An inventor proposes to make a heat engine using water/ice as the working substance, taking advantage of the fact that water expands as it freezes. A weight to be lifted is placed on top of a piston over a cylinder of water at 1°C. The system is then placed in thermal contact with a low-temperature reservoir at -1°C until the water freezes into ice, lifting the weight. The weight is then removed and the ice is melted by putting it in contact with a high-temperature reservoir at 1°C. The inventor is pleased with this device because it can seemingly perform an unlimited amount of work while absorbing only a finite amount of heat. Explain the flaw in the inventor's reasoning, and use the Clausius-Clapeyron relation to prove that the maximum efficiency of this engine is still given by the Carnot formula, 1 -Te/Th

The heat engine efficiency is given as

$e=\frac{W}{Q_h}$

Where,

W is the work done

Q_h is the heat absorbed from hot reservoir

The heat engine's working ingredient is either water or ice. The primary flaw in the inventor's logic is that when water freezes to ice, the weight it can lift is restricted.

The following is the reason for this. The weight of the water will exert pressure on the water, lowering the freezing point below - 1°C. As a result, the water did not freeze.

In a heat engine, the work done by the working substance is equal to the change in the potential energy of the water.

$W=∆PE$

Let h_i be the starting water level in the cylinder. The height of the water climbs to h_f once it freezes into ice. As a result, the change in water's potential energy is as follows.

$\Delta PE=mg\left(h_f-h_i\right)$

Where,

m is the mass of water

g is the acceleration due to gravity

Therefore,

$W=mg\left(h_f-h_i\right)$

Heat absorbed will be

$Q_h=L$

Where L is the latent heat of water into ice

Substitute mg (h_f -h_i)for W and L for Q_h, in the equation e= $e=\frac{W}{Q_h}$. to solve for e.

$e=\frac{mg\left(h_f-h_i\right)}{L}$

Using the Clausius-Clapeyron relationship, the pressure added to the system as the freezing point drops below T is as follows:
$\frac{dP}{dT}=\frac{L}{T_h\Delta V}$

Where

$∆V$is the change in volume of water

dT is the difference in temperature

Where, V_fis final volume and V_iis initial volume

$dT=T_h-T_c$

T_h is temperature of hot reservoir

Therefore, we have

role="math" localid="1646899721371" $$\begin{gathered} \frac{dP}{T_h-T_c}=\frac{L}{T_h\left(V_f-V_i\right)} \\ dP=\frac{L}{T_h\left(V_f-V_i\right)}\left(T_h-T_c\right) \end{gathered}$$

Cylinder's initial volume is

$V_i=A{h}_1$

Cylinder's final volume is

$V_f=A{h}_f$

Where, A is the area.

The extra pressure to the system is,

$dP=\frac{mg}{A}$

Where, mg is water weight

Substituting and solving

$$\begin{gathered} \frac{mg}{A}=\frac{L}{T_h\left(A{h}_f-A{h}_i\right)}\left(T_h-T_c\right) \\ \frac{mg\left(h_f-h_i\right)}{L}=\frac{T_h-T_c}{T_h} \end{gathered}$$

Substitute, $e=\frac{mg\left(h_f-h_i\right)}{L}$

role="math" localid="1646900275148" $$\begin{gathered} e=\frac{T_h-T_c}{T_h} \\ e=1-\frac{T_c}{T_h} \end{gathered}$$

Therefore, the maximum efficiency isrole="math" localid="1646900319016" $1-\frac{T_c}{T_h}$