91Ó°ÊÓ

Finding the slope of the boundary line using the Clausius-Clapeyron equation.

$\frac{dP}{dT}=\frac{\mathrm{Δ}S}{\mathrm{Δ}V}$

The change in entropy is represented by $∆S$, and the change in volume is represented by $∆V$.

When ice melts into water, the change in entropy is positive. In this procedure, the volume change is negative.

As a result, the phase boundary$\frac{\mathrm{Δ}S}{\mathrm{Δ}V}$ between water and ice must have a negative slope.

The volume of ice can be calculated as

$V_i=\frac{\text{mass of ice}}{\text{density of ice}}$

Substituting the values

$$\begin{gathered} V_i=\frac{1\mathrm{g}}{917\mathrm{kg}/{\mathrm{m}}^3} \\ Vi=\frac{1\mathrm{g}\left(\frac{1\mathrm{kg}}{{10}^3\mathrm{g}}\right)}{917\mathrm{kg}/{\mathrm{m}}^3} \\ {V}_i=1.0905×{10}^{-6}{\mathrm{m}}^{-3} \end{gathered}$$

The volume of the water can be calculated as

$V_w=\frac{\text{mass of water}}{\text{density of water}}$

Substituting the values

$$\begin{gathered} V_w=\frac{1\mathrm{g}}{1000\mathrm{Kg}/{\mathrm{m}}^3} \\ {V}_w=\frac{1\mathrm{g}\left(\frac{1\mathrm{kg}}{{10}^3\mathrm{g}}\right)}{{10}^3\mathrm{kg}/{\mathrm{m}}^3} \\ {V}_w==1.00×{10}^{-6}{\mathrm{m}}^{-3} \end{gathered}$$

Therefore the change in volume will be

$$\begin{gathered} \mathrm{Δ}V=V_W-V_i \\ =1·00×{10}^{-6}{\mathrm{m}}^{-3}-1.0905×{10}^{-6}{\mathrm{m}}^{-3} \\ =-0.0905×{10}^{-6}{\mathrm{m}}^{-3} \end{gathered}$$

To find the slope of phase boundary we will use Clausius Clapeyron equation:

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

Where,

L is the latent heat of water

T is the temperature

Substituting the values,

role="math" localid="1646865013176" $$\begin{gathered} \frac{dP}{dT}=\frac{333\mathrm{J}}{(273\mathrm{K})\left(-0.0905×{10}^{-6}{\mathrm{m}}^{-3}\right)} \\ =\left(-135×{10}^5\mathrm{Pa}/\mathrm{K}\right)\left(\frac{1×{10}^{-5}\mathrm{bar}/\mathrm{K}}{1\mathrm{Pa}/\mathrm{K}}\right) \\ =-135bar/K \end{gathered}$$

As a result, during ice melting, the temperature drops by one degree

(from 0 to -1°C).

To stay on the phase boundary, the pressure must increase by 135 bars.

When glacier ice is viewed as a fluid, the pressure rise along with the depth is

$P=\mathrm{ÒÏ}gh$

Where,

$\mathrm{ÒÏ}$is density

h is depth of the glacier

g is the acceleration due to gravity

Depth is given as:

$h=\frac{P}{\mathrm{ÒÏ}g}$

Substituting the values

$h=\frac{\left(135\mathrm{bar}\right)\left(\frac{1×{10}^5\mathrm{N}/{\mathrm{m}}^2}{1\mathrm{bar}}\right)}{\left(917\mathrm{Kg}/{\mathrm{m}}^3\right)\left(9·8\mathrm{m}/{\mathrm{s}}^2\right)}$

$h=1500\mathrm{m}$

Hence the depth is 1500m

Assume a 5 mm wide skater blade with a 20 cm long length.

The entire area in contact with the ice when the skater is leaning on the corner of the blade is:

$$\begin{gathered} A=\left(\text{length}\right)\text{(width}) \\ A=(20\mathrm{cm}\left)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)\right(5\mathrm{mm})\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right) \\ A={10}^{-3}{\mathrm{m}}^2 \end{gathered}$$

Assume the mass to be 100kg, hence the weight of skater will be

$$\begin{gathered} F=mg \\ F=(100\mathrm{kg})\left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ F=980N \end{gathered}$$

Pressure on the blade will be

$$\begin{gathered} P=\frac{F}{A} \\ P=\frac{980\mathrm{N}}{{10}^{-3}{\mathrm{m}}^2} \\ P=9.8×{10}^5\mathrm{N}/{\mathrm{m}}^2\left(\frac{1\mathrm{bar}}{1×{10}^5\mathrm{N}/{\mathrm{m}}^2}\right) \\ \mathrm{P}=9.8\mathrm{bar} \\ \mathrm{P}≈10\mathrm{bar} \end{gathered}$$

Thus the pressure will be 10 bar.

The following is the drop in melting temperature under this pressure:

$dT=-\frac{dP}{135\mathrm{bar}/\mathrm{K}}$

Substituting dP as 10 bars

$$\begin{gathered} dT=-\frac{10\mathrm{bar}}{135\mathrm{bar}/\mathrm{K}} \\ =-0.074K \end{gathered}$$

As a result, the temperature change at pressure for ice melting is -0.074 K. This is an extremely low temperature. As a result, the skater blades do not cause any ice melting.