heat capacity $C=900\,\mathrm{J/K}$ and obeys $C\Delta T=Q_{\mathrm{net}}$, find its temperature rise in 5.00 min.

Question 3 - Solution

Zero net heat can hide a large throughput

(a) Method and calculation

$$\Delta t=300\,\mathrm s,\quad Q_{\mathrm{in}}=80(300)=24000\,\mathrm J.$$$$Q_{\mathrm{out}}=24000\,\mathrm J,\quad Q_{\mathrm{net}}=0.$$Zero net heat does not imply that neither boundary transferred energy.

(b) Interpretation and further calculation

$$Q_{\mathrm{net}}=(80-65)(300)=4500\,\mathrm J.$$$$\Delta T=\frac{4500}{900}=5.00\,\mathrm K.$$Equal rates are the steady-temperature condition in this fixed-capacity model.

Physical check

The steady plate need not be in thermal equilibrium with either reservoir.