OPhO 2025, Финален кръг · част I — Задача 3. Theory Question 3

• (A.1) $T^* = T_m\left(1 - \frac{P\Delta v}{L_f}\right),\ P = \frac{Mg}{A_{skate}}$ ≈ $-0.088$ °C

  Official: $P = 1.18 \times 10^6$ Pa, $T^* = -0.088^{\circ}C$.

• (A.2) Only at temperatures very close to $T_m$.

• (B.1) $U = \frac{\pi}{12h^2}C\rho_{vap}\rho_{ice}$

  $\rho_{vap}, \rho_{ice}$ are number densities, assumed approximately constant.

• (B.2) $h = \left(\frac{\pi C\rho_{vap}\rho_{ice}T_m}{6L_f\Delta T}\right)^{1/3}$ ≈ $12$ nm

  For $\Delta T = 0.088$; the official solution adds "(Correct value: $h = 10.9$ nm.)" using the true vapour pressure 611 Pa.

• (B.3) 170 nm

  186 nm of ice, i.e. about 170 nm of water; answers within the same order of magnitude accepted. Pressure melting dominates very close to $0^{\circ}C$; intermolecular forces dominate as $T \to T_{m,p}$.

• (B.4) 5.25 N

  $f = \nu A\frac{v}{h}$ with $h = 170$ nm; too high, so other effects (frictional heating) must thicken the water layer.

• (C.1) $\frac{dh}{dx} = \frac{\nu v}{h\rho_w L_f}$

• (C.2) $\frac{dh}{dx} = -\frac{\kappa(T_m - T_0)}{\rho_w L_f\sqrt{\pi Dxv}}$

• (C.3) $h = \left(-C_2 + \sqrt{C_2^2 + 2C_1}\right)\sqrt{x},\ C_1 = \frac{\nu v}{\rho_w L_f},\ C_2 = \frac{\kappa(T_m - T_0)}{\rho_w L_f\sqrt{\pi Dv}}$

  Deepest layer about $0.94\ \mu$m, much larger than the nanoscale effects of part B.

• (C.4) $f = \nu wv\left(-C_2 + \sqrt{C_2^2 + 2C_1}\right)^{-1}(2\sqrt{L})$

  Near $0^{\circ}C$: $f = 2\nu wv\sqrt{L/(2C_1)} \approx 1.728$ N; at very low temperatures $f \propto T_m - T_0$.

• (C.5) Wax has a higher viscosity and higher thermal diffusivity and lower thermal conductivity; the liquid layer formed is too thick, the blade sinks into it and gets stuck.

• (D.1) $p = \frac{6\nu\dot{h}}{h^3}\left(\frac{w^2}{4} - y^2\right)$

• (D.2) $v > \frac{2P\left(-C_2 + \sqrt{C_2^2 + 2C_1}\right)^2 L^2}{\nu w^2}$ ≈ $2.33$ m/s

On Ice - Solution

(A.1) We wish to find the new melting point of the ice under the skate's pressure. This can be found via the Clausius-Clayperon relation:

(0.2) which gives the new melting point in the linear approximation as

(0.2) where $\Delta v$ is the change in specific volume and $P = \frac{Mg}{A_{skate}}$ is the pressure applied on the ice. Substituting in numerical values gives $P = 1.18 \times 10^6$ Pa, $T^* = -0.088^{\circ}C$. (0.2)

(A.2) Only at temperatures very close to $T_m$ will the pressure melting effect be significant. (0.2)

(B.1) The total potential energy of the two fluid layers due to intermolecular forces is expressed through integrating over both layers:

(0.2) Where $\rho_{vap}, \rho_{ice}$ are the number densities of the respective phases. Decomposing into Cartesian coordinates, fixing the ice atom at x=0, y=0 without loss of generality,

(0.4), 0.2 for integrating over the plane This assumes that both the gas and ice phase have approximately constant number densities. (0.2)

(B.2) Thermodynamic equilibrium occurs when the Gibbs free energy is minimised. (0.2) At the phase transition we have

so the bulk free energy of the liquid layer per unit area is

(0.2) where $\Delta T$ is the difference in ambient temperature from the melting point. The change in enthalpy is equal to the potential energy caused by the intermolecular forces. As a result the full expression for the Gibbs energy is

Minimising this with respect to $h$, we get

(0.2) Next, we need to find $\rho_{vap}$ and $\rho_{ice}$. The mass density of ice at $0^{\circ}C$ was given as $0.9168\frac{\mathrm{g}}{\mathrm{cm}^3}$, so using the molar mass of water $\mu = 18\frac{\mathrm{g}}{\mathrm{mol}}$ we get $\rho_{ice} = 3.07 \times 10^{28}\ \mathrm{m}^{-3}$. (0.2) For water vapour, we need the vapour pressure of water at $0^{\circ}C$. Again we use Clausius-Clayperon but this time with the latent heat of vaporisation, and noting that the specific volume of the gas is much larger than that of the liquid:

where $P_0$ is atmospheric pressure and $T_0 = 373$ K is the boiling point of water. Substituting in the appropriate values we get at $0^{\circ}C$ that $P = 819$ Pa. (Note: $L$ varies significantly over the range of 0 to $100^{\circ}C$, so this calculation is only approximate; the true value for vapour pressure at $0^{\circ}C$ is 611 Pa. Calculations using this value should not be penalised.) (0.2)

The number density of water vapour is therefore $\rho_{vap} = \frac{P_v}{k_B T} = 2.16 \times 10^{23}\ \mathrm{m}^{-3}$. (True value: $1.62 \times 10^{23}$.) Substituting the values, we get (0.1)

For the maximum value of $\Delta T = 0.088$, we get $h = 12.0$ nm. (Correct value: $h = 10.9$ nm.) (0.1)

(B.3) There are multiple valid ways to do this, and all answers within the same order of magnitude should be accepted. We consider the amount of heat transmitted by a medium with a temperature difference $\Delta T_m = 0.088$ K in the time the blade presses on the ice. The thermal diffusion lengthscale is $l_{diff} = \sqrt{\frac{DL}{v}} = \sqrt{\frac{\kappa L}{\rho cv}} = 3.37 \times 10^{-4}$ m. (0.2) This is the approximate depth to which the ice will be cooled. To find the heat transmitted per unit area, we use the Fourier law: $\dot{Q} = \kappa\frac{\Delta T}{l_{diff}} = 569\frac{\mathrm{W}}{\mathrm{m}^2}$. (0.2) This can melt a thickness of $h = \frac{\dot{Q}t}{\rho L_f} = 186$ nm of ice, corresponding to $170\,\mathrm{nm}$ of water due to increased density. (0.1) At these very high temperatures (very close to $0^{\circ}C$), pressure melting is more significant. However pressure melting has $h \propto T - T_{m,p}$ where $T_{m,p}$ is the melting point of the ice under the pressure, whereas the liquid layer formed by intermolecular forces has $h \propto \frac{1}{(T - T_m)^{-\frac{1}{3}}}$. This means that as $T \to T_{m,p}$ the intermolecular force will start to dominate, and below $0.088^{\circ}C$ pressure melting does not work at all. (0.2)

(B.4) We take the larger value of $h = 170$ nm. The frictional force originates from the shear viscosity of the water. The total force is $f = \nu A\frac{v}{h} = 5.25$ N. (0.2) This is not a lot, but we know that at $0^{\circ}C$ ice is extremely slippery, so a frictional force corresponding to approximately 1% of the body weight while gliding at a very slow speed is still too high. There must be other effects which thicken the water layer and lower the friction. (0.2) In fact, the intelligent reader will realise that the frictional heat generated per unit area is $9.6\frac{\mathrm{kW}}{\mathrm{m}^2}$, which is quite a lot higher than the $569\frac{\mathrm{W}}{\mathrm{m}^2}$ involved in heat transfers during pressure melting. This gives a hint that frictional heating is the dominant mechanism in skating, which will be explored in the next section.

(C.1) (Part C of this question is based on Lozowski and Szilder 2009, Lozowski and Szilder 2013.) Let $x$ be the distance along the blade under consideration. In a unit time the power dissipated at this position is

where $w = \frac{A_{skate}}{2L}$ is the width of the skate. (0.2) The power required to melt a layer of water of depth $dh$ is

equating these quantities gives (0.2)

(C.2) We first need to consider the thermal model given: as the blade comes into contact with the ice at time $t = 0$, the boundary layer temperature is instantaneously raised to $0^{\circ}C$. The thermal diffusion equation gives:

(0.2) In the vertical profile, this admits a similarity solution $T = f(z = \frac{y}{\sqrt{Dt}})$. (0.2) $f$ satisfies the differential equation

The solution to this equation is (0.1)

(0.1) or, in other words,

To find the constant $C$, we note that the boundary condition gives $T(y = 0) = T_m$ and $T(y \to \infty) = T_0$. We note that

(0.2) so $C = \frac{1}{\sqrt{2\pi}}(T_m - T_0)$. This means that the instantaneous heat flux at the surface into the ice per unit area is

where $t = \frac{x}{v}$ is the time the ice has spent in contact with the blade. (0.2) Equating this with (18) we obtain (0.1)

(C.3) The combined differential equation is (0.1)

We let the constants $C_1 = \frac{\nu v}{\rho_w L_f}$ and $C_2 = \frac{\kappa(T_m - T_0)}{\rho_w L_f\sqrt{\pi Dv}}$ so that

We note that, substituting $h = K_2 x^{\frac{1}{2}}$, we get

so $K_2 = -C_2 \pm \sqrt{C_2^2 + 2C_1}$. (0.3) With the initial condition $h(0) = 0$ we get

where we choose the positive sign to ensure the coefficient is always larger than 0. (0.1) If we substitute $\Delta T = 0.088$, we obtain $C_1 = 5.36 \times 10^{-12}$ and $C_2 = 3.03 \times 10^{-7}$, so the coefficient is $2.98 \times 10^{-6}$. (0.1) The deepest part of the liquid layer will have depth $0.94\ \mu$m. This is much larger than the nanoscale effects in part B. (0.1)

(C.4) The total frictional force is expressed by the integral (0.2)

At very high temperatures (i.e. very close to $0^{\circ}C$), we have $C_2 = 0$ so $f = 2\nu wv\sqrt{\frac{L}{2C_1}}(\approx 1.728\ \mathrm{N})$. (0.1) At very low temperatures, $C_2 \to \infty$ (0.1) so to lowest order in $\frac{C_1}{C_2}$ we have:

so the force is proportional to $T_m - T_0$. (0.2)

(C.5) Note that $\frac{C_1}{C_2} = \frac{\nu v\sqrt{\pi Dv}}{\kappa(T_m - T_0)}$. Wax has a higher viscosity and higher thermal diffusivity, as well as lower thermal conductivity. The liquid layer formed is too thick, the blade sinks into it and gets stuck. (Other physical answers are also accepted, including explanations that the blade pressure inhibits melting, wax is a soft material which plastically deforms, etc.) (0.4)

(D.1) The governing equations of the fluid flow are

(these can be derived from differential element analysis if needed, but it is acceptable to quote them as standard equations of motion.) (0.2) We approximate that the vertical variation of pressure is negligible so that $v_y$ can be written as

(0.2) substituting in the boundary conditions $v_y(0) = v_y(h) = 0$. Substituting this expression in the continuity equation, we obtain

(0.2) where we have also used the fact that $\frac{\partial^2 v_z}{\partial z^2} = 0$, and integrated the expression over $z$ from 0 to h. This gives (0.2)

(D.2) The upward force experienced per unit length by the blade due to the pressure from the excess flow is the expression in (D.1) integrated over the width of the blade: (0.2)

In other words, the thinning of the liquid layer $\dot{h} = -v\frac{dh}{dx}$ is determined by sufficient pressure to support the weight of the skater, so that

with $P$ as defined in (A.1). (0.2) For this to be smaller than the other effects, we need $\frac{Ph^3}{\nu w^2 v} < \frac{h}{2x}$ or

where we have replaced $x$ with $L$ to represent the maximum value. (0.2) For the given numerical values, this is $2.33\frac{\mathrm{m}}{\mathrm{s}}$.