NBPhO 2025

Part (i) — 2.0 / 2.0 pts

Criterion	Points	Result
Realize that the pressure inside the cylinder equals the saturated vapour pressure of water at temperature $T_0$	0.8	✓ “the gap therefore fills with saturated water vapour at $T_0 = 90\,°\text{C}$, at the pressure $p_{\mathrm{sat}}(T_0)$”, with an explicit aside that there is no air in the gap
Read the density $\rho$ from the graph, in the range $[400,\,440]\,\text{g\,m}^{-3}$	0.2	✓ $\rho_{\mathrm{sat}}(90\,°\text{C}) \approx 0.42\,\text{kg\,m}^{-3} = 420\,\text{g\,m}^{-3}$
Use the ideal gas law to find an expression for the pressure $p_1$ at temperature $T_0$	0.4	✓ $p_{\mathrm{sat}} = \rho_{\mathrm{sat}} R T_0/\mu \approx 70\,\text{kPa}$
Correct expression for the force: $S(p_0 - p_1)$	0.4	✓ $F_{\min} = (p_0 - p_{\mathrm{sat}}(T_0))\,S$ derived from a signed force balance on the piston
Correct numerical answer	0.2	✓ $F_{\min} \approx 300\,\text{N}$

Part (ii) — 2.0 / 2.0 pts

Criterion	Points	Result
Read the vapour density $\rho_1$ from the graph ($\rho_1 \in [390,\,420]\,\text{g\,m}^{-3}$)	0.2	✓ $\rho_{\mathrm{sat}}(89\,°\text{C}) \approx 0.40\,\text{kg\,m}^{-3} = 400\,\text{g\,m}^{-3}$
Correct expression for mass of water vapour	0.3	✓ $m_v = \rho_{\mathrm{sat}}(T_1)\,V = \rho_{\mathrm{sat}}(T_1)\,a S$
Correct expression for the latent heat ($m_v L$)	0.3	✓ “latent heat absorbed: $m_v L$”, with explicit identification of $L$ as the enthalpy of vaporisation
Correct expression for heat lost by water ($(m - m_v)\cdot c\cdot(T_0 - T_1)$)	0.3	✓ $(m - m_v)\,c\,\Delta T$ written verbatim before the $m_v \ll m$ simplification
Expression for energy conservation	0.4	✓ $(m - m_v)\,c\,\Delta T = m_v L$
Correct expression for mass of water $m$	0.3	✓ $m \approx \rho_{\mathrm{sat}}(T_1)\,a S L / (c\,\Delta T)$
Correct numerical answer $m \in [630,\,680]\,\text{g}$ (with correct dimension)	0.2	✓ $m \approx 0.65\,\text{kg} = 650\,\text{g}$

Part (iii) — 2.9 / 3.0 pts

Criterion	Points	Result
Heat going away from skin (up) = heat going to skin (down) at equilibrium	0.4	✓ “In: heat conducted down from the laminar layer, $q$ per unit area. Out: latent heat carried by evaporating water, $L\,j$ per unit area. In steady state these must be equal”
Heat flux down $\kappa\,dT/dx$	0.2	✓ $q = \kappa(T - T_s)/d$ from the linear $T(x)$ profile
Heat flux up magnitude $\frac{DL\mu}{R}\frac{d}{dx}\frac{P}{T}$ (or equivalent)	0.5	✓ Solution-2 form $L\,j = L D\,d\rho_v/dx$, derived via $\rho_v = m_{\mathrm{H_2O}} n$ so the mass flux is $j = D\,d\rho_v/dx$; the per-NB-note $\rho$-substitution subsumes the $LJm$, $m=\mu/N_A$ and $n = P/(Tk_B)$ subitems
Direction of the heat flow opposite to $dn/dx$ (explicit or via the minus sign)	0.1	✓ “Heat flux (Fourier), magnitude, downward toward the skin” and “Going upward (water leaves the skin and reaches the drier bulk)” — the two fluxes called out as anti-parallel
$P = rp$ (boundary condition at the bulk and at the skin)	0.1	✓ "$\rho_v(d) = r\,\rho_{\mathrm{sat}}(T)$" at the bulk and ”$\rho_v(0) = \rho_{\mathrm{sat}}(T_s)$… the wetness pins the local humidity to $100\%$” at the skin
Integrating correctly to $\kappa(T - T_s) = (DL\mu/R)\,[\,p(T_s)/T_s - r\,p(T)/T\,]$ (or equivalent in $\rho$)	0.4	✓ steady-state plus linear profiles give $\kappa(T-T_s) = LD\,(\rho_{\mathrm{sat}}(T_s) - r\,\rho_{\mathrm{sat}}(T))$ — the $\rho$-form is the post-substitution equivalent the NB note explicitly accepts
$\rho = p\mu/RT$	0.1	✓ used in part (i) at $90\,°\text{C}$ and again in the $110\,°\text{C}$ cross-check; the part-(iii) derivation works in $\rho$ from the start so the conversion is built into the boundary conditions
Reading $\rho$ correctly ($\rho_1 \in [800,\,815]\,\text{g\,m}^{-3}$ at $110\,°\text{C}$)	0.2	~ 0.1 / 0.2 — Claude reports $\rho_{\mathrm{sat}}(110\,°\text{C}) \approx 830\,\text{g\,m}^{-3}$, marginally above the keyed band; the in-text ideal-gas cross-check at $p_{\mathrm{sat}} = 143\,\text{kPa}$ would actually yield $808\,\text{g\,m}^{-3}$ (in band), so the 830 is a slightly aggressive chart extrapolation past the chart’s right edge at $105\,°\text{C}$
Graphical method — noticing $\rho(T_s) = r\rho(T) + (\kappa/DL)(T - T_s)$ defines a straight line in $(T, \rho)$	0.8	✓ explicit “straight line in $T_s$, of slope $-\kappa/(LD)$, passing through the point $(T_s, \rho) = (T, \rho_\infty)$ on the chart”; the numerical-tabulation root-find is one of the “any other valid numerical method” routes the scheme explicitly accepts
Correct final result $T_s \in [36,\,47]\,°\text{C}$	0.2	✓ $T_s \approx 43\,°\text{C}$

Discrepancies

$\rho_{\mathrm{sat}}(110\,°\text{C})$: Claude $830\,\text{g\,m}^{-3}$ vs official band $[800, 815]\,\text{g\,m}^{-3}$. The downstream impact is small: Claude’s intercept $\rho_\infty = r\,\rho_{\mathrm{sat}}(T) = 0.03 \cdot 830 \approx 25\,\text{g\,m}^{-3}$ versus the official’s $24.3\,\text{g\,m}^{-3}$ — a $\sim$2.5% difference at the right end of the line that swings by $\sim$34 g/m³ across the relevant $T$-range. Final $T_s$: Claude $43\,°\text{C}$ vs official $41.5\,°\text{C}$, both well inside the [36, 47] band; the 1.5°C gap is exactly what the intercept shift predicts.

Overall score: 6.9 / 7.0 pts

Full marks on parts (i) and (ii); a single 0.1 pt dock in part (iii) for an extrapolated chart read of $\rho_{\mathrm{sat}}(110\,°\text{C})$ that is marginally above the keyed band.

Numerical answers match the official key on every part: $F_{\min} \approx 300\,\text{N}$ (exact match), $m \approx 650\,\text{g}$ (exact match), $T_s \approx 43\,°\text{C}$ vs official $41.5\,°\text{C}$ (within the [36, 47] tolerance band, with the 1.5°C gap traceable to the marginally high $\rho(110\,°\text{C})$ extrapolation).

Commentary

Where this solution goes beyond the grading scheme. The “Overview” front-loads the three structural ideas — (1) liquid–vapour mechanical contact pins the gas pressure to $p_{\mathrm{sat}}$, (2) $L$ is enthalpy and already includes $p\,\Delta V$, (3) conduction and diffusion both scale as $1/d$ so the layer thickness cancels — and quotes the small parameter $1 - p_{\mathrm{sat}}/p_0 \approx 0.3$ that controls part (i). Part (i) follows the boxed $300\,\text{N}$ with a “physics remark” contrasting the vaporisation route with the alternative of breaking a tensile water column (several MPa for clean water in clean tubing) and three consistency checks: the $T_0 \to 100\,°\text{C}$ boiling-point limit where $F_{\min} \to 0$ (matching the everyday observation that boiling water lifts pot lids), the $T_0 \to 0\,°\text{C}$ cold limit where $F_{\min} \to p_0 S \approx 1000\,\text{N}$ (almost as hard as pulling against a hard vacuum), and an explicit dimensional check. Part (ii) carries an explicit $p_{\mathrm{sat}} V \approx 200\,\text{J}$ versus $m_v L \approx 2700\,\text{J}$ comparison that quantifies the $\sim$7% double-counting hazard for a student who treats $L$ as an internal-energy change, plus an a posteriori check $m_v/m \approx 2 \times 10^{-3}$ that justifies dropping $m_v$ on the right-hand side, an order-of-magnitude estimate (“a litre of water cools by $1\,\text{K}$ when it loses $\sim 4200\,\text{J}$”), and a $4\%$ sensitivity calculation showing how using $\rho(90\,°\text{C})$ instead of $\rho(89\,°\text{C})$ would shift the answer to $0.68\,\text{kg}$. Part (iii) develops the wet-bulb equation pedagogically: first the explicit linear profiles for $T(x)$ and $\rho_v(x)$ across the laminar layer, then the energy balance with the careful note that the bulk humidity $r$ does not enter at $x = 0$ because wetness pins the local humidity to $100\%$. The “memorable check” — comparing dry sauna ($r \to 0$, line passes through $(T, 0)$, $T_s$ drops further) with steam sauna ($r \to 1$, line passes through the saturation curve, $T_s \to T = 110\,°\text{C}$, fatal) — frames the answer in everyday physical experience. The Lewis number is named ($\kappa/(\rho_{\mathrm{air}} c_p^{\mathrm{air}} D) \approx 1$ for water vapour in air) and the answer is contextualised against meteorological wet-bulb data (dry desert at $40\,°\text{C}$ has $T_{wb} \approx 20\,°\text{C}$). A plausibility check connects $43\,°\text{C}$ to protein-denaturation onset at $\sim 60\,°\text{C}$, explaining why the experience is painful but survivable.

Where the official solution is sharper. Three places. (1) The official actually overlays the line on the saturation chart and reads the intersection from the printed graph (/figures/nbpho-2025/sol02-fig1.png), which is the most direct realisation of the “graphical method” worth 0.8 pts. Claude works the same equation through a tabulated numerical root-find with an ASCII sketch — equivalent in physical content and accepted by the scheme, but visually less convincing than a real overlay on the supplied chart. (2) The official gives a tighter, less aggressive read of $\rho_{\mathrm{sat}}(110\,°\text{C}) \approx 810\,\text{g\,m}^{-3}$ (matching exactly the steam-table-based ideal-gas computation $p_{\mathrm{sat}}(110\,°\text{C}) \approx 143\,\text{kPa}$, $\rho = p\mu/(RT) \approx 808\,\text{g\,m}^{-3}$), placing it inside the keyed [800, 815] band; Claude’s 830 sits 1.85% above the upper band edge, and his own in-text ideal-gas cross-check, if computed exactly, would have produced 808 — the inconsistency cost the 0.1 pt dock. (3) Part (ii): the official’s energy balance includes a one-line justification ”$\mu L \gg 4R(T_1 - T_0)$ … [so we may] neglect the dependence of $L$ on temperature and the heat capacity of water vapours”, a clean inequality that shows in two seconds why both omissions are safe. Claude justifies dropping $m_v$ from $(m - m_v)$ and warns against double-counting $p\,\Delta V$, but never explicitly checks whether $L(T)$-variation and the vapour heat capacity contribute at the percent level in the same way — the omission is implicit in his ”$L$ is enthalpy, full stop” framing, where the official spells out the inequality.