CLASS 11 · CHAPTER 10 · HEAT

Thermal Properties of Matter

Pour heat into a block of ice and its temperature climbs, right up until it starts to melt. Then, for a while, nothing on the thermometer moves at all, even though heat is still pouring in. Where is it going?

01 · See it

Watch it happen

Heat added: 0 kJ (to 1 kg of ice, starting at -20°C)

Solid (ice), -20.0°C

The shaded bands are where temperature stops rising completely, all the incoming heat is going into breaking bonds, not raising T. Notice how much wider the boiling band is than the melting band: turning water into steam takes almost seven times more heat than melting the same mass of ice.

Drag the heat slider up from zero. Temperature climbs steadily while the sample is entirely solid, then goes flat the instant melting starts, every joule going into breaking the ice’s crystal structure, none of it raising the thermometer. Push further and it climbs again as liquid water, then flattens a second time at 100°C while it boils. Notice the second flat stretch is far wider than the first: turning water into steam costs much more energy than melting the same mass of ice.

02 · Derive it

Where the formula comes from

Within a single phase, a substance’s temperature rise is proportional to the heat supplied, the proportionality constant is its specific heat capacity $s$ :

Specific heat capacity

\textcolor{#e08a1e}{Q} = m\,s\,\Delta T

This is exactly what produces each sloped segment of the graph above. But at a melting or boiling point, temperature stops responding to heat altogether, every joule instead goes into changing the substance’s state, not its temperature. The heat required per unit mass for that is the latent heat $L$ :

Latent heat

\textcolor{#e08a1e}{Q} = mL

This is why calorimetry problems (ice dropped into warm water, for instance) always split into pieces: a latent-heat term for any phase change that happens along the way, plus an $ms\Delta T$ term for every stretch where only temperature is changing, all set equal by heat lost = heat gained.

A related but separate idea: when a solid is heated without changing phase, it also expands. For a linear expansion coefficient $\alpha_l$ , the corresponding volume expansion coefficient works out to $\alpha_v = 3\alpha_l$ , since length, breadth, and height each stretch by the same fractional amount.

03 · Break it

Where the shortcut stops working

It’s tempting to treat “heat” and “temperature” as interchangeable, in everyday speech they usually are. In physics they can’t be: the entire width of each flat plateau in the simulation above is heat flowing in with zero change in temperature. If the two words meant the same thing, that plateau couldn’t exist.

A second, very practical trap: assuming a harder boil means a hotter liquid. Turn the stove to maximum and water at sea level still won’t exceed 100°C, it’ll simply boil away faster. Every extra joule is paying the liquid-to-vapour latent heat for more of the water, not raising what’s left above its boiling point. The only way to raise boiling water’s temperature past 100°C is to raise the pressure above it, which is exactly how a pressure cooker cooks food faster.

And the steam-versus-boiling-water comparison is worth taking seriously: both sit at 100°C, but steam is carrying an enormous amount of extra energy as latent heat, ready to dump the instant it condenses. Same temperature, very different amount of heat delivered.

04 · Master it

Apply it under exam conditions

Q1. How much heat is needed to convert 10 g of ice at -10°C completely to steam at 100°C? (s_ice = 2.06 kJ/kg·K, s_water = 4.186 kJ/kg·K, L_f = 333 kJ/kg, L_v = 2260 kJ/kg)

Q_1 = ms_{ice}\Delta T = (0.01)(2.06)(10) = 0.206\text{ kJ}

Q_2 = mL_f = (0.01)(333) = 3.33\text{ kJ}

Q_3 = ms_{water}\Delta T = (0.01)(4.186)(100) = 4.186\text{ kJ}

Q_4 = mL_v = (0.01)(2260) = 22.6\text{ kJ}

Q_{total} = 0.206+3.33+4.186+22.6 \approx \textcolor{#e08a1e}{30.3\text{ kJ}}

Q2. 50 g of ice at 0°C is dropped into 200 g of water at 25°C in an insulated container. Find the final temperature. (s_water = 4186 J/kg·K, L_f = 333,000 J/kg)

Heat lost by the water equals heat gained melting and then warming the ice:

m_w s_w (25 - T_f) = m_i L_f + m_i s_w T_f

(0.2)(4186)(25-T_f) = (0.05)(333{,}000) + (0.05)(4186)T_f

20{,}930 - 837.2\,T_f = 16{,}650 + 209.3\,T_f \;\Rightarrow\; T_f \approx \textcolor{#e08a1e}{4.1\text{°C}}

05 · FAQs

Quick answers

If I crank up the flame under boiling water, doesn't the water get hotter?+

No, not past 100°C at normal atmospheric pressure. A bigger flame makes the water boil faster, more bubbles, more vapour produced per second, but the liquid itself stays pinned at its boiling point. All that extra heat goes into vaporising more water per second, not raising its temperature.

Why does sweating cool the body down?+

Evaporation is a phase change, liquid sweat turning to vapour, and that takes a large amount of latent heat. That heat is drawn directly from your skin, cooling it, exactly the same physics as the flat plateau in the simulation above, just running in reverse: heat leaving a liquid at constant temperature as it turns to vapour.

Why does a steam burn hurt more than a splash of boiling water at the same 100°C?+

Temperature alone doesn't tell you how much heat something delivers. Steam at 100°C carries an extra 2260 kJ/kg of latent heat on top of that, heat it releases the instant it condenses back to liquid on your skin. Boiling water at 100°C has already spent that latent heat; it can only cool down from 100°C, it can't also condense.

Is temperature just another name for heat?+

No, and the plateau in the simulation is exactly why that distinction matters. Heat is energy in transit, flowing because of a temperature difference. Temperature is a measure of the substance's own thermal state. During melting or boiling, heat keeps flowing in continuously, but temperature holds still, the clearest proof the two aren't the same quantity.

Why is the ice region of the graph steeper than the water region?+

Slope on this graph is 1/(mass × specific heat capacity), so a smaller specific heat capacity means a steeper rise for the same heat input. Ice's specific heat capacity (about 2.06 kJ/kg·K) is roughly half water's (about 4.186 kJ/kg·K), so the same amount of heat raises ice's temperature about twice as fast as it would raise liquid water's.

Related concepts

Physics doesn’t stay inside chapter boundaries. Neither should you.

Matter

Mechanical Properties of Fluids

Squeeze a flowing fluid through a narrow throat and watch its own pressure drop in real time.

Heat

Kinetic Theory

Watch molecules bounce in a box and see that only temperature, never pressure or volume, changes their speed.

Heat

Thermodynamics

Compress the same gas two ways and watch it either stay cool or heat up like a diesel engine.