Temperature and Heat — Question 1

Question 1

Heat is a transfer, not a stored substance

A rigid object receives $900\,\mathrm J$ through thermal contact and loses $250\,\mathrm J$ through another thermal contact. No other energy crosses its boundary. Its initial internal energy is $4.00\,\mathrm{kJ}$. For this situation use $\Delta U=Q_{\mathrm{net}}$, with heat into the object positive.

1. Find net heat transfer and final internal energy.

2. Can the object be said to contain 650 J of heat? Repeat the final-energy calculation after shifting the arbitrary zero of internal energy upward by 10 kJ.

Question 1 - Solution

Heat is a transfer, not a stored substance

(a) Method and calculation

$$Q_{\mathrm{net}}=Q_{\mathrm{in}}-Q_{\mathrm{out}}=900-250=650\,\mathrm J.$$$$U_f=U_i+\Delta U=4000+650=4650\,\mathrm J.$$

(b) Interpretation and further calculation

Heat labels energy in transit because of a temperature difference; $U$ labels a state quantity. With $U^*=U+10000\,\mathrm J$,$$U_i^*=14000\,\mathrm J,\quad U_f^*=14650\,\mathrm J.$$$$\Delta U^*=650\,\mathrm J=\Delta U.$$The transferred energy is independent of the reference chosen for stored energy.

Physical check

The positive net transfer raises internal energy; changing its zero changes neither transfer nor difference.