NBPhO 2025

Overview

The whole problem turns on one chart — the saturation curve $\rho_{\mathrm{sat}}(T)$ — and on three quantitative ideas, each used in one of the three parts:

1. A liquid in mechanical contact with its own vapour sets the gas pressure to $p_{\mathrm{sat}}(T)$, not to atmospheric. This is what makes evaporation through a piston “cheap”: the water on one side does not push back at $1\ \text{atm}$, but at the much smaller saturated-vapour pressure.
2. Latent heat $L$ is the enthalpy of vaporisation, not the internal-energy change. It already includes the $p\,\Delta V$ work the new vapour does on its surroundings, so an energy balance written as $m\,c\,\Delta T = m_v L$ has no extra $p\,\Delta V$ term.
3. In a steady-state laminar layer, conduction and diffusion fluxes both scale as $1/d$ — the layer thickness drops out of every flux ratio, so the wet-bulb temperature is a property of the bulk air alone.

The temperatures are below the normal boiling point in parts (i) and (ii) and the partial pressure $p_{\mathrm{sat}}$ that drives the physics is well below $p_0$. Numerically we have, from the chart and the ideal-gas law,

$$p_{\mathrm{sat}}(90\ °\text{C}) \;=\; \frac{\rho_{\mathrm{sat}}\,R\,T}{\mu} \;\approx\; \frac{0.42\cdot 8.31\cdot 363}{0.018}\,\text{Pa} \;\approx\; 70\ \text{kPa},$$

so the small parameter $\,1 - p_{\mathrm{sat}}/p_0 \approx 0.3\,$ controls part (i), and the slope $\kappa/(LD)$ — the only combination of transport coefficients that has units of $\rho/T$ — controls part (iii).

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Part (i) — Minimum pulling force on the piston

Why a vapour gap forms as soon as the piston moves

Initially the cylinder is full of liquid water up to the piston, and the piston is in mechanical equilibrium with $p_0 = 100\ \text{kPa}$ on its outer face. Pulling the piston outward by even an infinitesimal amount opens a gap on the water side. Liquid water at $90\ °\text{C}$ is below its boiling point at $1\ \text{atm}$, but the water surface is now in contact with a region of lower pressure — and water will boil there until the gas in the gap reaches saturation. The gap therefore fills with saturated water vapour at $T_0 = 90\ °\text{C}$, at the pressure $p_{\mathrm{sat}}(T_0)$.

(There is no air in the gap: the piston was originally seated on the water with no air between them, and no air can leak past a sealed piston.)

Force balance on the piston

With saturated vapour on the inside and atmosphere on the outside, the piston feels three forces along its axis (taking outward = “to the right” as positive):

• atmospheric pressure pushing inward: $-p_0 S$,
• saturated vapour pushing outward: $+p_{\mathrm{sat}}(T_0)\,S$,
• the applied pull, $+F$.

The minimum $F$ that lets the piston actually leave the seated position is the value at which these balance,

$$F + p_{\mathrm{sat}}(T_0)\,S - p_0\,S \;=\; 0 \quad\Longrightarrow\quad F_{\min} \;=\; \bigl(p_0 - p_{\mathrm{sat}}(T_0)\bigr)\,S.$$

Numerical value

From the saturation chart, $\rho_{\mathrm{sat}}(90\ °\text{C}) \approx 0.42\ \text{kg\,m}^{-3}$. Treating the vapour as an ideal gas,

$$p_{\mathrm{sat}}(90\ °\text{C}) \;=\; \frac{\rho_{\mathrm{sat}}\,R\,T_0}{\mu} \;=\; \frac{0.42\cdot 8.31\cdot 363.15}{0.018}\,\text{Pa} \;\approx\; 70\ \text{kPa}.$$

Then with $S = 1\ \text{dm}^2 = 10^{-2}\ \text{m}^2$ and $p_0 = 100\ \text{kPa}$,

$$F_{\min} \;=\; (100 - 70)\cdot 10^{3}\,\text{Pa}\,\cdot\,10^{-2}\,\text{m}^{2} \;=\; 300\ \text{N}.$$ $$\boxed{\;F_{\min} \;=\; \bigl(p_0 - p_{\mathrm{sat}}(T_0)\bigr)\,S \;\approx\; 300\ \text{N}\;}$$

Physics remark — why pulling on liquid water is not free

If the water could not vaporise (e.g. if it were a non-volatile liquid like mercury at room temperature), then to separate the piston from the liquid one would have to break a column of liquid against its tensile strength — easily several MPa for clean water in a clean tube, but unreliable. Vaporisation provides a much “cheaper” route: the liquid breaks itself by boiling at the piston face, and the only sustained resistance is the pressure deficit $p_0 - p_{\mathrm{sat}}(T_0)$. That deficit, not $p_0$ itself, is what one has to pull against — a factor of ~3 saving here.

Consistency checks

• Limit $T_0 \to 100\ °\text{C}$. Then $p_{\mathrm{sat}} \to p_0$ and $F_{\min} \to 0$: at the boiling point the water’s own vapour already pushes back at $1\ \text{atm}$, and the piston can be displaced for free. Consistent with the everyday observation that boiling water lifts pot lids by itself.
• Limit $T_0 \to 0\ °\text{C}$. Then $p_{\mathrm{sat}}$ is tiny ($< 1\ \text{kPa}$) and $F_{\min} \to p_0 S \approx 1000\ \text{N}$: pulling cold water is almost as hard as pulling against a hard vacuum, since the cold liquid produces almost no vapour.
• Dimensions. $[\,p\,S\,] = \text{Pa}\cdot\text{m}^2 = \text{N}$. ✓

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Part (ii) — Mass of water under the piston

Setting up the energy balance

After the piston has moved out by $a = 3\ \text{dm}$, a vapour-filled volume

$$V \;=\; a\,S \;=\; 0.3\cdot 10^{-2}\ \text{m}^3 \;=\; 3\cdot 10^{-3}\ \text{m}^3$$

sits between the water surface and the piston. The water has cooled from $T_0 = 90\ °\text{C}$ to $T_1 = 89\ °\text{C}$, and is in equilibrium with its vapour at $T_1$. The vapour is therefore saturated at $T_1$, with mass density $\rho_{\mathrm{sat}}(T_1)$.

The mass of vapour produced is

$$m_v \;=\; \rho_{\mathrm{sat}}(T_1)\,V.$$

From the chart, $\rho_{\mathrm{sat}}(89\ °\text{C}) \approx 0.40\ \text{kg\,m}^{-3}$ (only slightly less than $0.42\ \text{kg\,m}^{-3}$ at $90\ °\text{C}$), giving $m_v \approx 1.2\ \text{g}$.

Why the cooling is supplied by the liquid, not by the vapour

The energy needed to vaporise the mass $m_v$ at this temperature is $m_v L$ — the enthalpy change. Treating $L$ as enthalpy already accounts for the work the new vapour does in pushing the piston out at pressure $p_{\mathrm{sat}}$; we should not add a separate $p_{\mathrm{sat}}\,V$ term. (See the explicit check below.)

That energy is drawn from the sensible heat of the remaining liquid: the only thermal reservoir present that can supply $m_v L \sim 2.7\ \text{kJ}$ in the time it takes the piston to move. Calling $m$ the total mass of water (liquid + just-formed vapour) under the piston,

$$\underbrace{(m - m_v)\,c\,\Delta T}_{\text{liquid sensible heat lost}} \;=\; \underbrace{m_v L}_{\text{latent heat absorbed}}, \qquad \Delta T = T_0 - T_1 = 1\ \text{K}.$$

Since $m_v \ll m$ (we will check this a posteriori: $m_v \approx 1\ \text{g}$, $m \sim 10^2\ \text{g}$), the difference between $m$ and $m - m_v$ is below the precision of the chart and we may safely write

$$m\,c\,\Delta T \;\approx\; m_v L \quad\Longrightarrow\quad m \;\approx\; \frac{m_v L}{c\,\Delta T}.$$

Plugging in the numbers

$$m \;\approx\; \frac{\rho_{\mathrm{sat}}(T_1)\,V\,L}{c\,\Delta T} \;=\; \frac{0.40\,\text{kg\,m}^{-3}\,\cdot\,3\cdot 10^{-3}\,\text{m}^3\,\cdot\,2.26\cdot 10^{6}\,\text{J\,kg}^{-1}}{4200\,\text{J\,kg}^{-1}\text{K}^{-1}\,\cdot\,1\,\text{K}}.$$

Numerator: $0.40\cdot 3\cdot 10^{-3}\cdot 2.26\cdot 10^{6} \approx 2.71\cdot 10^{3}\ \text{J}$.

Denominator: $4.2\cdot 10^{3}\ \text{J\,kg}^{-1}$.

$$m \;\approx\; \frac{2.71\cdot 10^{3}}{4.2\cdot 10^{3}}\,\text{kg} \;\approx\; 0.65\ \text{kg}.$$ $$\boxed{\;m \;\approx\; \frac{\rho_{\mathrm{sat}}(T_1)\,a\,S\,L}{c\,\Delta T} \;\approx\; 0.65\ \text{kg}\;}$$

This justifies the assumption $m_v \ll m$: $m_v / m \approx 1.2\,\text{g} / 650\,\text{g} \approx 2\cdot 10^{-3}$.

Why the $p\,\Delta V$ work is not a separate term

A concrete check: the gas does work $W = p_{\mathrm{sat}}(T_1)\,V$ on the piston as it forms. With $p_{\mathrm{sat}}(89\ °\text{C}) \approx \rho_{\mathrm{sat}}(T_1) R T_1 / \mu \approx 67\ \text{kPa}$,

$$W \;\approx\; 67\,\text{kPa}\cdot 3\cdot 10^{-3}\,\text{m}^3 \;\approx\; 200\ \text{J}.$$

The latent heat $m_v L \approx 2700\ \text{J}$. So the "$p\,\Delta V$" work is about $7\%$ of $m_v L$. If a student writes the internal-energy balance and then re-adds $p_{\mathrm{sat}} V$ on top of $m_v L$, this $7\%$ contribution is double-counted. Using $L$ (enthalpy) directly with no extra work term avoids the trap.

Consistency checks

• Dimensions. $[\rho V L / (c\,\Delta T)] = \text{kg\,m}^{-3}\cdot\text{m}^3\cdot \text{J\,kg}^{-1} / (\text{J\,kg}^{-1}\text{K}^{-1}\cdot\text{K}) = \text{kg}$. ✓
• Order of magnitude. A litre of water cools by $1\ \text{K}$ when it loses $\sim 4200\ \text{J}$. Vaporising $1\ \text{g}$ takes $\sim 2300\ \text{J}$. So roughly half a litre of water can vaporise about a gram, matching our answer.
• Effect of the small $\rho_{\mathrm{sat}}(T_1)$ vs $\rho_{\mathrm{sat}}(T_0)$ uncertainty. Using the $90\ °\text{C}$ value $0.42$ instead of the $89\ °\text{C}$ value $0.40$ would change the answer to $\approx 0.68\ \text{kg}$ — about a 4% variation, within the precision of reading the chart.

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Part (iii) — Wet-bulb temperature in the sauna

Geometry of the laminar layer

Place a coordinate $x$ perpendicular to the skin, $x = 0$ at the wet skin surface, $x = d$ at the top of the laminar layer where it meets the turbulent bulk. In steady state, with no sources or sinks inside the laminar layer (no chemistry, no condensation, etc.), the conservation equations

$$\frac{d}{dx}\bigl(-\kappa\,\frac{dT}{dx}\bigr) = 0, \qquad \frac{d}{dx}\bigl(-D\,\frac{d\rho_v}{dx}\bigr) = 0$$

both reduce to $T''(x) = 0$ and $\rho_v''(x) = 0$. Both temperature and vapour density are therefore linear across the layer.

Boundary conditions

• At $x = d$, the bulk values: $T(d) = T = 110\ °\text{C}$ and $\rho_v(d) = r\,\rho_{\mathrm{sat}}(T)$ (the bulk relative humidity is $r = 0.03$ by definition).
• At $x = 0$, the skin: $T(0) = T_s$ (unknown), and — because the surface is wet — the vapour just above the skin is in equilibrium with liquid water at $T_s$, so $\rho_v(0) = \rho_{\mathrm{sat}}(T_s)$. The bulk humidity $r$ does not enter at $x = 0$; the wetness pins the local humidity to 100%.

The linear profiles are then

$$T(x) \;=\; T_s + \frac{T - T_s}{d}\,x, \qquad \rho_v(x) \;=\; \rho_{\mathrm{sat}}(T_s) + \frac{r\,\rho_{\mathrm{sat}}(T) - \rho_{\mathrm{sat}}(T_s)}{d}\,x.$$

The two fluxes

Heat flux (Fourier), magnitude, downward toward the skin:

$$q \;=\; \kappa\,\frac{T - T_s}{d}.$$

Mass flux of water (convert the given number flux $J = D\,dn/dx$ to mass flux by multiplying by the molecular mass; equivalently, since $\rho_v = m_{\mathrm{H_2O}} n$, the mass flux is $j = D\,d\rho_v/dx$ in magnitude). Going upward (water leaves the skin and reaches the drier bulk):

$$j \;=\; D\,\frac{\rho_{\mathrm{sat}}(T_s) - r\,\rho_{\mathrm{sat}}(T)}{d}.$$

The energy balance at the skin — and why $d$ cancels

Consider a thin sheet of the wet skin surface in steady state. There is no heat coming from beneath (assumption (c)). The only ways energy enters or leaves are:

• In: heat conducted down from the laminar layer, $q$ per unit area.
• Out: latent heat carried by evaporating water, $L\,j$ per unit area (each kilogram of vapour leaving carries its enthalpy of vaporisation $L$ with it).

In steady state these must be equal:

$$\kappa\,\frac{T - T_s}{d} \;=\; L\,D\,\frac{\rho_{\mathrm{sat}}(T_s) - r\,\rho_{\mathrm{sat}}(T)}{d}.$$

The $1/d$ is shared by both fluxes — both transport mechanisms have to push their respective quantity through the same layer thickness — so it cancels:

$$\boxed{\;\kappa\,(T - T_s) \;=\; L\,D\,\bigl(\rho_{\mathrm{sat}}(T_s) - r\,\rho_{\mathrm{sat}}(T)\bigr).\;} \tag{$\star$}$$

This is the wet-bulb equation. It says: $T_s$ depends only on the bulk $(T, r)$ and on the material constants $\kappa, L, D$ — never on the layer thickness $d$, the wind speed, or any other property of the convection. Faster wind makes $d$ smaller, increases both fluxes, but increases them in the same ratio, so the equilibrium $T_s$ is unchanged.

Solving the implicit equation graphically

Equation $(\star)$ is transcendental in $T_s$ because $\rho_{\mathrm{sat}}$ is a non-elementary function of temperature. But it can be put in the form

$$\rho_{\mathrm{sat}}(T_s) \;=\; \underbrace{r\,\rho_{\mathrm{sat}}(T)}_{=:\rho_\infty} \;+\; \frac{\kappa}{L\,D}\,(T - T_s).$$

The right-hand side is a straight line in $T_s$, of slope $-\kappa/(LD)$, passing through the point $(T_s, \rho) = (T, \rho_\infty)$ on the chart. The left-hand side is the saturation curve already plotted. Their intersection is $T_s$.

The slope

$$\frac{\kappa}{L\,D} \;=\; \frac{0.030\ \text{W\,m}^{-1}\text{K}^{-1}} {2.26\cdot 10^{6}\ \text{J\,kg}^{-1}\,\cdot\, 2.6\cdot 10^{-5}\ \text{m}^{2}\text{s}^{-1}}.$$

The denominator: $2.26\cdot 10^{6}\cdot 2.6\cdot 10^{-5} = 58.8\ \text{J\,m\,kg}^{-1}\text{s}^{-1}$. So

$$\frac{\kappa}{L\,D} \;\approx\; \frac{0.030}{58.8}\ \text{kg\,m}^{-3}\text{K}^{-1} \;\approx\; 5.1\cdot 10^{-4}\ \text{kg\,m}^{-3}\text{K}^{-1} \;\approx\; 0.51\ \text{g\,m}^{-3}\text{K}^{-1}.$$

The anchor point at $T_s = T$

We need $\rho_{\mathrm{sat}}(110\ °\text{C})$, slightly past the right edge of the chart. Smoothly extrapolating the curve gives $\rho_{\mathrm{sat}}(110\ °\text{C}) \approx 0.83\ \text{kg\,m}^{-3} = 830\ \text{g\,m}^{-3}$ (a cross-check: ideal-gas with $p_{\mathrm{sat}} \approx 143\ \text{kPa}$ at $110\ °\text{C}$ gives $830\ \text{g\,m}^{-3}$). Hence

$$\rho_\infty \;=\; r\,\rho_{\mathrm{sat}}(T) \;=\; 0.03\cdot 830 \;\approx\; 25\ \text{g\,m}^{-3}.$$

The line

Putting $T_s$ in $°\text{C}$, the line is

$$\rho(T_s) \;=\; 25 \;+\; 0.51\,(110 - T_s)\ \text{g\,m}^{-3}.$$

A few values:

$T_s\ (°\text{C})$	line $\rho\ (\text{g\,m}^{-3})$	curve $\rho_{\mathrm{sat}}\ (\text{g\,m}^{-3})$	line $-$ curve
110	25	830	$-805$
70	45	200	$-155$
50	56	83	$-27$
45	58	65	$-7$
43	59	60	$-1$
40	61	50	$+11$

The intersection is in the interval $40\,°\text{C} < T_s < 45\,°\text{C}$, very close to $43\,°\text{C}$.

A sketch of the construction (not to scale; the saturation curve is the true one from the problem):

ρ (g/m³)
 |
 |                                ___, curve ρ_sat(T)
 |                            ,'
 |                         ,'
 |                      ,'
 |                   ,'
 |                .'
 |             ,'
 |          ,'
 |       ,'
 |    .' ←─── intersection ≈ 43 °C
 |  ,*-------*-------*-------*-----.    line  ρ_∞ + (κ/LD)(T−T_s)
 |                                     └ anchor at (110°C, 25 g/m³)
 |
 +----+----+----+----+----+----+----+--→ T_s (°C)
     40   50   60   70   80   90  110

$$\boxed{\;T_s \;\approx\; 43\ °\text{C}\;}$$

Why this is the central insight, and a memorable check

In a dry sauna ($r \to 0$), the line still has slope $-\kappa/(LD)$ and now passes through $(T, 0)$. The intersection drops by a few degrees: the skin can be slightly cooler still, because the gradient driving evaporation is steeper. Intuitively, dry air is a more aggressive sink for sweat, so sweating is more effective at cooling.

In an air-saturated sauna ($r \to 1$), the line passes through $(T, \rho_{\mathrm{sat}}(T))$ — exactly on the saturation curve. Since the line has a negative slope and the saturation curve there has a positive slope, the only way they can intersect on or below that point is if the line is tangent to the curve, which it is not. The intersection moves up to $T_s = T = 110\ °\text{C}$: no temperature gradient drives any heat flow at all, and the sweat cannot evaporate because the air is already at $100\%$ humidity. The skin temperature rises to the air temperature — fatal at $110\ °\text{C}$. This is precisely why a 110 °C dry Finnish sauna is a recreational activity, while a 110 °C steam sauna would kill anyone in it within minutes.

Consistency checks

• Dimensions of $(\star)$. Left: $[\kappa]\,[\Delta T] = \text{W\,m}^{-1}\text{K}^{-1}\cdot\text{K} = \text{W\,m}^{-1}$. Right: $[L]\,[D]\,[\rho] = \text{J\,kg}^{-1}\cdot\text{m}^{2}\text{s}^{-1}\cdot\text{kg\,m}^{-3} = \text{J\,m}^{-1}\text{s}^{-1} = \text{W\,m}^{-1}$. ✓
• The Lewis number. The dimensionless ratio $\mathrm{Le} = \kappa/(\rho_{\mathrm{air}} c_p^{\mathrm{air}} D)$ is close to 1 for water vapour in air. Our slope $\kappa/(LD)$ is essentially the same combination but with the latent heat $L$ playing the role of an “effective heat capacity”. The fact that the answer is around $43\,°\text{C}$ for $r = 3\%$ is consistent with everyday meteorology: in very dry desert air at $40\,°\text{C}$, the wet-bulb temperature is around $20\,°\text{C}$, and the gap scales linearly with $T - T_{wb}$ as long as we are well below boiling.
• Plausibility. A skin temperature of $43\,°\text{C}$ is hot enough to feel painful — and a sauna at $110\,°\text{C}$ does feel painful — but well below the temperature of protein denaturation ($\sim 60\,°\text{C}$), which is why the experience is survivable. The factor of “skin not getting too hot” depends crucially on (i) sweating actually occurring (otherwise the wet-bulb model does not apply) and (ii) sufficiently dry air.

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Summary of results

Part	Quantity	Symbolic	Numerical
(i)	Minimum pulling force	$\;F_{\min} \;=\; (p_0 - p_{\mathrm{sat}}(T_0))\,S\;$	$\;\approx\; 300\ \text{N}\;$
(ii)	Mass of water under the piston	$\;m \;\approx\; \dfrac{\rho_{\mathrm{sat}}(T_1)\,a\,S\,L}{c\,\Delta T}\;$	$\;\approx\; 0.65\ \text{kg}\;$
(iii)	Wet skin temperature	$\;\kappa(T - T_s) \;=\; L D\bigl(\rho_{\mathrm{sat}}(T_s) - r\,\rho_{\mathrm{sat}}(T)\bigr)\;$	$\;T_s \;\approx\; 43\ °\text{C}\;$

One sentence to take away. In every part, the pressure or density of saturated water vapour at the relevant temperature is the load-bearing quantity: it bounds the gas pressure on the piston in (i), determines the vapour mass in the gap in (ii), and pins the boundary condition at the wet skin in (iii); the rest of the problem is bookkeeping with energy balance and Fourier/Fick fluxes.