Why Steam at 100 °C Burns Worse Than Boiling Water

On a sunny afternoon at the beach, the sand scorches your feet while the sea stays cool. Both receive the same sunlight, but sand needs far less heat to warm up than water.

This part covers specific heat, calorimetry, changes of state, and latent heat. Take the specific heat of water as $4186$ J/kg K, latent heat of fusion of ice as $3.34\times 10^5$ J/kg and latent heat of vaporisation of water as $2.26\times 10^6$ J/kg.

**Specific heat capacity is the heat needed to raise the temperature of $1$ kg by $1$ K, $s=\frac{Q}{m\Delta T}$; molar specific heat is the heat per mole, $C=\frac{Q}{\mu \Delta T}$; and for a gas $C_p>C_v$ because at constant pressure part of the heat does work as the gas expands.

Worked example 1 — water.** Heating $2$ kg of water from $25$ °C to $75$ °C:

$$Q=ms\Delta T=2\times 4186\times 50=418600\text{ J}\approx 4.19\times 10^5\text{ J}$$

Worked example 2 — molar value. Since $1$ mol of water is $0.018$ kg, its molar specific heat is $4186\times 0.018\approx 75$ J/mol K.

**$C_p$ and $C_v$.** At constant volume all the heat raises internal energy. At constant pressure the gas also pushes its surroundings back, so more heat is needed for the same rise; for an ideal gas the difference is $R$ per mole.

An everyday example. Coastal towns have milder weather than inland ones, because the sea's high specific heat makes it warm and cool slowly.

The substance. Water's unusually high specific heat makes it an excellent coolant in vehicle radiators and power plants.

**In an isolated system, the heat lost by the hotter bodies equals the heat gained by the colder bodies, so $m_1{s}_1(T_1-T)=m_2{s}_2(T-T_2)$ at the final common temperature $T$.

Worked example 1 — unknown temperature.** $1$ kg of water at $80$ °C is mixed with $3$ kg at $20$ °C:

$$1(80-T)=3(T-20)\Rightarrow 80+60=4T\Rightarrow T=35\text{ °C}$$

Worked example 2 — unknown specific heat. A $0.2$ kg metal block at $100$ °C is dropped into $0.5$ kg of water at $20$ °C, and the mixture settles at $25$ °C.

$$0.2\times s\times 75=0.5\times 4186\times 5\Rightarrow s=\frac{10465}{15}\approx 698\text{ J/kg K}$$

An everyday example. Mixing a bucket of hot water with cold water for a bath settles at a temperature in between, closer to whichever you used more of.

The substance. Heat lost equals heat gained only if no heat escapes — the calorimeter itself also absorbs some.

During melting or boiling, heat supplied breaks bonds between molecules rather than raising temperature, so a temperature-time graph shows flat plateaus at the melting and boiling points, with sloping sections in between.

Changes of state: melting (solid to liquid), vaporisation (liquid to gas), condensation, freezing, and sublimation (solid straight to gas, as with camphor).

Worked example — building a heating curve. A $500$ W heater warms $0.5$ kg of ice from $-10$ °C to steam ($s_{ice}=2100$ J/kg K).

- **Ice $-10$ °C to $0$ °C**: $0.5\times 2100\times 10=10500$ J, taking $21$ s (slope)
- **Melting at $0$ °C**: $0.5\times 3.34\times 10^5=1.67\times 10^5$ J, taking $334$ s (plateau)
- **Water $0$ °C to $100$ °C**: $0.5\times 4186\times 100=209300$ J, taking about $419$ s (slope)
- **Boiling at $100$ °C**: $0.5\times 2.26\times 10^6=1.13\times 10^6$ J, taking $2260$ s (plateau)

The boiling plateau is by far the longest, and the slope is steeper for ice than for water because ice has a lower specific heat.

An everyday example. A pressure cooker raises the boiling point by raising the pressure, so food cooks faster at a higher temperature.

The substance. Boiling point falls as pressure falls, which is why cooking takes longer at high altitudes.

**Latent heat is the heat needed per kilogram to change state at constant temperature, $Q=mL$; in mixing problems, include $mL$ for every mass that melts or condenses, and first check whether all of it actually changes state.

Worked example 1 — ice in water.** $0.1$ kg of ice at $0$ °C is dropped into $0.5$ kg of water at $40$ °C.

Check first: melting all the ice needs $0.1\times 3.34\times 10^5=33400$ J, while cooling the water to $0$ °C would release $0.5\times 4186\times 40=83720$ J. All the ice melts. Then

$$0.5\times 4186(40-T)=33400+0.1\times 4186\times T\Rightarrow T\approx 20.0\text{ °C}$$

Worked example 2 — why steam burns. $1$ g of steam at $100$ °C cooling to skin temperature $37$ °C gives

$$2260+1\times 4.186\times 63\approx 2524\text{ J}$$

while $1$ g of boiling water gives only about $264$ J — nearly ten times less.

An everyday example. Steam from a boiling pressure cooker's whistle can scald far worse than a splash of boiling water.

The substance. **If there is too little heat to melt all the ice, the final temperature is exactly $0$ °C** with some ice left over.