Question

The density of ice is 917 kg/m*.

(a) Use the Clausius-Clapeyron relation to explain why the slope of the phase boundary between water and ice is negative.

(b) How much pressure would you have to put on an ice cube to make it melt at -1°C?

(c) ApprOximately how deep under a glacier would you have to be before the weight of the ice above gives the pressure you found in part (b)? (Note that the pressure can be greater at some locations, as where the glacier flows over a protruding rock.)

(d) Make a rough estimate of the pressure under the blade of an ice skate, and calculate the melting temperature of ice at this pressure. Some authors have claimed that skaters glide with very little friction because the increased pressure under the blade melts the ice to create a thin layer of water. What do you think of this explanation?

Short answer

The phase boundary between water and ice must have a negative slope.

During ice melting, the temperature drops by one degree (from 0 to -1°C).

Hence the depth is 1500m.

Thus the pressure will be 10 bar.

The temperature change at pressure for ice melting is -0.074 K.

Step-by-step solution

Given information

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

$\frac{dP}{dT}=\frac{\Delta S}{\Delta V}$

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

Part (a) Step 2: Explanation 

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$\frac{\Delta S}{\Delta V}$ between water and ice must have a negative slope.

Part (b) Step 3: Explanation

The volume of ice can be calculated as

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

Substituting the values

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

The volume of the water can be calculated as

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

Substituting the values

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

Therefore the change in volume will be

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

Part (b) Step 4 : Calculation 

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

$\frac{dP}{dT}=\frac{L}{T\Delta V}$

Where,

L is the latent heat of water

T is the temperature

Substituting the values,

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

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.

Part (c) Step 5: Explanation

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

$P=\rho gh$

Where,

$\rho$is density

h is depth of the glacier

g is the acceleration due to gravity

Depth is given as:

$h=\frac{P}{\rho g}$

Substituting the values

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

$h=1500\mathrm{m}$

Hence the depth is 1500m

Part (d) Step 6: Explanation

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

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

Thus the pressure will be 10 bar.

Explanation

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.