Phase Changes — Why Does Boiling Water Stay at 100°C? 🧊💧💨

Phase Changes — Why Does Boiling Water Stay at 100°C? 🧊💧💨

A kitchen experiment 🍲: a pot of water on the stove, thermometer inside. Temperature rises… 80… 90… 100°C, then it starts to boil. Leave the flame on. Weirdly, the temperature stops climbing — it just sits at 100°C no matter how long you wait! 🤔 Where does all that stove energy go? The answer is a beautiful concept called latent heat — the very same concept that explains why sweating cools us down, why a steam burn is worse than a boiling-water burn, and why dry ice skips liquid and turns straight into gas.

The core idea in one paragraph 📌

Matter’s three main states — solid, liquid, gas — can flip between one another by absorbing or releasing heat: melting/freezing, vaporization/condensation, sublimation/deposition. The big secret: during a phase change, temperature stays constant because all the heat goes into breaking intermolecular bonds, not into speeding molecules up. The energy required is \( Q = m L \), where \( L \) is the specific latent heat (fusion: \( L_f \); vaporization: \( L_v \)). Water is exceptionally energetic: \( L_v \approx 2260\,\text{kJ/kg} \) — that’s why sweating cools us so effectively.

Four key formulas 📐

\[ \boxed{Q_\text{heat/cool} = m c \Delta T} \]

Temperature change without phase change — the formula from the previous section.

\[ \boxed{Q_\text{melt/freeze} = m L_f} \]

Solid ↔ liquid at the melting point. Positive for melting, negative for freezing.

\[ \boxed{Q_\text{vaporize/condense} = m L_v} \]

Liquid ↔ gas at the boiling point. For water, \( L_v \) is about \( 5.4\times L_f \).

\[ \boxed{Q_\text{sublime} = m L_s \approx m(L_f + L_v)} \]

Direct solid ↔ gas without passing through liquid — as in dry ice (solid CO₂) or naphthalene.

Melting / boiling points and latent heats 📊

Substance Melt (°C) Boil (°C) \( L_f \) (kJ/kg) \( L_v \) (kJ/kg)
Helium — −269 — 21
Liquid nitrogen −210 −196 25.7 200
Ethanol −114 78.4 108 855
Mercury −39 357 11.4 295
Water 0 100 334 2260 ✨
Lead 327 1749 24.5 866
Silver 962 2162 105 2323
Copper 1085 2562 205 4726
Iron 1538 2861 247 6088
Tungsten 3422 5555 285 4482

Note: water’s \( L_v \) (~2260 kJ/kg) is higher than most metals‘ — that anomaly is what lets our body cool itself with sweat and lets steam engines produce useful power.

Why does temperature stay flat? 🔒

Imagine a locked hall with 100 people in a queue, one person exiting every second. Until everyone leaves, the “average number of people inside” is dropping, but the exit rate (proportional to incoming energy) is constant. Analogously, each joule you feed to boiling water peels a batch of molecules off the surface — breaking hydrogen bonds — instead of speeding up remaining ones. Temperature can’t rise as long as there’s still something to detach.

Heating curve for 1 kg of water from \( -20 \)°C to \( 120 \)°C steam:

T (°C)
120 ┤                              ⋯⋯⋯
100 ┤                    ▬▬▬▬▬▬▬▬━━━  ← vaporize: L_v = 2260 kJ (long!)
    ┤              ▬▬━━━━
  0 ┤    ▬▬━━━━ ← melt: L_f = 334 kJ (short)
-20 ┤━━━
    └──────────────────────────────→ heat supplied (kJ)

Three sloping segments (\( mc\Delta T \)) and two flat plateaus (phase change with \( mL \)).

Example 1: Total heat to turn 1 kg of ice into steam 🧊➡️💨

\( m = 1\,\text{kg} \) ice at \( -20\,°\text{C} \) → steam at \( 120\,°\text{C} \). Five steps:

Step Formula Value
Warm the ice (\( -20 \to 0 \)) \( mc_\text{ice}\Delta T = 1 \times 2100 \times 20 \) 42 kJ
Melt the ice (\( 0° \)C) \( mL_f = 1 \times 334 \) 334 kJ
Warm the water (\( 0 \to 100 \)) \( mc_\text{w}\Delta T = 1 \times 4186 \times 100 \) 418.6 kJ
Vaporize (\( 100° \)C) \( mL_v = 1 \times 2260 \) 2260 kJ
Warm the steam (\( 100 \to 120 \)) \( mc_\text{steam}\Delta T = 1 \times 2010 \times 20 \) 40.2 kJ
Total 3094.8 kJ

Notice: vaporization alone consumes ~73% of the energy! That’s why water is such a workhorse in every steam-based process.

Example 2: How much ice to cool a drink? 🥤

\( m_w = 500\,\text{g} \) of water at 25°C. We want to reach 5°C. How many grams of 0°C ice?

Heat given up by water: \( Q_w = 0.5 \times 4186 \times (25-5) = 41{,}860\,\text{J} \)

Heat absorbed by ice (melt + heat the melted water):
\[ Q_\text{ice} = m_i L_f + m_i c_w (5 – 0) = m_i (334{,}000 + 4186 \times 5) = m_i \times 354{,}930 \]

From \( Q_w = Q_\text{ice} \): \( m_i = 41{,}860 / 354{,}930 \approx 118\,\text{g} \) ≈ 4 standard ice cubes 🧊.

Example 3: How much does sweat cool the body? 💦

Assume on a hot day the body loses 500 g of sweat that fully evaporates. How much heat is rejected?

\[ Q = m L_v = 0.5 \times 2260 = 1130\,\text{kJ} \approx 270\,\text{kcal} \]

One medium burger’s worth of energy expelled just as cooling! That’s why desert sands scorch but desert dwellers survive.

Triple point and critical point — science bonus 🧭

Python analysis 🐍

1) Heating curve for water from \( -20 \) to \( 120° \)C

import numpy as np, matplotlib.pyplot as plt

c_ice, c_w, c_steam = 2100, 4186, 2010     # J/(kg·K)
L_f, L_v = 334_000, 2_260_000              # J/kg
m = 1.0

segments = [
    (m*c_ice*20,        -20,   0),  # heat ice
    (m*L_f,               0,   0),  # melt
    (m*c_w*100,           0, 100),  # heat liquid
    (m*L_v,             100, 100),  # vaporize
    (m*c_steam*20,      100, 120),  # heat steam
]

Q_cum, T_pts, Q_pts = 0, [-20], [0]
for Q, T0, T1 in segments:
    Q_cum += Q
    T_pts.append(T1)
    Q_pts.append(Q_cum / 1000)      # kJ

plt.plot(Q_pts, T_pts, "o-")
plt.xlabel("Heat supplied (kJ)"); plt.ylabel("Temperature (°C)")
plt.title("Heating 1 kg of water from -20 to +120°C")
plt.grid(alpha=0.3); plt.show()
print(f"Total energy: {Q_cum/1000:.1f} kJ")

2) Calorimetry with phase change — ice in water

def ice_in_water(m_ice, m_water, T_water_C):
    """Drop ice into warm water; return final T or leftover ice."""
    c_w, L_f = 4186, 334_000
    Q_available = m_water * c_w * (T_water_C - 0)   # heat water gives down to 0°C
    Q_needed    = m_ice * L_f                       # heat to melt all the ice

    if Q_available < Q_needed:
        m_melted = Q_available / L_f
        return {"state": "some ice remains",
                "T_final": 0.0,
                "ice_left_g": (m_ice - m_melted) * 1000}
    else:
        m_total = m_water + m_ice
        # All ice melted; final T from balance
        Tf = (Q_available - Q_needed) / (m_total * c_w)
        return {"state": "all ice melted", "T_final": Tf, "ice_left": 0}

print(ice_in_water(m_ice=0.05,  m_water=0.5, T_water_C=25))
print(ice_in_water(m_ice=0.500, m_water=0.5, T_water_C=25))  # too much ice → leftover

3) How sweat keeps body temperature down

# Toy model: body generates heat at a constant rate; sweat evaporates.
# Compare with-vs-without-sweat body temperature over one hour.

P_metabolism = 120       # W (resting)
m_body = 70              # kg
c_body = 3500            # J/(kg·K)
L_v = 2.26e6             # J/kg
sweat_rate = 0.5 / 3600  # 0.5 L/h → kg/s

import numpy as np
t = np.linspace(0, 3600, 200)     # 1 hour
dT_no_sweat = P_metabolism * t / (m_body * c_body)
# With sweat: rejected power = sweat_rate * L_v
P_net = P_metabolism - sweat_rate * L_v
dT_with_sweat = P_net * t / (m_body * c_body)

import matplotlib.pyplot as plt
plt.plot(t/60, 37 + dT_no_sweat,   label="No sweat (hypothetical)", color="red")
plt.plot(t/60, 37 + dT_with_sweat, label="Normal sweat",            color="steelblue")
plt.xlabel("Time (min)"); plt.ylabel("Body temperature (°C)")
plt.axhline(41, color="orange", ls="--", label="Heat-stroke danger")
plt.title("Why sweating keeps us alive")
plt.legend(); plt.grid(alpha=0.3); plt.show()

print(f"Rejected power at 0.5 L/h sweat: {sweat_rate * L_v:.1f} W (~10× resting metabolism!)")

Take-home summary 🎁

Three states and six phase transitions (melt/freeze, vaporize/condense, sublime/deposit). During any phase change, temperature is constant because all heat goes into breaking molecular bonds ⇒ \( Q = mL \), not \( Q = mc\Delta T \). Water is uniquely energetic (\( L_v \approx 2260\,\text{kJ/kg} \)) — that’s what powers our sweat, steam engines, and the planet’s climate system. The heating curve has 5 segments: 3 sloped + 2 flat. Dry ice (solid CO₂) skips liquid entirely — hence the name. Triple point: the only \( T,P \) where all three phases coexist in equilibrium 🎯.

“Nice to know” box: Why is a steam burn worse than a boiling-water burn? ⚠️

Both are at 100°C — so why is steam more dangerous? Because in addition to the sensible heat, steam carries a huge latent heat of ~2260 kJ/kg. When steam hits your skin, it condenses (back to liquid), releasing all that latent heat right on you. For comparison: 10 g of 100°C water cooling to skin temperature (37°C) delivers ~2.6 kJ. But 10 g of 100°C steam first condenses (~22.6 kJ released), then cools to 37°C — a total of ~25 kJ, roughly 10× the thermal injury ✨. The same physics that keeps us alive through sweat can, in the kitchen with a moment of inattention, cause a third-degree burn.

Test yourself 📝

References and further exploration 📚

Articles and reference

Videos (YouTube)

External simulators

On this site 🔗

Next up we dissect the three ways heat moves: conduction, convection, radiation 🔥 — why a wooden pan-handle stays cool, why radiators sit low in the room, and why the Sun’s heat reaches us across the vacuum of space. See you there! 👋

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