USAPhO 2023, Полуфинал — A3. The Motive Power of Ice

Question A3 — The Motive Power of Ice

In the Carnot cycle, a gas is heated at constant temperature $T_H$ and cooled at constant temperature $T_C$. Furthermore, no other heat transfer occurs, and all other steps of the cycle are reversible. The laws of thermodynamics state that any such cycle must have efficiency $\eta = W/Q_{\text{in}} = 1-(T_C/T_H)$. Below we will explore two other heat engines, which recover this efficiency in certain limits.

a. Consider the following heat engine involving one mole of ideal monatomic gas. The gas begins at temperature $T_0$, pressure $P_0$, and volume $V_0$, and undergoes four reversible steps: (1) expanded at constant pressure until its temperature rises to $(1+\beta)T_0$; (2) expanded at constant temperature until its pressure falls to $P_0/\alpha$; (3) contracted at constant pressure until its temperature falls back to $T_0$; (4) contracted at constant temperature until its pressure rises back to $P_0$.

i. Which steps require heat to be transferred to the gas? For each such step, give the total heat input in terms of $P_0$, $V_0$, $\alpha$, and $\beta$.

Solution: Heat is added to the gas in the first two steps. In the first step, we have heating at constant pressure, which has molar heat capacity $c_p = 5R/2$, so

where we used the ideal gas law in the final step. In the second step, the heat added to the gas compensates for the work done while it expands, so

ii. Under what conditions on $\alpha$ and $\beta$ would we expect the efficiency of this heat engine to approach that of a Carnot cycle working between the same maximum and minimum temperatures?

Solution: Since all the steps are reversible, we recover the Carnot efficiency when almost all the heat transfer happens at the same temperature, i.e. when $Q_1 \ll Q_2$. This holds when

We should also consider the other two steps, which have heat transfer
$|Q_3| = c_p|\Delta T| = \frac{5}{2}R\beta T_0 = \frac{5}{2}\beta P_0V_0$
and

We have $|Q_3| \ll |Q_4|$ when

which is stricter than the previous condition. Thus, we recover the Carnot efficiency when $\beta \ll \ln\alpha$.

iii. Find the efficiency of this heat engine for general $\alpha$ and $\beta$.

Solution: We calculate the net work throughout all four steps,

where the first and third terms are just the usual $P\Delta V$ work, and the second and fourth use the form done in an isothermal process. Simplifying gives

Therefore, the efficiency of the process is

One way to check the answer is to note that the Carnot efficiency would be $\beta/(1+\beta)$. Our general result reduces to this efficiency when the second term in the denominator is negligible, which is precisely the condition identified in part (b).

b.

Now consider a heat engine built around the freezing and melting of water, which occurs at a pressure-dependent temperature $T_c(P)$. Initially, a volume of $V$ of water is squeezed underneath a piston, so that it experiences a total pressure $P_1$, and the water is on the edge of freezing, with temperature $T_c(P_1)$. The engine then undergoes four reversible steps: (1) a mass is slowly placed on the piston, raising the total pressure to $P_2$; (2) the water is cooled to temperature $T_c(P_2)$ and frozen; (3) the mass is slowly removed from the piston, lowering the pressure back to $P_1$; (4) the ice is heated back to temperature $T_c(P_1)$ and melted. Assume that water and ice are incompressible, with fixed densities $\rho_w$ and $\rho_i$.

i. What is the net work done by this engine, in terms of $P_1$, $P_2$, $V$, and the densities?

Solution: The system expands at a pressure $P_2$ and contracts at a pressure $P_1$, so the net work done is

Concretely, this work used to raise the mass, so it could also be calculated as $Mg\Delta H$.

ii. Assume the latent heat per unit mass $L$ to melt ice is large, so that freezing and melting account for essentially all of the heat transfer in the cycle. What is the efficiency of the engine, in terms of $P_1$, $P_2$, $L$, and the densities?

Solution: The heat transferred in is used to melt the ice, $Q_{\text{in}} = \rho_w V L$, so

Note that the efficiency can also be expressed as $\eta = 1-\frac{Q_{\text{out}}}{Q_{\text{in}}}$, but in order to find $Q_{\text{out}}$ directly, we would need to know how the latent heat varies with pressure, which isn't given in the problem.

iii. Since we assumed all heat transfer occurs during melting or freezing, this cycle has the same efficiency as a Carnot cycle. In the limit where $P_1$ and $P_2$ are very close, use this fact to infer an expression for $dT_c/dP$ in terms of $T_c$, $L$, and the densities.

Solution: The Carnot efficiency, for the same high and low temperatures, is

Combining this with the result of part (b) gives

and taking the limit of $P_1\approx P_2$, the left-hand side becomes a derivative, so

This is equivalent to the well-known Clausius–Clapeyron equation, and this heat engine was first devised by the brothers Thomson and Kelvin.