91Ó°ÊÓ

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.

$\mathrm{Δ}PE=mg\left(h_f-h_i\right)$

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:
$\frac{dP}{dT}=\frac{L}{T_h\mathrm{Δ}V}$

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" $1-\frac{T_c}{T_h}$