Temperature and Heat — Question 10

Question 10

Two heating routes reveal why heat is not a state variable

A rigid body has $U(T)-U(300\,\mathrm K)=100(T-300)\,\mathrm J$, with $T$ in kelvins. Route A raises it from 300 K to 310 K solely by thermal contact. Route B reaches the same final state using 400 J of electrical work input and thermal contact. For these stationary routes use $\Delta U=Q+W_{\mathrm{in}}$.

1. Find $Q_A$ and $Q_B$ and compare their common internal-energy change.

2. Return from 310 K to 300 K solely by thermal cooling. For the cycle using route B followed by this return, find net heat and net work input.

Question 10 - Solution

Two heating routes reveal why heat is not a state variable

(a) Method and calculation

$$\Delta U=100(310-300)=1000\,\mathrm J.$$$$Q_A=1000\,\mathrm J,\qquad Q_B=1000-400=600\,\mathrm J.$$Equal endpoint states determine $\Delta U$ but do not determine $Q$.

(b) Interpretation and further calculation

$$Q_{\mathrm{return}}=-1000\,\mathrm J,\quad Q_{\mathrm{cycle}}=600-1000=-400\,\mathrm J.$$$$W_{\mathrm{in,cycle}}=400\,\mathrm J,\quad\Delta U_{\mathrm{cycle}}=-400+400=0.$$A complete return restores state variables, while the signed heat integral need not vanish.

Physical check

The work supplied on route B is ultimately released as additional net heat to the surroundings over the cycle.