NBPhO 2025

Part i (2 points)

Water is in a good approximation incompressible; hence, when the piston starts moving, the growing volume must be filled by gas which can only be the water vapours. Thus, the water starts boiling: these vapours must be in equilibrium with water, hence the vapour pressure must be equal to the pressure inside the piston. We can read from the graph that at $T_0$, the vapour density is $\rho = 420\,\text{g\,m}^{-3}$; this corresponds to the pressure $p_1 = \rho RT/\mu = 70\,\text{kPa}$. With atmospheric pressure $p_0 = 100\,\text{kPa}$, the force needed to pull the piston is $S(p_0 - p_1) = 300\,\text{N}$.

Grading

• Realize that the pressure inside the cylinder equals the saturated vapour pressure of water at temperature $T_0$: 0.8 pts
• Read the density $\rho$ from the graph, in the range $[400,\,440]\,\text{g\,m}^{-3}$: 0.2 pts
• Use the ideal gas law to find an expression for the pressure $p_1$ at temperature $T_0$: 0.4 pts
• Correct expression for the force: $S(p_0 - p_1)$: 0.4 pts
• Correct numerical answer: 0.2 pts

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Part ii (2 points)

When the piston is pulled by displacement $a$, creating new volume $V_\text{new} = S \times a$, the water partially evaporates to fill this volume with vapour and the remaining liquid water cools from temperature $T_0$ to $T_1$. The mass of vapour $m_v$ needed to fill the new volume can be calculated using the vapour density $\rho_1 = 405\,\text{g\,m}^{-3}$ as $m_v = Sa\rho_1$. For the heat balance, the energy needed for evaporation must come from the cooling of the remaining liquid water:

$$m_v L = (m - m_v) \cdot c \cdot (T_0 - T_1). \tag{1}$$

Here we have neglected the dependence of $L$ on temperature and the heat capacity of water vapours, because $\mu L \gg 4R(T_1 - T_0)$ (but we have not neglected the work done by the piston, because $L$ is the enthalpy of evaporation and already includes $p\Delta V$). Similarly, since $L \gg c(T_1 - T_0)$, we can neglect $m_v$ in the right-hand-side and express

$$m = \frac{m_v L}{c(T_0 - T_1)} = \frac{\rho_1 SaL}{c(T_0 - T_1)} = 650\,\text{g}. \tag{2}$$

Grading

• Read the vapour density $\rho_1$ from the graph ($\rho_1 \in [390,\,420]\,\text{g\,m}^{-3}$): 0.2 pts
• Correct expression for mass of water vapour: 0.3 pts
• Correct expression for the latent heat ($m_v L$): 0.3 pts
• Correct expression for heat lost by water ($(m - m_v)\cdot c\cdot(T_0 - T_1)$): 0.3 pts
• Expression for energy conservation: 0.4 pts
• Correct expression for mass of water $m$: 0.3 pts
• Correct numerical answer $m \in [630,\,680]\,\text{g}$ (with correct dimension): 0.2 pts

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Part iii (3 points)

At thermal equilibrium, there is as much heat flux to the skin as there is heat loss due to evaporation. The former (per area) equals $\kappa\frac{dT}{dx}$ and the latter (per area) equals $-LJm$ where $m$ is the mass of one molecule, which we find to be $m = \mu/N_A$ to get $J\mu L/N_A$. Note that the minus sign comes from the fact that the particles diffuse from higher density areas to lower density areas. Now from the ideal gas law $n = P/(Tk_B) = PN_A/(TR)$ to get

$$J = -D\frac{d}{dx}\frac{rp}{Tk_B} = -D\frac{d}{dx}\frac{PN_A}{RT}.$$

Now the pressure of the water vapour is related to $r$ through $P = rp$, where $p$ denotes the saturation pressure of vapour. So,

$$\kappa\frac{dT}{dx} = -\frac{DL\mu}{R}\frac{d}{dx}\frac{rp}{T},$$

where $p = p(T)$ denotes the water vapour saturation pressure at the local air temperature; hence by integrating over $x$ we obtain

$$\kappa(T - T_s) = \frac{DL\mu}{R}\left[\frac{p(T_s)}{T_s} - \frac{rp(T)}{T}\right],$$

where the index $s$ denotes quantities evaluated at the skin surface. Also, we have used the fact that $r_s = 1$, because at the skin surface, the air is in direct contact with water (due to sweating, skin is wet), so that $pr_s = p(T_s)$. Substituting $\rho = \frac{p\mu}{RT}$ we obtain

$$\rho(T_s) = r\rho(T) + \frac{\kappa}{DL}(T - T_s).$$

Here we evaluate from the graph $r\rho(T) = 24.3\,\text{g\,m}^{-3}$ and $\frac{\kappa}{DL} = 0.51\,\text{g\,m}^{-3}\text{K}^{-1}$. Now we can draw this straight line onto the provided graph to find the intersection point at $T_s = 41.5\,°\text{C}$.

Solution 2 by Eppu Leinonen: One can also work directly with $\rho$ through the fact that $n = N/V = MN_A/(\mu V) = \rho N_A/\mu$. Then the heat flux magnitude will directly become

$$LJm = LmD\frac{dn}{dx} = LmD\frac{N_A}{\mu}\frac{d\rho}{dx} = LD\frac{d\rho_v}{dx},$$

where $\rho_v$ is the density of the water vapour. Then with correct signs we get

$$\kappa\frac{dT}{dx} = -LD\frac{d\rho_v}{dx}$$

from which by integrating and using $\rho_v = r\rho$ we get

$$\rho(T_s) = r\rho(T) + \frac{\kappa}{DL}(T - T_s)$$

and the solution proceeds the same way as in Solution 1.

Grading

NB! The $\rho = p\mu/RT$ substitution can be done earlier so the schemes below represent only the relevant observations which can be done with $\rho$ already. Equivalent forms will give points (i.e. if using $k$ and $N_A$ instead of $R$ in the middle steps).

• Heat going away from skin (up) = heat going to skin (down) at equilibrium: 0.4 pts
• Heat flux down $\kappa\frac{dT}{dx}$: 0.2 pts
• Heat flux up magnitude $\frac{DL\mu}{R}\frac{d}{dx}\frac{P}{T}$: 0.5 pts
  • Magnitude of heat flux up is $LJm$: 0.3 pts
  • $m = \mu/N_A$: 0.1 pts
  • $n = P/(Tk_B)$: 0.1 pts
• Deducing that the direction of the heat flow is opposite to $\frac{dn}{dx}$ (explicitly mentioned or with the existence of the minus sign in the equations): 0.1 pts
• $P = rp$: 0.1 pts
• $\kappa(T - T_s) = \frac{DL\mu}{R}\left[\frac{p(T_s)}{T_s} - \frac{rp(T)}{T}\right]$ (integrating correctly): 0.4 pts
  • Or doing a change from $d \to \Delta$ in the derivatives, which has to be motivated properly (for Fick’s law one must state that $J$ is constant due to conservation of particles).
• $\rho = p\mu/RT$: 0.1 pts
• Reading $\rho$ correctly ($\rho_1 \in [800,\,815]\,\text{g\,m}^{-3}$): 0.2 pts
• Graphical method: 0.8 pts
  • Noticing that $\rho(T_s) = r\rho(T) + \frac{\kappa}{DL}(T - T_s)$ defines a straight line in $(T,\,\rho)$: 0.8 pts
  • Any other valid numerical method that is explained is accepted.
• Correct final result $T \in [36,\,47]\,°\text{C}$: 0.2 pts